Python Cookbook, 3rd Edition

Chapter 1. Data Structures and Algorithms

Python provides a variety of useful built-in data structures, such as lists, sets, and dictionaries. For the most part, the use of these structures is straightforward. However, common questions concerning searching, sorting, ordering, and filtering often arise. Thus, the goal of this chapter is to discuss common data structures and algorithms involving data. In addition, treatment is given to the various data structures contained in the collections module.

1.1. Unpacking a Sequence into Separate Variables

Problem

You have an N-element tuple or sequence that you would like to unpack into a collection of N variables.

Solution

Any sequence (or iterable) can be unpacked into variables using a simple assignment operation. The only requirement is that the number of variables and structure match the sequence. For example:

>>> p = (4, 5)
>>> x, y = p
>>> x
4
>>> y
5
>>>

>>> data = [ 'ACME', 50, 91.1, (2012, 12, 21) ]
>>> name, shares, price, date = data
>>> name
'ACME'
>>> date
(2012, 12, 21)

>>> name, shares, price, (year, mon, day) = data
>>> name
'ACME'
>>> year
2012
>>> mon
12
>>> day
21
>>>

If there is a mismatch in the number of elements, you’ll get an error. For example:

>>> p = (4, 5)
>>> x, y, z = p
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ValueError: need more than 2 values to unpack
>>>

Discussion

Unpacking actually works with any object that happens to be iterable, not just tuples or lists. This includes strings, files, iterators, and generators. For example:

>>> s = 'Hello'
>>> a, b, c, d, e = s
>>> a
'H'
>>> b
'e'
>>> e
'o'
>>>

When unpacking, you may sometimes want to discard certain values. Python has no special syntax for this, but you can often just pick a throwaway variable name for it. For example:

>>> data = [ 'ACME', 50, 91.1, (2012, 12, 21) ]
>>> _, shares, price, _ = data
>>> shares
50
>>> price
91.1
>>>

However, make sure that the variable name you pick isn’t being used for something else already.

1.2. Unpacking Elements from Iterables of Arbitrary Length

Problem

You need to unpack N elements from an iterable, but the iterable may be longer than N elements, causing a “too many values to unpack” exception.

Solution

Python “star expressions” can be used to address this problem. For example, suppose you run a course and decide at the end of the semester that you’re going to drop the first and last homework grades, and only average the rest of them. If there are only four assignments, maybe you simply unpack all four, but what if there are 24? A star expression makes it easy:

def drop_first_last(grades):
    first, *middle, last = grades
    return avg(middle)

As another use case, suppose you have user records that consist of a name and email address, followed by an arbitrary number of phone numbers. You could unpack the records like this:

>>> record = ('Dave', '[e-mail]', '773-555-1212', '847-555-1212')
>>> name, email, *phone_numbers = record
>>> name
'Dave'
>>> email
'[e-mail]'
>>> phone_numbers
['773-555-1212', '847-555-1212']
>>>

It’s worth noting that the phone_numbers variable will always be a list, regardless of how many phone numbers are unpacked (including none). Thus, any code that uses phone_numbers won’t have to account for the possibility that it might not be a list or perform any kind of additional type checking.

The starred variable can also be the first one in the list. For example, say you have a sequence of values representing your company’s sales figures for the last eight quarters. If you want to see how the most recent quarter stacks up to the average of the first seven, you could do something like this:

*trailing_qtrs, current_qtr = sales_record
trailing_avg = sum(trailing_qtrs) / len(trailing_qtrs)
return avg_comparison(trailing_avg, current_qtr)

Here’s a view of the operation from the Python interpreter:

>>> *trailing, current = [10, 8, 7, 1, 9, 5, 10, 3]
>>> trailing
[10, 8, 7, 1, 9, 5, 10]
>>> current
3

Discussion

Extended iterable unpacking is tailor-made for unpacking iterables of unknown or arbitrary length. Oftentimes, these iterables have some known component or pattern in their construction (e.g. “everything after element 1 is a phone number”), and star unpacking lets the developer leverage those patterns easily instead of performing acrobatics to get at the relevant elements in the iterable.

It is worth noting that the star syntax can be especially useful when iterating over a sequence of tuples of varying length. For example, perhaps a sequence of tagged tuples:

records = [
     ('foo', 1, 2),
     ('bar', 'hello'),
     ('foo', 3, 4),
]

def do_foo(x, y):
    print('foo', x, y)

def do_bar(s):
    print('bar', s)

for tag, *args in records:
    if tag == 'foo':
        do_foo(*args)
    elif tag == 'bar':
        do_bar(*args)

Star unpacking can also be useful when combined with certain kinds of string processing operations, such as splitting. For example:

>>> line = 'nobody:*:-2:-2:Unprivileged User:/var/empty:/usr/bin/false'
>>> uname, *fields, homedir, sh = line.split(':')
>>> uname
'nobody'
>>> homedir
'/var/empty'
>>> sh
'/usr/bin/false'
>>>

Sometimes you might want to unpack values and throw them away. You can’t just specify a bare * when unpacking, but you could use a common throwaway variable name, such as _ or ign (ignored). For example:

>>> record = ('ACME', 50, 123.45, (12, 18, 2012))
>>> name, *_, (*_, year) = record
>>> name
'ACME'
>>> year
2012
>>>

There is a certain similarity between star unpacking and list-processing features of various functional languages. For example, if you have a list, you can easily split it into head and tail components like this:

>>> items = [1, 10, 7, 4, 5, 9]
>>> head, *tail = items
>>> head
1
>>> tail
[10, 7, 4, 5, 9]
>>>

One could imagine writing functions that perform such splitting in order to carry out some kind of clever recursive algorithm. For example:

>>> def sum(items):
...     head, *tail = items
...     return head + sum(tail) if tail else head
...
>>> sum(items)
36
>>>

However, be aware that recursion really isn’t a strong Python feature due to the inherent recursion limit. Thus, this last example might be nothing more than an academic curiosity in practice.

1.3. Keeping the Last N Items

Problem

You want to keep a limited history of the last few items seen during iteration or during some other kind of processing.

Solution

Keeping a limited history is a perfect use for a collections.deque. For example, the following code performs a simple text match on a sequence of lines and yields the matching line along with the previous N lines of context when found:

from collections import deque

def search(lines, pattern, history=5):
    previous_lines = deque(maxlen=history)
    for line in lines:
        if pattern in line:
            yield line, previous_lines
        previous_lines.append(line)

# Example use on a file
if __name__ == '__main__':
    with open('somefile.txt') as f:
        for line, prevlines in search(f, 'python', 5):
            for pline in prevlines:
                print(pline, end='')
            print(line, end='')
            print('-'*20)

Discussion

When writing code to search for items, it is common to use a generator function involving yield, as shown in this recipe’s solution. This decouples the process of searching from the code that uses the results. If you’re new to generators, see Recipe 4.3.

