import itertools

Ask a programmer to review 10 lines of code and they’ll find 10 issues. Ask them to review 500 and they’ll say it looks good.

Hey pips!

>>> import itertools

What’s itertools, you might ask?

It’s a standard library which comes with Python. It provides many functions for handling iterables of any kind – hence the name “iter” tools!

Recall that an iterable is any object that can be iterated over (you’ll most commonly be using list). Like many other built-in functions we’ve been looking at, the functions in the itertools library all return a special accessor iterator, rather than a list. You can be assured these standard library functions are super-fast and very memory efficient.

So, suppose you wanted to join 2 iterables together and iterate over them at once. With lists we could do

>>> for each in [1, 2, 3] + [4, 5, 6]:
        ...

But for other iterables, this could end up problematic for a number of reasons. Maybe you need to guarantee a particular order, or they’re different types, or they just don’t support the + operator.

itertools.chain() solves this by wrapping all those individual iterables in 1 iterator:

>>> from itertools import chain

>>> it_1 = (0, 1)
>>> it_2 = {7, 6, 5}
>>> it_3 = range(70, 145, 25)

>>> for each in chain(it_1, it_2, it_3):
        print(each)
0
1
5
6
7
70
95
120

Now imagine you want to iterate over this in chunks of 3. You could do this with something like

>>> it = list(chain(it_1, it_2, it_3))

>>> for i in range(int(len(it) / 3)):
        first  = it[3*i]
        second = it[3*i + 1]
        third  = it[3*i + 2]

It works, but it’s a little cumbersome, especially for much larger batches. itertools.batched() is what you’re looking for:

>>> from itertools import batched

>>> for first, second, third in batched(it, 3):
        print(first + second + third)
6
83
215

The second argument to batched() there is the batch size. And the great thing about using the builtin is that it handles edge cases for you! If the last batch has fewer than 3 items, it will still batch them without complaint. (Of course, this is Python, so you can also disable this behaviour by using batched(strict = True))

[!Note] This is a relatively new function, only added in Python 3.12. You may need to bump your Python version to use it.

Where itertools gets really cool is the combinatorics functions. These are insanely helpful for Olympiad/competitive programming problems.

For instance, you can generate all the permutations of some number of objects with itertools.permutations():

>>> from itertools import permutations as perms

>>> choices = [0, 1, "s"]
>>> list(perms(choices))
[(0, 1, 's'), (0, 's', 1), (1, 0, 's'), (1, 's', 0), ('s', 0, 1), ('s', 1, 0)]

Or if you want only a subset:

>>> list(perms(choices, 2))
[(0, 1), (0, 's'), (1, 0), (1, 's'), ('s', 0), ('s', 1)]

Or maybe these are fruit in a bag, and you’re picking a pair without replacement:

>>> from itertools import combinations as combs

>>> fruits = ["apples", "bananas", "jackfruit", "guava"]
>>> list(combs(fruits, 2))
[('apples', 'bananas'), ('apples', 'jackfruit'), ('apples', 'guava'), ('bananas', 'jackfruit'), ('bananas', 'guava'), ('jackfruit', 'guava')]

Pretty awesome, right? This table in the Python docs^↗ summarises the different combinatorics functions perfectly:

Make sure to check the docs for itertools! There’s a ton of useful and intriguing functions in there waiting to be used ;)

Challenge

I have a farm with 66 banana trees. The trees come in 4 exotic varieties:

• 21 trees produce single bananas, worth £0.20
• 20 trees produce boomerang bananas, worth £0.26
• 13 trees produce golden bananas, worth £0.71
• 12 trees produce banana bunches, worth £0.91

(You can confirm with my good friend Quincy if you’re skeptical.)

Every harvest, I collect the produce of 33 randomly selected trees. Each tree has the same chance of being selected. What’s the probability that on a particular harvest I collect more than £20 worth of bananas?

A decimal value is expected, but props if you manage to find an exact value…

[!Tip] You can use itertools @.@

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