Python’s “Itertools”
Python has an itertools module, which provides a core set of fast, memory-efficient tools for creating iterators. We will briefly showcase a few itertools here. The majority of these functions create generators, thus we will have to iterate over them in order to explicitly demonstrate their use. It is hard to overstate the utility of this module - it is strongly recommended that you take some time to see what it has in store.
There are three built-in functions, range, enumerate, and zip, that belong in itertools, but they are so useful that they are made accessible immediately and do not need to be imported. It is essential that range, enumerate, and zip become tools that you are comfortable using.
range
Generate a sequence of integers in the specified “range”:
# will generate 0.. 1.. 2.. ... 8.. 9
>>> range(10)
range(0, 10)
>>> list(range(10))
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
# will generate 0.. 3.. 6.. 9
>>> range(0, 10, 3)
range(0, 10, 3)
>>> list(range(0, 10, 3))
[0, 3, 6, 9]
enumerate
Enumerates the items in an iterable: yielding a tuple containing the iteration count (starting with 0) and the corresponding item from the the iterable.
# will generate (0, 'apple').. (1, 'banana').. (2, 'cat').. (3, 'dog')]
>>> enumerate(["apple", "banana", "cat", "dog"])
<enumerate at 0x23e3557b3f0>
>>> list(enumerate(["apple", "banana", "cat", "dog"]))
[(0, 'apple'), (1, 'banana'), (2, 'cat'), (3, 'dog')]
zip
Zips together the corresponding elements of several iterables into tuples. This is valuable for “pairing” corresponding items across multiple iterables.
>>> names = ["Angie", "Brian", "Cassie", "David"]
>>> exam_1_scores = [90, 82, 79, 87]
>>> exam_2_scores = [95, 84, 72, 91]
# will generate ('Angie', 90, 95).. ('Brian', 82, 84).. ('Cassie', 79, 72).. ('David', 87, 91)]
>>> zip(names, exam_1_scores, exam_2_scores)
<zip at 0x20de1082608>
>>> list(zip(names, exam_1_scores, exam_2_scores))
[('Angie', 90, 95), ('Brian', 82, 84), ('Cassie', 79, 72), ('David', 87, 91)]
The following are some of the many useful tools provided by the itertools module:
itertools.chain
Chains together multiple iterables, end-to-end, forming a single iterable:
>>> from itertools import chain
>>> gen_1 = range(0, 5, 2) # 0.. 2.. 4
>>> gen_2 = (i**2 for i in range(3, 6)) # 9.. 16.. 25
>>> iter_3 = ["moo", "cow"]
>>> iter_4 = "him"
# will generate: 0.. 2.. 4.. 9.. 16.. 25.. 'moo'.. 'cow'.. 'h'.. 'i'.. 'm'
>>> chain(gen_1, gen_2, iter_3, iter_4)
<itertools.chain at 0x20de109ec18>
itertools.combinations Generate all length-n tuples storing “combinations” of items from an iterable:
>>> from itertools import combinations
# will generate: (0, 1, 2).. (0, 1, 3).. (0, 2, 3).. (1, 2, 3)
>>> combinations([0, 1, 2, 3], 3) # generate all length-3 combinations from [0, 1, 2, 3]
<itertools.combinations at 0x20de10a7728>
Reading Comprehension: Itertools I
Using the itertools.combinations function, find the probability that two randomly drawn items from the list ["apples", "bananas", "pears", "pears", "oranges"] would yield a combination of “apples” and “pears”.
Reading Comprehension: Itertools II
Given the list x_vals = [0.1, 0.3, 0.6, 0.9], create a generator, y_gen, that will generate the y-value \(y = x^2\) for each value of \(x\). Then, using zip, create a list of the \((x, y)\) pairs, each pair stored in a tuple.
Reading Comprehension: Solutions
Itertools I: Solution
>>> from itertools import combinations
>>> ls = ["apples", "bananas", "pears", "pears", "oranges"]
>>> comb_ls = list(combinations(ls, 2))
>>> comb_ls.count(("apples", "pears")) / len(comb_ls)
0.2
Itertools II: Solution
>>> x_vals = [0.1, 0.3, 0.6, 0.9]
>>> y_gen = (x**2 for x in x_vals)
>>> list(zip(x_vals, y_gen))
[(0.1, 0.01), (0.3, 0.09), (0.6, 0.36), (0.9, 0.81)]
In this instance, the use of zip is a bit contrived. We could have foregone creating y_gen by just using the following list-comprehension:
>>> x_vals = [0.1, 0.3, 0.6, 0.9]
>>> [(x, x**2) for x in x_vals]
[(0.1, 0.01), (0.3, 0.09), (0.6, 0.36), (0.9, 0.81)]