Using deque(maxlen=N) creates a fixed-sized queue. When new items are added and the queue is full, the oldest item is automatically removed. For example:

>>> q = deque(maxlen=3)
>>> q.append(1)
>>> q.append(2)
>>> q.append(3)
>>> q
deque([1, 2, 3], maxlen=3)
>>> q.append(4)
>>> q
deque([2, 3, 4], maxlen=3)
>>> q.append(5)
>>> q
deque([3, 4, 5], maxlen=3)

Although you could manually perform such operations on a list (e.g., appending, deleting, etc.), the queue solution is far more elegant and runs a lot faster.

More generally, a deque can be used whenever you need a simple queue structure. If you don’t give it a maximum size, you get an unbounded queue that lets you append and pop items on either end. For example:

>>> q = deque()
>>> q.append(1)
>>> q.append(2)
>>> q.append(3)
>>> q
deque([1, 2, 3])
>>> q.appendleft(4)
>>> q
deque([4, 1, 2, 3])
>>> q.pop()
3
>>> q
deque([4, 1, 2])
>>> q.popleft()
4

Adding or popping items from either end of a queue has O(1) complexity. This is unlike a list where inserting or removing items from the front of the list is O(N).

1.4. Finding the Largest or Smallest N Items

Problem

You want to make a list of the largest or smallest N items in a collection.

Solution

The heapq module has two functions—nlargest() and nsmallest()—that do exactly what you want. For example:

import heapq

nums = [1, 8, 2, 23, 7, -4, 18, 23, 42, 37, 2]
print(heapq.nlargest(3, nums))  # Prints [42, 37, 23]
print(heapq.nsmallest(3, nums)) # Prints [-4, 1, 2]

Both functions also accept a key parameter that allows them to be used with more complicated data structures. For example:

portfolio = [
   {'name': 'IBM', 'shares': 100, 'price': 91.1},
   {'name': 'AAPL', 'shares': 50, 'price': 543.22},
   {'name': 'FB', 'shares': 200, 'price': 21.09},
   {'name': 'HPQ', 'shares': 35, 'price': 31.75},
   {'name': 'YHOO', 'shares': 45, 'price': 16.35},
   {'name': 'ACME', 'shares': 75, 'price': 115.65}
]

cheap = heapq.nsmallest(3, portfolio, key=lambda s: s['price'])
expensive = heapq.nlargest(3, portfolio, key=lambda s: s['price'])

Discussion

If you are looking for the N smallest or largest items and N is small compared to the overall size of the collection, these functions provide superior performance. Underneath the covers, they work by first converting the data into a list where items are ordered as a heap. For example:

>>> nums = [1, 8, 2, 23, 7, -4, 18, 23, 42, 37, 2]
>>> import heapq
>>> heap = list(nums)
>>> heapq.heapify(heap)
>>> heap
[-4, 2, 1, 23, 7, 2, 18, 23, 42, 37, 8]
>>>

The most important feature of a heap is that heap[0] is always the smallest item. Moreover, subsequent items can be easily found using the heapq.heappop() method, which pops off the first item and replaces it with the next smallest item (an operation that requires O(log N) operations where N is the size of the heap). For example, to find the three smallest items, you would do this:

>>> heapq.heappop(heap)
-4
>>> heapq.heappop(heap)
1
>>> heapq.heappop(heap)
2

The nlargest() and nsmallest() functions are most appropriate if you are trying to find a relatively small number of items. If you are simply trying to find the single smallest or largest item (N=1), it is faster to use min() and max(). Similarly, if N is about the same size as the collection itself, it is usually faster to sort it first and take a slice (i.e., use sorted(items)[:N] or sorted(items)[-N:]). It should be noted that the actual implementation of nlargest() and nsmallest() is adaptive in how it operates and will carry out some of these optimizations on your behalf (e.g., using sorting if N is close to the same size as the input).

Although it’s not necessary to use this recipe, the implementation of a heap is an interesting and worthwhile subject of study. This can usually be found in any decent book on algorithms and data structures. The documentation for the heapq module also discusses the underlying implementation details.

1.5. Implementing a Priority Queue

Problem

You want to implement a queue that sorts items by a given priority and always returns the item with the highest priority on each pop operation.

Solution

The following class uses the heapq module to implement a simple priority queue:

import heapq

class PriorityQueue:
    def __init__(self):
        self._queue = []
        self._index = 0

    def push(self, item, priority):
        heapq.heappush(self._queue, (-priority, self._index, item))
        self._index += 1

    def pop(self):
        return heapq.heappop(self._queue)[-1]

Here is an example of how it might be used:

>>> class Item:
...     def __init__(self, name):
...         self.name = name
...     def __repr__(self):
...         return 'Item({!r})'.format(self.name)
...
>>> q = PriorityQueue()
>>> q.push(Item('foo'), 1)
>>> q.push(Item('bar'), 5)
>>> q.push(Item('spam'), 4)
>>> q.push(Item('grok'), 1)
>>> q.pop()
Item('bar')
>>> q.pop()
Item('spam')
>>> q.pop()
Item('foo')
>>> q.pop()
Item('grok')
>>>

Observe how the first pop() operation returned the item with the highest priority. Also observe how the two items with the same priority (foo and grok) were returned in the same order in which they were inserted into the queue.

Discussion

The core of this recipe concerns the use of the heapq module. The functions heapq.heappush() and heapq.heappop() insert and remove items from a list _queue in a way such that the first item in the list has the smallest priority (as discussed in Recipe 1.4). The heappop() method always returns the “smallest” item, so that is the key to making the queue pop the correct items. Moreover, since the push and pop operations have O(log N) complexity where N is the number of items in the heap, they are fairly efficient even for fairly large values of N.

In this recipe, the queue consists of tuples of the form (-priority, index, item). The priority value is negated to get the queue to sort items from highest priority to lowest priority. This is opposite of the normal heap ordering, which sorts from lowest to highest value.

The role of the index variable is to properly order items with the same priority level. By keeping a constantly increasing index, the items will be sorted according to the order in which they were inserted. However, the index also serves an important role in making the comparison operations work for items that have the same priority level.

