Itertools in Python 3, By Example

What Is Itertools and Why Should You Use It?

According to the itertools docs, it is a “module [that] implements a number of iterator building blocks inspired by constructs from APL, Haskell, and SML… Together, they form an ‘iterator algebra’ making it possible to construct specialized tools succinctly and efficiently in pure Python.”

Loosely speaking, this means that the functions in itertools “operate” on iterators to produce more complex iterators. Consider, for example, the built-in zip() function, which takes any number of iterables as arguments and returns an iterator over tuples of their corresponding elements:

>>> list(zip([1, 2, 3], ['a', 'b', 'c']))
[(1, 'a'), (2, 'b'), (3, 'c')]

How, exactly, does zip() work?

[1, 2, 3] and ['a', 'b', 'c'], like all lists, are iterable, which means they can return their elements one at a time. Technically, any Python object that implements the .__iter__() or .__getitem__() methods is iterable. (See the Python 3 docs glossary for a more detailed explanation.)

The iter() built-in function, when called on an iterable, returns an iterator object for that iterable:

>>> iter([1, 2, 3, 4])
<list_iterator object at 0x7fa80af0d898>

Under the hood, the zip() function works, in essence, by calling iter() on each of its arguments, then advancing each iterator returned by iter() with next() and aggregating the results into tuples. The iterator returned by zip() iterates over these tuples.

The map() built-in function is another “iterator operator” that, in its simplest form, applies a single-parameter function to each element of an iterable one element at a time:

>>> list(map(len, ['abc', 'de', 'fghi']))
[3, 2, 4]

The map() function works by calling iter() on its second argument, advancing this iterator with next() until the iterator is exhausted, and applying the function passed to its first argument to the value returned by next() at each step. In the above example, len() is called on each element of ['abc', 'de', 'fghi'] to return an iterator over the lengths of each string in the list.

Since iterators are iterable, you can compose zip() and map() to produce an iterator over combinations of elements in more than one iterable. For example, the following sums corresponding elements of two lists:

>>> list(map(sum, zip([1, 2, 3], [4, 5, 6])))
[5, 7, 9]

This is what is meant by the functions in itertools forming an “iterator algebra.” itertools is best viewed as a collection of building blocks that can be combined to form specialized “data pipelines” like the one in the example above.

Historical Note: In Python 2, the built-in zip() and map() functions do not return an iterator, but rather a list. To return an iterator, the izip() and imap() functions of itertools must be used. In Python 3, izip() and imap() have been removed from itertools and replaced the zip() and map() built-ins. So, in a way, if you have ever used zip() or map() in Python 3, you have already been using itertools!

There are two main reasons why such an “iterator algebra” is useful: improved memory efficiency (via lazy evaluation) and faster execution time. To see this, consider the following problem:

Given a list of values inputs and a positive integer n, write a function that splits inputs into groups of length n. For simplicity, assume that the length of the input list is divisible by n. For example, if inputs = [1, 2, 3, 4, 5, 6] and n = 2, your function should return [(1, 2), (3, 4), (5, 6)].

Taking a naive approach, you might write something like this:

def naive_grouper(inputs, n):
    num_groups = len(inputs) // n
    return [tuple(inputs[i*n:(i+1)*n]) for i in range(num_groups)]

When you test it, you see that it works as expected:

>>> nums = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
>>> naive_grouper(nums, 2)
[(1, 2), (3, 4), (5, 6), (7, 8), (9, 10)]

What happens when you try to pass it a list with, say, 100 million elements? You will need a whole lot of available memory! Even if you have enough memory available, your program will hang for a while until the output list is populated.

To see this, store the following in a script called naive.py:

def naive_grouper(inputs, n):
    num_groups = len(inputs) // n
    return [tuple(inputs[i*n:(i+1)*n]) for i in range(num_groups)]


for _ in naive_grouper(range(100000000), 10):
    pass

From the console, you can use the time command (on UNIX systems) to measure memory usage and CPU user time. Make sure you have at least 5GB of free memory before executing the following:

$ time -f "Memory used (kB): %M\nUser time (seconds): %U" python3 naive.py
Memory used (kB): 4551872
User time (seconds): 11.04

Note: On Ubuntu, you may need to run /usr/bin/time instead of time for the above example to work.