To elaborate on that, instances of Item in the example can’t be ordered. For example:

>>> a = Item('foo')
>>> b = Item('bar')
>>> a < b
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: unorderable types: Item() < Item()
>>>

If you make (priority, item) tuples, they can be compared as long as the priorities are different. However, if two tuples with equal priorities are compared, the comparison fails as before. For example:

>>> a = (1, Item('foo'))
>>> b = (5, Item('bar'))
>>> a < b
True
>>> c = (1, Item('grok'))
>>> a < c
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: unorderable types: Item() < Item()
>>>

By introducing the extra index and making (priority, index, item) tuples, you avoid this problem entirely since no two tuples will ever have the same value for index (and Python never bothers to compare the remaining tuple values once the result of comparison can be determined):

>>> a = (1, 0, Item('foo'))
>>> b = (5, 1, Item('bar'))
>>> c = (1, 2, Item('grok'))
>>> a < b
True
>>> a < c
True
>>>

If you want to use this queue for communication between threads, you need to add appropriate locking and signaling. See Recipe 12.3 for an example of how to do this.

The documentation for the heapq module has further examples and discussion concerning the theory and implementation of heaps.

1.6. Mapping Keys to Multiple Values in a Dictionary

Problem

You want to make a dictionary that maps keys to more than one value (a so-called “multidict”).

Solution

A dictionary is a mapping where each key is mapped to a single value. If you want to map keys to multiple values, you need to store the multiple values in another container such as a list or set. For example, you might make dictionaries like this:

d = {
   'a' : [1, 2, 3],
   'b' : [4, 5]
}

e = {
   'a' : {1, 2, 3},
   'b' : {4, 5}
}

The choice of whether or not to use lists or sets depends on intended use. Use a list if you want to preserve the insertion order of the items. Use a set if you want to eliminate duplicates (and don’t care about the order).

To easily construct such dictionaries, you can use defaultdict in the collections module. A feature of defaultdict is that it automatically initializes the first value so you can simply focus on adding items. For example:

from collections import defaultdict

d = defaultdict(list)
d['a'].append(1)
d['a'].append(2)
d['b'].append(4)
...

d = defaultdict(set)
d['a'].add(1)
d['a'].add(2)
d['b'].add(4)
...

One caution with defaultdict is that it will automatically create dictionary entries for keys accessed later on (even if they aren’t currently found in the dictionary). If you don’t want this behavior, you might use setdefault() on an ordinary dictionary instead. For example:

d = {}    # A regular dictionary
d.setdefault('a', []).append(1)
d.setdefault('a', []).append(2)
d.setdefault('b', []).append(4)
...

However, many programmers find setdefault() to be a little unnatural—not to mention the fact that it always creates a new instance of the initial value on each invocation (the empty list [] in the example).

Discussion

In principle, constructing a multivalued dictionary is simple. However, initialization of the first value can be messy if you try to do it yourself. For example, you might have code that looks like this:

d = {}
for key, value in pairs:
    if key not in d:
         d[key] = []
    d[key].append(value)

Using a defaultdict simply leads to much cleaner code:

d = defaultdict(list)
for key, value in pairs:
    d[key].append(value)

This recipe is strongly related to the problem of grouping records together in data processing problems. See Recipe 1.15 for an example.

1.7. Keeping Dictionaries in Order

Problem

You want to create a dictionary, and you also want to control the order of items when iterating or serializing.

Solution

To control the order of items in a dictionary, you can use an OrderedDict from the collections module. It exactly preserves the original insertion order of data when iterating. For example:

from collections import OrderedDict

d = OrderedDict()
d['foo'] = 1
d['bar'] = 2
d['spam'] = 3
d['grok'] = 4

# Outputs "foo 1", "bar 2", "spam 3", "grok 4"
for key in d:
    print(key, d[key])

An OrderedDict can be particularly useful when you want to build a mapping that you may want to later serialize or encode into a different format. For example, if you want to precisely control the order of fields appearing in a JSON encoding, first building the data in an OrderedDict will do the trick:

>>> import json
>>> json.dumps(d)
'{"foo": 1, "bar": 2, "spam": 3, "grok": 4}'
>>>

Discussion

An OrderedDict internally maintains a doubly linked list that orders the keys according to insertion order. When a new item is first inserted, it is placed at the end of this list. Subsequent reassignment of an existing key doesn’t change the order.

Be aware that the size of an OrderedDict is more than twice as large as a normal dictionary due to the extra linked list that’s created. Thus, if you are going to build a data structure involving a large number of OrderedDict instances (e.g., reading 100,000 lines of a CSV file into a list of OrderedDict instances), you would need to study the requirements of your application to determine if the benefits of using an OrderedDict outweighed the extra memory overhead.

1.8. Calculating with Dictionaries

Problem

You want to perform various calculations (e.g., minimum value, maximum value, sorting, etc.) on a dictionary of data.

Solution

Consider a dictionary that maps stock names to prices:

prices = {
   'ACME': 45.23,
   'AAPL': 612.78,
   'IBM': 205.55,
   'HPQ': 37.20,
   'FB': 10.75
}

In order to perform useful calculations on the dictionary contents, it is often useful to invert the keys and values of the dictionary using zip(). For example, here is how to find the minimum and maximum price and stock name:

min_price = min(zip(prices.values(), prices.keys()))
# min_price is (10.75, 'FB')

max_price = max(zip(prices.values(), prices.keys()))
# max_price is (612.78, 'AAPL')

Similarly, to rank the data, use zip() with sorted(), as in the following:

prices_sorted = sorted(zip(prices.values(), prices.keys()))
# prices_sorted is [(10.75, 'FB'), (37.2, 'HPQ'),
#                   (45.23, 'ACME'), (205.55, 'IBM'),
#                   (612.78, 'AAPL')]

When doing these calculations, be aware that zip() creates an iterator that can only be consumed once. For example, the following code is an error:

prices_and_names = zip(prices.values(), prices.keys())
print(min(prices_and_names))   # OK
print(max(prices_and_names))   # ValueError: max() arg is an empty sequence

Discussion

If you try to perform common data reductions on a dictionary, you’ll find that they only process the keys, not the values. For example:

min(prices)   # Returns 'AAPL'
max(prices)   # Returns 'IBM'

This is probably not what you want because you’re actually trying to perform a calculation involving the dictionary values. You might try to fix this using the values() method of a dictionary:

min(prices.values())  # Returns 10.75
max(prices.values())  # Returns 612.78

Unfortunately, this is often not exactly what you want either. For example, you may want to know information about the corresponding keys (e.g., which stock has the lowest price?).

You can get the key corresponding to the min or max value if you supply a key function to min() and max(). For example:

min(prices, key=lambda k: prices[k])  # Returns 'FB'
max(prices, key=lambda k: prices[k])  # Returns 'AAPL'

However, to get the minimum value, you’ll need to perform an extra lookup step. For example:

min_value = prices[min(prices, key=lambda k: prices[k])]

The solution involving zip() solves the problem by “inverting” the dictionary into a sequence of (value, key) pairs. When performing comparisons on such tuples, the value element is compared first, followed by the key. This gives you exactly the behavior that you want and allows reductions and sorting to be easily performed on the dictionary contents using a single statement.

It should be noted that in calculations involving (value, key) pairs, the key will be used to determine the result in instances where multiple entries happen to have the same value. For instance, in calculations such as min() and max(), the entry with the smallest or largest key will be returned if there happen to be duplicate values. For example:

>>> prices = { 'AAA' : 45.23, 'ZZZ': 45.23 }
>>> min(zip(prices.values(), prices.keys()))
(45.23, 'AAA')
>>> max(zip(prices.values(), prices.keys()))
(45.23, 'ZZZ')
>>>

1.9. Finding Commonalities in Two Dictionaries

Problem

You have two dictionaries and want to find out what they might have in common (same keys, same values, etc.).

Solution

Consider two dictionaries:

a = {
   'x' : 1,
   'y' : 2,
   'z' : 3
}

b = {
   'w' : 10,
   'x' : 11,
   'y' : 2
}

To find out what the two dictionaries have in common, simply perform common set operations using the keys() or items() methods. For example:

# Find keys in common
a.keys() & b.keys()   # { 'x', 'y' }

# Find keys in a that are not in b
a.keys() - b.keys()   # { 'z' }

# Find (key,value) pairs in common
a.items() & b.items() # { ('y', 2) }

These kinds of operations can also be used to alter or filter dictionary contents. For example, suppose you want to make a new dictionary with selected keys removed. Here is some sample code using a dictionary comprehension:

# Make a new dictionary with certain keys removed
c = {key:a[key] for key in a.keys() - {'z', 'w'}}
# c is {'x': 1, 'y': 2}

Discussion

A dictionary is a mapping between a set of keys and values. The keys() method of a dictionary returns a keys-view object that exposes the keys. A little-known feature of keys views is that they also support common set operations such as unions, intersections, and differences. Thus, if you need to perform common set operations with dictionary keys, you can often just use the keys-view objects directly without first converting them into a set.

The items() method of a dictionary returns an items-view object consisting of (key, value) pairs. This object supports similar set operations and can be used to perform operations such as finding out which key-value pairs two dictionaries have in common.

Although similar, the values() method of a dictionary does not support the set operations described in this recipe. In part, this is due to the fact that unlike keys, the items contained in a values view aren’t guaranteed to be unique. This alone makes certain set operations of questionable utility. However, if you must perform such calculations, they can be accomplished by simply converting the values to a set first.

1.10. Removing Duplicates from a Sequence while Maintaining Order

Problem

You want to eliminate the duplicate values in a sequence, but preserve the order of the remaining items.

Solution

If the values in the sequence are hashable, the problem can be easily solved using a set and a generator. For example:

def dedupe(items):
    seen = set()
    for item in items:
        if item not in seen:
            yield item
            seen.add(item)

Here is an example of how to use your function:

>>> a = [1, 5, 2, 1, 9, 1, 5, 10]
>>> list(dedupe(a))
[1, 5, 2, 9, 10]
>>>

This only works if the items in the sequence are hashable. If you are trying to eliminate duplicates in a sequence of unhashable types (such as dicts), you can make a slight change to this recipe, as follows:

def dedupe(items, key=None):
    seen = set()
    for item in items:
        val = item if key is None else key(item)
        if val not in seen:
            yield item
            seen.add(val)

Here, the purpose of the key argument is to specify a function that converts sequence items into a hashable type for the purposes of duplicate detection. Here’s how it works:

>>> a = [ {'x':1, 'y':2}, {'x':1, 'y':3}, {'x':1, 'y':2}, {'x':2, 'y':4}]
>>> list(dedupe(a, key=lambda d: (d['x'],d['y'])))
[{'x': 1, 'y': 2}, {'x': 1, 'y': 3}, {'x': 2, 'y': 4}]
>>> list(dedupe(a, key=lambda d: d['x']))
[{'x': 1, 'y': 2}, {'x': 2, 'y': 4}]
>>>

This latter solution also works nicely if you want to eliminate duplicates based on the value of a single field or attribute or a larger data structure.

Discussion

If all you want to do is eliminate duplicates, it is often easy enough to make a set. For example:

>>> a
[1, 5, 2, 1, 9, 1, 5, 10]
>>> set(a)
{1, 2, 10, 5, 9}
>>>

However, this approach doesn’t preserve any kind of ordering. So, the resulting data will be scrambled afterward. The solution shown avoids this.

The use of a generator function in this recipe reflects the fact that you might want the function to be extremely general purpose—not necessarily tied directly to list processing. For example, if you want to read a file, eliminating duplicate lines, you could simply do this:

with open(somefile,'r') as f:
    for line in dedupe(f):
        ...

The specification of a key function mimics similar functionality in built-in functions such as sorted(), min(), and max(). For instance, see Recipes 1.8 and 1.13.

1.11. Naming a Slice

Problem

Your program has become an unreadable mess of hardcoded slice indices and you want to clean it up.

Solution

Suppose you have some code that is pulling specific data fields out of a record string with fixed fields (e.g., from a flat file or similar format):

######    0123456789012345678901234567890123456789012345678901234567890'
record = '....................100          .......513.25     ..........'
cost = int(record[20:32]) * float(record[40:48])

Instead of doing that, why not name the slices like this?

SHARES = slice(20,32)
PRICE  = slice(40,48)

cost = int(record[SHARES]) * float(record[PRICE])

In the latter version, you avoid having a lot of mysterious hardcoded indices, and what you’re doing becomes much clearer.

Discussion

As a general rule, writing code with a lot of hardcoded index values leads to a readability and maintenance mess. For example, if you come back to the code a year later, you’ll look at it and wonder what you were thinking when you wrote it. The solution shown is simply a way of more clearly stating what your code is actually doing.