The list and tuple implementation in naive_grouper() requires approximately 4.5GB of memory to process range(100000000). Working with iterators drastically improves this situation. Consider the following:

def better_grouper(inputs, n):
    iters = [iter(inputs)] * n
    return zip(*iters)

There’s a lot going on in this little function, so let’s break it down with a concrete example. The expression [iters(inputs)] * n creates a list of n references to the same iterator:

>>> nums = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
>>> iters = [iter(nums)] * 2
>>> list(id(itr) for itr in iters)  # IDs are the same.
[139949748267160, 139949748267160]

Next, zip(*iters) returns an iterator over pairs of corresponding elements of each iterator in iters. When the first element, 1, is taken from the “first” iterator, the “second” iterator now starts at 2 since it is just a reference to the “first” iterator and has therefore been advanced one step. So, the first tuple produced by zip() is (1, 2).

At this point, “both” iterators in iters start at 3, so when zip() pulls 3 from the “first” iterator, it gets 4 from the “second” to produce the tuple (3, 4). This process continues until zip() finally produces (9, 10) and “both” iterators in iters are exhausted:

>>> nums = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
>>> list(better_grouper(nums, 2))
[(1, 2), (3, 4), (5, 6), (7, 8), (9, 10)]

The better_grouper() function is better for a couple of reasons. First, without the reference to the len() built-in, better_grouper() can take any iterable as an argument (even infinite iterators). Second, by returning an iterator rather than a list, better_grouper() can process enormous iterables without trouble and uses much less memory.

Store the following in a file called better.py and run it with time from the console again:

def better_grouper(inputs, n):
    iters = [iter(inputs)] * n
    return zip(*iters)


for _ in better_grouper(range(100000000), 10):
    pass

$ time -f "Memory used (kB): %M\nUser time (seconds): %U" python3 better.py
Memory used (kB): 7224
User time (seconds): 2.48

That’s a whopping 630 times less memory used than naive.py in less than a quarter of the time!

Now that you’ve seen what itertools is (“iterator algebra”) and why you should use it (improved memory efficiency and faster execution time), let’s take a look at how to take better_grouper() to the next level with itertools.

The grouper Recipe

The problem with better_grouper() is that it doesn’t handle situations where the value passed to the second argument isn’t a factor of the length of the iterable in the first argument:

>>> nums = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
>>> list(better_grouper(nums, 4))
[(1, 2, 3, 4), (5, 6, 7, 8)]

The elements 9 and 10 are missing from the grouped output. This happens because zip() stops aggregating elements once the shortest iterable passed to it is exhausted. It would make more sense to return a third group containing 9 and 10.

To do this, you can use itertools.zip_longest(). This function accepts any number of iterables as arguments and a fillvalue keyword argument that defaults to None. The easiest way to get a sense of the difference between zip() and zip_longest() is to look at some example output:

>>> import itertools as it
>>> x = [1, 2, 3, 4, 5]
>>> y = ['a', 'b', 'c']
>>> list(zip(x, y))
[(1, 'a'), (2, 'b'), (3, 'c')]
>>> list(it.zip_longest(x, y))
[(1, 'a'), (2, 'b'), (3, 'c'), (4, None), (5, None)]

With this in mind, replace zip() in better_grouper() with zip_longest():

import itertools as it


def grouper(inputs, n, fillvalue=None):
    iters = [iter(inputs)] * n
    return it.zip_longest(*iters, fillvalue=fillvalue)

Now you get a better result:

>>> nums = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
>>> print(list(grouper(nums, 4)))
[(1, 2, 3, 4), (5, 6, 7, 8), (9, 10, None, None)]

The grouper() function can be found in the Recipes section of the itertools docs. The recipes are an excellent source of inspiration for ways to use itertools to your advantage.

Et tu, Brute Force?

Here’s a common interview-style problem:

You have three $20 dollar bills, five $10 dollar bills, two $5 dollar bills, and five $1 dollar bills. How many ways can you make change for a $100 dollar bill?

To “brute force” this problem, you just start listing off the ways there are to choose one bill from your wallet, check whether any of these makes change for $100, then list the ways to pick two bills from your wallet, check again, and so on and so forth.

But you are a programmer, so naturally you want to automate this process.

First, create a list of the bills you have in your wallet:

bills = [20, 20, 20, 10, 10, 10, 10, 10, 5, 5, 1, 1, 1, 1, 1]

A choice of k things from a set of n things is called a combination, and itertools has your back here. The itertools.combinations() function takes two arguments—an iterable inputs and a positive integer n—and produces an iterator over tuples of all combinations of n elements in inputs.

For example, to list the combinations of three bills in your wallet, just do:

>>> list(it.combinations(bills, 3))
 [(20, 20, 20), (20, 20, 10), (20, 20, 10), ... ]

To solve the problem, you can loop over the positive integers from 1 to len(bills), then check which combinations of each size add up to $100:

>>> makes_100 = []
>>> for n in range(1, len(bills) + 1):
...     for combination in it.combinations(bills, n):
...         if sum(combination) == 100:
...             makes_100.append(combination)

If you print out makes_100, you will notice there are a lot of repeated combinations. This makes sense because you can make change for $100 with three $20 dollar bills and four $10 bills, but combinations() does this with the first four $10 dollars bills in your wallet; the first, third, fourth and fifth $10 dollar bills; the first, second, fourth and fifth $10 bills; and so on.

To remove duplicates from makes_100, you can convert it to a set:

>>> set(makes_100)
{(20, 20, 10, 10, 10, 10, 10, 5, 1, 1, 1, 1, 1),
 (20, 20, 10, 10, 10, 10, 10, 5, 5),
 (20, 20, 20, 10, 10, 10, 5, 1, 1, 1, 1, 1),
 (20, 20, 20, 10, 10, 10, 5, 5),
 (20, 20, 20, 10, 10, 10, 10)}

So, there are five ways to make change for a $100 bill with the bills you have in your wallet.

Here’s a variation on the same problem:

How many ways are there to make change for a $100 bill using any number of $50, $20, $10, $5, and $1 dollar bills?

In this case, you don’t have a pre-set collection of bills, so you need a way to generate all possible combinations using any number of bills. For this, you’ll need the itertools.combinations_with_replacement() function.

It works just like combinations(), accepting an iterable inputs and a positive integer n, and returns an iterator over n-tuples of elements from inputs. The difference is that combinations_with_replacement() allows elements to be repeated in the tuples it returns.

For example:

>>> list(it.combinations_with_replacement([1, 2], 2))
[(1, 1), (1, 2), (2, 2)]

Compare that to combinations():

>>> list(it.combinations([1, 2], 2))
[(1, 2)]

Here’s what the solution to the revised problem looks like:

>>> bills = [50, 20, 10, 5, 1]
>>> make_100 = []
>>> for n in range(1, 101):
...     for combination in it.combinations_with_replacement(bills, n):
...         if sum(combination) == 100:
...             makes_100.append(combination)

In this case, you do not need to remove any duplicates since combinations_with_replacement() won’t produce any:

>>> len(makes_100)
343

If you run the above solution, you may notice that it takes a while for the output to display. That is because it has to process 96,560,645 combinations!

Another “brute force” itertools function is permutations(), which accepts a single iterable and produces all possible permutations (rearrangements) of its elements:

>>> list(it.permutations(['a', 'b', 'c']))
[('a', 'b', 'c'), ('a', 'c', 'b'), ('b', 'a', 'c'),
 ('b', 'c', 'a'), ('c', 'a', 'b'), ('c', 'b', 'a')]

Any iterable of three elements will have six permutations, and the number of permutations of longer iterables grows extremely fast. In fact, an iterable of length n has n! permutations, where

To put this in perspective, here’s a table of these numbers for n = 1 to n = 10:

The phenomenon of just a few inputs producing a large number of outcomes is called a combinatorial explosion and is something to keep in mind when working with combinations(), combinations_with_replacement(), and permutations().

It is usually best to avoid brute force algorithms, although there are times you may need to use one (for example, if the correctness of the algorithm is critical, or every possible outcome must be considered). In that case, itertools has you covered.

>>> combinations([1, 2, 3], 2)
(1, 2), (1, 3), (2, 3)

>>> combinations_with_replacement([1, 2], 2)
(1, 1), (1, 2), (2, 2)

>>> permutations('abc')
('a', 'b', 'c'), ('a', 'c', 'b'), ('b', 'a', 'c'),
('b', 'c', 'a'), ('c', 'a', 'b'), ('c', 'b', 'a')

Sequences of Numbers

With itertools, you can easily generate iterators over infinite sequences. In this section, you will explore numeric sequences, but the tools and techniques seen here are by no means limited to numbers.

>>> def evens():
...     """Generate even integers, starting with 0."""
...     n = 0
...     while True:
...         yield n
...         n += 2
...
>>> evens = evens()
>>> list(next(evens) for _ in range(5))
[0, 2, 4, 6, 8]

>>> def odds():
...     """Generate odd integers, starting with 1."""
...     n = 1
...     while True:
...         yield n
...         n += 2
...