In general, the built-in slice() creates a slice object that can be used anywhere a slice is allowed. For example:

>>> items = [0, 1, 2, 3, 4, 5, 6]
>>> a = slice(2, 4)
>>> items[2:4]
[2, 3]
>>> items[a]
[2, 3]
>>> items[a] = [10,11]
>>> items
[0, 1, 10, 11, 4, 5, 6]
>>> del items[a]
>>> items
[0, 1, 4, 5, 6]

If you have a slice instance s, you can get more information about it by looking at its s.start, s.stop, and s.step attributes, respectively. For example:

>>> a = slice(5, 50, 2)
>>> a.start
5
>>> a.stop
50
>>> a.step
2
>>>

In addition, you can map a slice onto a sequence of a specific size by using its indices(size) method. This returns a tuple (start, stop, step) where all values have been suitably limited to fit within bounds (as to avoid IndexError exceptions when indexing). For example:

>>> s = 'HelloWorld'
>>> a.indices(len(s))
(5, 10, 2)
>>> for i in range(*a.indices(len(s))):
...     print(s[i])
...
W
r
d
>>>

1.12. Determining the Most Frequently Occurring Items in a Sequence

Problem

You have a sequence of items, and you’d like to determine the most frequently occurring items in the sequence.

Solution

The collections.Counter class is designed for just such a problem. It even comes with a handy most_common() method that will give you the answer.

To illustrate, let’s say you have a list of words and you want to find out which words occur most often. Here’s how you would do it:

words = [
   'look', 'into', 'my', 'eyes', 'look', 'into', 'my', 'eyes',
   'the', 'eyes', 'the', 'eyes', 'the', 'eyes', 'not', 'around', 'the',
   'eyes', "don't", 'look', 'around', 'the', 'eyes', 'look', 'into',
   'my', 'eyes', "you're", 'under'
]

from collections import Counter
word_counts = Counter(words)
top_three = word_counts.most_common(3)
print(top_three)
# Outputs [('eyes', 8), ('the', 5), ('look', 4)]

Discussion

As input, Counter objects can be fed any sequence of hashable input items. Under the covers, a Counter is a dictionary that maps the items to the number of occurrences. For example:

>>> word_counts['not']
1
>>> word_counts['eyes']
8
>>>

If you want to increment the count manually, simply use addition:

>>> morewords = ['why','are','you','not','looking','in','my','eyes']
>>> for word in morewords:
...     word_counts[word] += 1
...
>>> word_counts['eyes']
9
>>>

Or, alternatively, you could use the update() method:

>>> word_counts.update(morewords)
>>>

A little-known feature of Counter instances is that they can be easily combined using various mathematical operations. For example:

>>> a = Counter(words)
>>> b = Counter(morewords)
>>> a
Counter({'eyes': 8, 'the': 5, 'look': 4, 'into': 3, 'my': 3, 'around': 2,
         "you're": 1, "don't": 1, 'under': 1, 'not': 1})
>>> b
Counter({'eyes': 1, 'looking': 1, 'are': 1, 'in': 1, 'not': 1, 'you': 1,
         'my': 1, 'why': 1})

>>> # Combine counts
>>> c = a + b
>>> c
Counter({'eyes': 9, 'the': 5, 'look': 4, 'my': 4, 'into': 3, 'not': 2,
         'around': 2, "you're": 1, "don't": 1, 'in': 1, 'why': 1,
         'looking': 1, 'are': 1, 'under': 1, 'you': 1})

>>> # Subtract counts
>>> d = a - b
>>> d
Counter({'eyes': 7, 'the': 5, 'look': 4, 'into': 3, 'my': 2, 'around': 2,
         "you're": 1, "don't": 1, 'under': 1})
>>>

Needless to say, Counter objects are a tremendously useful tool for almost any kind of problem where you need to tabulate and count data. You should prefer this over manually written solutions involving dictionaries.

1.13. Sorting a List of Dictionaries by a Common Key

Problem

You have a list of dictionaries and you would like to sort the entries according to one or more of the dictionary values.

Solution

Sorting this type of structure is easy using the operator module’s itemgetter function. Let’s say you’ve queried a database table to get a listing of the members on your website, and you receive the following data structure in return:

rows = [
    {'fname': 'Brian', 'lname': 'Jones', 'uid': 1003},
    {'fname': 'David', 'lname': 'Beazley', 'uid': 1002},
    {'fname': 'John', 'lname': 'Cleese', 'uid': 1001},
    {'fname': 'Big', 'lname': 'Jones', 'uid': 1004}
]

It’s fairly easy to output these rows ordered by any of the fields common to all of the dictionaries. For example:

from operator import itemgetter

rows_by_fname = sorted(rows, key=itemgetter('fname'))
rows_by_uid = sorted(rows, key=itemgetter('uid'))

print(rows_by_fname)
print(rows_by_uid)

The preceding code would output the following:

[{'fname': 'Big', 'uid': 1004, 'lname': 'Jones'},
 {'fname': 'Brian', 'uid': 1003, 'lname': 'Jones'},
 {'fname': 'David', 'uid': 1002, 'lname': 'Beazley'},
 {'fname': 'John', 'uid': 1001, 'lname': 'Cleese'}]

[{'fname': 'John', 'uid': 1001, 'lname': 'Cleese'},
 {'fname': 'David', 'uid': 1002, 'lname': 'Beazley'},
 {'fname': 'Brian', 'uid': 1003, 'lname': 'Jones'},
 {'fname': 'Big', 'uid': 1004, 'lname': 'Jones'}]

The itemgetter() function can also accept multiple keys. For example, this code

rows_by_lfname = sorted(rows, key=itemgetter('lname','fname'))
print(rows_by_lfname)

Produces output like this:

[{'fname': 'David', 'uid': 1002, 'lname': 'Beazley'},
 {'fname': 'John', 'uid': 1001, 'lname': 'Cleese'},
 {'fname': 'Big', 'uid': 1004, 'lname': 'Jones'},
 {'fname': 'Brian', 'uid': 1003, 'lname': 'Jones'}]

Discussion

In this example, rows is passed to the built-in sorted() function, which accepts a keyword argument key. This argument is expected to be a callable that accepts a single item from rows as input and returns a value that will be used as the basis for sorting. The itemgetter() function creates just such a callable.

The operator.itemgetter() function takes as arguments the lookup indices used to extract the desired values from the records in rows. It can be a dictionary key name, a numeric list element, or any value that can be fed to an object’s __getitem__() method. If you give multiple indices to itemgetter(), the callable it produces will return a tuple with all of the elements in it, and sorted() will order the output according to the sorted order of the tuples. This can be useful if you want to simultaneously sort on multiple fields (such as last and first name, as shown in the example).

The functionality of itemgetter() is sometimes replaced by lambda expressions. For example:

rows_by_fname = sorted(rows, key=lambda r: r['fname'])
rows_by_lfname = sorted(rows, key=lambda r: (r['lname'],r['fname']))

This solution often works just fine. However, the solution involving itemgetter() typically runs a bit faster. Thus, you might prefer it if performance is a concern.

Last, but not least, don’t forget that the technique shown in this recipe can be applied to functions such as min() and max(). For example:

>>> min(rows, key=itemgetter('uid'))
{'fname': 'John', 'lname': 'Cleese', 'uid': 1001}
>>> max(rows, key=itemgetter('uid'))
{'fname': 'Big', 'lname': 'Jones', 'uid': 1004}
>>>

1.14. Sorting Objects Without Native Comparison Support

Problem

You want to sort objects of the same class, but they don’t natively support comparison operations.

Solution

The built-in sorted() function takes a key argument that can be passed a callable that will return some value in the object that sorted will use to compare the objects. For example, if you have a sequence of User instances in your application, and you want to sort them by their user_id attribute, you would supply a callable that takes a User instance as input and returns the user_id. For example:

>>> class User:
...     def __init__(self, user_id):
...         self.user_id = user_id
...     def __repr__(self):
...         return 'User({})'.format(self.user_id)
...
>>> users = [User(23), User(3), User(99)]
>>> users
[User(23), User(3), User(99)]
>>> sorted(users, key=lambda u: u.user_id)
[User(3), User(23), User(99)]
>>>

Instead of using lambda, an alternative approach is to use operator.attrgetter():

>>> from operator import attrgetter
>>> sorted(users, key=attrgetter('user_id'))
[User(3), User(23), User(99)]
>>>

Discussion

The choice of whether or not to use lambda or attrgetter() may be one of personal preference. However, attrgetter() is often a tad bit faster and also has the added feature of allowing multiple fields to be extracted simultaneously. This is analogous to the use of operator.itemgetter() for dictionaries (see Recipe 1.13). For example, if User instances also had a first_name and last_name attribute, you could perform a sort like this:

by_name = sorted(users, key=attrgetter('last_name', 'first_name'))

It is also worth noting that the technique used in this recipe can be applied to functions such as min() and max(). For example:

>>> min(users, key=attrgetter('user_id')
User(3)
>>> max(users, key=attrgetter('user_id')
User(99)
>>>

1.15. Grouping Records Together Based on a Field

Problem

You have a sequence of dictionaries or instances and you want to iterate over the data in groups based on the value of a particular field, such as date.

Solution

The itertools.groupby() function is particularly useful for grouping data together like this. To illustrate, suppose you have the following list of dictionaries:

rows = [
    {'address': '5412 N CLARK', 'date': '07/01/2012'},
    {'address': '5148 N CLARK', 'date': '07/04/2012'},
    {'address': '5800 E 58TH', 'date': '07/02/2012'},
    {'address': '2122 N CLARK', 'date': '07/03/2012'},
    {'address': '5645 N RAVENSWOOD', 'date': '07/02/2012'},
    {'address': '1060 W ADDISON', 'date': '07/02/2012'},
    {'address': '4801 N BROADWAY', 'date': '07/01/2012'},
    {'address': '1039 W GRANVILLE', 'date': '07/04/2012'},
]

Now suppose you want to iterate over the data in chunks grouped by date. To do it, first sort by the desired field (in this case, date) and then use itertools.groupby():

from operator import itemgetter
from itertools import groupby

# Sort by the desired field first
rows.sort(key=itemgetter('date'))

# Iterate in groups
for date, items in groupby(rows, key=itemgetter('date')):
    print(date)
    for i in items:
        print('    ', i)

This produces the following output:

    07/01/2012
         {'date': '07/01/2012', 'address': '5412 N CLARK'}
         {'date': '07/01/2012', 'address': '4801 N BROADWAY'}
    07/02/2012
         {'date': '07/02/2012', 'address': '5800 E 58TH'}
         {'date': '07/02/2012', 'address': '5645 N RAVENSWOOD'}
         {'date': '07/02/2012', 'address': '1060 W ADDISON'}
    07/03/2012
         {'date': '07/03/2012', 'address': '2122 N CLARK'}
    07/04/2012
         {'date': '07/04/2012', 'address': '5148 N CLARK'}
         {'date': '07/04/2012', 'address': '1039 W GRANVILLE'}

Discussion

The groupby() function works by scanning a sequence and finding sequential “runs” of identical values (or values returned by the given key function). On each iteration, it returns the value along with an iterator that produces all of the items in a group with the same value.

An important preliminary step is sorting the data according to the field of interest. Since groupby() only examines consecutive items, failing to sort first won’t group the records as you want.

If your goal is to simply group the data together by dates into a large data structure that allows random access, you may have better luck using defaultdict() to build a multidict, as described in Recipe 1.6. For example:

from collections import defaultdict
rows_by_date = defaultdict(list)
for row in rows:
    rows_by_date[row['date']].append(row)

This allows the records for each date to be accessed easily like this:

>>> for r in rows_by_date['07/01/2012']:
...     print(r)
...
{'date': '07/01/2012', 'address': '5412 N CLARK'}
{'date': '07/01/2012', 'address': '4801 N BROADWAY'}
>>>

For this latter example, it’s not necessary to sort the records first. Thus, if memory is no concern, it may be faster to do this than to first sort the records and iterate using groupby().

1.16. Filtering Sequence Elements

Problem

You have data inside of a sequence, and need to extract values or reduce the sequence using some criteria.

Solution

The easiest way to filter sequence data is often to use a list comprehension. For example:

>>> mylist = [1, 4, -5, 10, -7, 2, 3, -1]
>>> [n for n in mylist if n > 0]
[1, 4, 10, 2, 3]
>>> [n for n in mylist if n < 0]
[-5, -7, -1]
>>>

One potential downside of using a list comprehension is that it might produce a large result if the original input is large. If this is a concern, you can use generator expressions to produce the filtered values iteratively. For example:

>>> pos = (n for n in mylist if n > 0)
>>> pos
<generator object <genexpr> at 0x1006a0eb0>
>>> for x in pos:
...     print(x)
...
1
4
10
2
3
>>>

Sometimes, the filtering criteria cannot be easily expressed in a list comprehension or generator expression. For example, suppose that the filtering process involves exception handling or some other complicated detail. For this, put the filtering code into its own function and use the built-in filter() function. For example:

values = ['1', '2', '-3', '-', '4', 'N/A', '5']

def is_int(val):
    try:
        x = int(val)
        return True
    except ValueError:
        return False

ivals = list(filter(is_int, values))
print(ivals)
# Outputs ['1', '2', '-3', '4', '5']

filter() creates an iterator, so if you want to create a list of results, make sure you also use list() as shown.

Discussion

List comprehensions and generator expressions are often the easiest and most straightforward ways to filter simple data. They also have the added power to transform the data at the same time. For example:

>>> mylist = [1, 4, -5, 10, -7, 2, 3, -1]
>>> import math
>>> [math.sqrt(n) for n in mylist if n > 0]
[1.0, 2.0, 3.1622776601683795, 1.4142135623730951, 1.7320508075688772]
>>>

One variation on filtering involves replacing the values that don’t meet the criteria with a new value instead of discarding them. For example, perhaps instead of just finding positive values, you want to also clip bad values to fit within a specified range. This is often easily accomplished by moving the filter criterion into a conditional expression like this:

>>> clip_neg = [n if n > 0 else 0 for n in mylist]
>>> clip_neg
[1, 4, 0, 10, 0, 2, 3, 0]
>>> clip_pos = [n if n < 0 else 0 for n in mylist]
>>> clip_pos
[0, 0, -5, 0, -7, 0, 0, -1]
>>>

Another notable filtering tool is itertools.compress(), which takes an iterable and an accompanying Boolean selector sequence as input. As output, it gives you all of the items in the iterable where the corresponding element in the selector is True. This can be useful if you’re trying to apply the results of filtering one sequence to another related sequence. For example, suppose you have the following two columns of data:

addresses = [
    '5412 N CLARK',
    '5148 N CLARK',
    '5800 E 58TH',
    '2122 N CLARK',
    '5645 N RAVENSWOOD',
    '1060 W ADDISON',
    '4801 N BROADWAY',
    '1039 W GRANVILLE',
]

counts = [ 0, 3, 10, 4, 1, 7, 6, 1]

Now suppose you want to make a list of all addresses where the corresponding count value was greater than 5. Here’s how you could do it:

>>> from itertools import compress
>>> more5 = [n > 5 for n in counts]
>>> more5
[False, False, True, False, False, True, True, False]
>>> list(compress(addresses, more5))
['5800 E 58TH', '4801 N BROADWAY', '1039 W GRANVILLE']
>>>

The key here is to first create a sequence of Booleans that indicates which elements satisfy the desired condition. The compress() function then picks out the items corresponding to True values.

Like filter(), compress() normally returns an iterator. Thus, you need to use list() to turn the results into a list if desired.

1.17. Extracting a Subset of a Dictionary

Problem

You want to make a dictionary that is a subset of another dictionary.

Solution

This is easily accomplished using a dictionary comprehension. For example:

prices = {
   'ACME': 45.23,
   'AAPL': 612.78,
   'IBM': 205.55,
   'HPQ': 37.20,
   'FB': 10.75
}

# Make a dictionary of all prices over 200
p1 = { key:value for key, value in prices.items() if value > 200 }

# Make a dictionary of tech stocks
tech_names = { 'AAPL', 'IBM', 'HPQ', 'MSFT' }
p2 = { key:value for key,value in prices.items() if key in tech_names }

Discussion

Much of what can be accomplished with a dictionary comprehension might also be done by creating a sequence of tuples and passing them to the dict() function. For example:

p1 = dict((key, value) for key, value in prices.items() if value > 200)

However, the dictionary comprehension solution is a bit clearer and actually runs quite a bit faster (over twice as fast when tested on the prices dictionary used in the example).

Sometimes there are multiple ways of accomplishing the same thing. For instance, the second example could be rewritten as:

# Make a dictionary of tech stocks
tech_names = { 'AAPL', 'IBM', 'HPQ', 'MSFT' }
p2 = { key:prices[key] for key in prices.keys() & tech_names }

However, a timing study reveals that this solution is almost 1.6 times slower than the first solution. If performance matters, it usually pays to spend a bit of time studying it. See Recipe 14.13 for specific information about timing and profiling.

1.18. Mapping Names to Sequence Elements

Problem

You have code that accesses list or tuple elements by position, but this makes the code somewhat difficult to read at times. You’d also like to be less dependent on position in the structure, by accessing the elements by name.

Solution

collections.namedtuple() provides these benefits, while adding minimal overhead over using a normal tuple object. collections.namedtuple() is actually a factory method that returns a subclass of the standard Python tuple type. You feed it a type name, and the fields it should have, and it returns a class that you can instantiate, passing in values for the fields you’ve defined, and so on. For example:

>>> from collections import namedtuple
>>> Subscriber = namedtuple('Subscriber', ['addr', 'joined'])
>>> sub = Subscriber('[e-mail]', '2012-10-19')
>>> sub
Subscriber(addr='[e-mail]', joined='2012-10-19')
>>> sub.addr
'[e-mail]'
>>> sub.joined
'2012-10-19'
>>>

Although an instance of a namedtuple looks like a normal class instance, it is interchangeable with a tuple and supports all of the usual tuple operations such as indexing and unpacking. For example:

>>> len(sub)
2
>>> addr, joined = sub
>>> addr
'[e-mail]'
>>> joined
'2012-10-19'
>>>

A major use case for named tuples is decoupling your code from the position of the elements it manipulates. So, if you get back a large list of tuples from a database call, then manipulate them by accessing the positional elements, your code could break if, say, you added a new column to your table. Not so if you first cast the returned tuples to namedtuples.

To illustrate, here is some code using ordinary tuples:

def compute_cost(records):
    total = 0.0
    for rec in records:
        total += rec[1] * rec[2]
    return total

References to positional elements often make the code a bit less expressive and more dependent on the structure of the records. Here is a version that uses a namedtuple:

from collections import namedtuple

Stock = namedtuple('Stock', ['name', 'shares', 'price'])

def compute_cost(records):
    total = 0.0
    for rec in records:
        s = Stock(*rec)
        total += s.shares * s.price
    return total

Naturally, you can avoid the explicit conversion to the Stock namedtuple if the records sequence in the example already contained such instances.

Discussion

One possible use of a namedtuple is as a replacement for a dictionary, which requires more space to store. Thus, if you are building large data structures involving dictionaries, use of a namedtuple will be more efficient. However, be aware that unlike a dictionary, a namedtuple is immutable. For example:

>>> s = Stock('ACME', 100, 123.45)
>>> s
Stock(name='ACME', shares=100, price=123.45)
>>> s.shares = 75
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
AttributeError: can't set attribute
>>>

If you need to change any of the attributes, it can be done using the _replace() method of a namedtuple instance, which makes an entirely new namedtuple with specified values replaced. For example:

>>> s = s._replace(shares=75)
>>> s
Stock(name='ACME', shares=75, price=123.45)
>>>

A subtle use of the _replace() method is that it can be a convenient way to populate named tuples that have optional or missing fields. To do this, you make a prototype tuple containing the default values and then use _replace() to create new instances with values replaced. For example:

from collections import namedtuple

Stock = namedtuple('Stock', ['name', 'shares', 'price', 'date', 'time'])

# Create a prototype instance
stock_prototype = Stock('', 0, 0.0, None, None)

# Function to convert a dictionary to a Stock
def dict_to_stock(s):
    return stock_prototype._replace(**s)

Here is an example of how this code would work:

>>> a = {'name': 'ACME', 'shares': 100, 'price': 123.45}
>>> dict_to_stock(a)
Stock(name='ACME', shares=100, price=123.45, date=None, time=None)
>>> b = {'name': 'ACME', 'shares': 100, 'price': 123.45, 'date': '12/17/2012'}
>>> dict_to_stock(b)
Stock(name='ACME', shares=100, price=123.45, date='12/17/2012', time=None)
>>>

Last, but not least, it should be noted that if your goal is to define an efficient data structure where you will be changing various instance attributes, using namedtuple is not your best choice. Instead, consider defining a class using __slots__ instead (see Recipe 8.4).

1.19. Transforming and Reducing Data at the Same Time

Problem

You need to execute a reduction function (e.g., sum(), min(), max()), but first need to transform or filter the data.

Solution

A very elegant way to combine a data reduction and a transformation is to use a generator-expression argument. For example, if you want to calculate the sum of squares, do the following:

nums = [1, 2, 3, 4, 5]
s = sum(x * x for x in nums)

Here are a few other examples:

# Determine if any .py files exist in a directory
import os
files = os.listdir('dirname')
if any(name.endswith('.py') for name in files):
    print('There be python!')
else:
    print('Sorry, no python.')

# Output a tuple as CSV
s = ('ACME', 50, 123.45)
print(','.join(str(x) for x in s))

# Data reduction across fields of a data structure
portfolio = [
   {'name':'GOOG', 'shares': 50},
   {'name':'YHOO', 'shares': 75},
   {'name':'AOL', 'shares': 20},
   {'name':'SCOX', 'shares': 65}
]
min_shares = min(s['shares'] for s in portfolio)

Discussion

The solution shows a subtle syntactic aspect of generator expressions when supplied as the single argument to a function (i.e., you don’t need repeated parentheses). For example, these statements are the same:

s = sum((x * x for x in nums))    # Pass generator-expr as argument
s = sum(x * x for x in nums)      # More elegant syntax

Using a generator argument is often a more efficient and elegant approach than first creating a temporary list. For example, if you didn’t use a generator expression, you might consider this alternative implementation:

nums = [1, 2, 3, 4, 5]
s = sum([x * x for x in nums])

This works, but it introduces an extra step and creates an extra list. For such a small list, it might not matter, but if nums was huge, you would end up creating a large temporary data structure to only be used once and discarded. The generator solution transforms the data iteratively and is therefore much more memory-efficient.

Certain reduction functions such as min() and max() accept a key argument that might be useful in situations where you might be inclined to use a generator. For example, in the portfolio example, you might consider this alternative:

# Original: Returns 20
min_shares = min(s['shares'] for s in portfolio)

# Alternative: Returns {'name': 'AOL', 'shares': 20}
min_shares = min(portfolio, key=lambda s: s['shares'])

1.20. Combining Multiple Mappings into a Single Mapping

Problem

You have multiple dictionaries or mappings that you want to logically combine into a single mapping to perform certain operations, such as looking up values or checking for the existence of keys.

Solution

Suppose you have two dictionaries:

a = {'x': 1, 'z': 3 }
b = {'y': 2, 'z': 4 }

Now suppose you want to perform lookups where you have to check both dictionaries (e.g., first checking in a and then in b if not found). An easy way to do this is to use the ChainMap class from the collections module. For example:

from collections import ChainMap
c = ChainMap(a,b)
print(c['x'])      # Outputs 1  (from a)
print(c['y'])      # Outputs 2  (from b)
print(c['z'])      # Outputs 3  (from a)

Discussion

A ChainMap takes multiple mappings and makes them logically appear as one. However, the mappings are not literally merged together. Instead, a ChainMap simply keeps a list of the underlying mappings and redefines common dictionary operations to scan the list. Most operations will work. For example:

>>> len(c)
3
>>> list(c.keys())
['x', 'y', 'z']
>>> list(c.values())
[1, 2, 3]
>>>

If there are duplicate keys, the values from the first mapping get used. Thus, the entry c['z'] in the example would always refer to the value in dictionary a, not the value in dictionary b.

Operations that mutate the mapping always affect the first mapping listed. For example:

>>> c['z'] = 10
>>> c['w'] = 40
>>> del c['x']
>>> a
{'w': 40, 'z': 10}
>>> del c['y']
Traceback (most recent call last):
...
KeyError: "Key not found in the first mapping: 'y'"
>>>

A ChainMap is particularly useful when working with scoped values such as variables in a programming language (i.e., globals, locals, etc.). In fact, there are methods that make this easy:

>>> values = ChainMap()
>>> values['x'] = 1
>>> # Add a new mapping
>>> values = values.new_child()
>>> values['x'] = 2
>>> # Add a new mapping
>>> values = values.new_child()
>>> values['x'] = 3
>>> values
ChainMap({'x': 3}, {'x': 2}, {'x': 1})
>>> values['x']
3
>>> # Discard last mapping
>>> values = values.parents
>>> values['x']
2
>>> # Discard last mapping
>>> values = values.parents
>>> values['x']
1
>>> values
ChainMap({'x': 1})
>>>

As an alternative to ChainMap, you might consider merging dictionaries together using the update() method. For example:

>>> a = {'x': 1, 'z': 3 }
>>> b = {'y': 2, 'z': 4 }
>>> merged = dict(b)
>>> merged.update(a)
>>> merged['x']
1
>>> merged['y']
2
>>> merged['z']
3
>>>

This works, but it requires you to make a completely separate dictionary object (or destructively alter one of the existing dictionaries). Also, if any of the original dictionaries mutate, the changes don’t get reflected in the merged dictionary. For example:

>>> a['x'] = 13
>>> merged['x']
1

A ChainMap uses the original dictionaries, so it doesn’t have this behavior. For example:

>>> a = {'x': 1, 'z': 3 }
>>> b = {'y': 2, 'z': 4 }
>>> merged = ChainMap(a, b)
>>> merged['x']
1
>>> a['x'] = 42
>>> merged['x']   # Notice change to merged dicts
42
>>>