87 lines
2.4 KiB
Markdown
87 lines
2.4 KiB
Markdown
---
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title: Itertools
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---
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Itertools is a python module of functions that return generators, objects that only function when iterated over.
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Some examples of itertool functions include but not limited to: chain(), imap(), product(), and compress().
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### Iterators terminating on the shortest input sequence:
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#### chain()
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The `chain()` function takes several iterators as arguments and returns a single iterator that produces the contents of all of them as though they came from one sequence.
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```py
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import itertools
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list(itertools.chain([1, 2], [3, 4]))
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# Output
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# [1, 2, 3, 4]
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```
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#### islice()
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The `islice()` function returns an iterator which returns selected items from the input iterator, by index. It takes the same arguments as the slice operator for lists: start, stop, and step. Start and stop are optional.
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```py
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import itertools
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list(itertools.islice(count(), 5))
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# Output
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# [0,1, 2, 3, 4]
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```
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#### izip()
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`izip()` returns an iterator that combines the elements of several iterators into tuples. It works like the built-in function `zip()`, except that it returns an iterator instead of a list.
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```py
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import itertools
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list(izip([1, 2, 3], ['a', 'b', 'c']))
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# Output
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# [(1, 'a'),(2, 'b'),(3, 'c')]
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```
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### compress()
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It make an iterator that filters elements from data returning only those that have a corresponding element in selectors that evaluates to True. Stops when either the data or selectors iterables has been exhausted.
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```py
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import itertools
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itertools.compress('ABCDEF', [1,0,1,0,1,1])
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# Output
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# A C E F
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```
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### Combinatoric iterators
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#### product()
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`product()` returns a Cartesian product, equivalent to a nested for-loop. In comparison, the usual `zip()` function, which returns the convolution.
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```py
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from itertools import product
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list(product([1,2,3],[3,4]))
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# Output
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# [(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)]
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A = [[1,2,3],[3,4,5]]
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list(product(*A))
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# Output
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# [(1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5), (3, 3), (3, 4), (3, 5)]
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B = [[1,2,3],[3,4,5],[7,8]]
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list(product(*B))
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# Output
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# [(1, 3, 7), (1, 3, 8), (1, 4, 7), (1, 4, 8), (1, 5, 7), (1, 5, 8), (2, 3, 7), (2, 3, 8), (2, 4, 7), (2, 4, 8), (2, 5, 7), (2, 5, 8), (3, 3, 7), (3, 3, 8), (3, 4, 7), (3, 4, 8), (3, 5, 7), (3, 5, 8)]
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```
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#### More Information
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- [Python Itertools Docs](https://docs.python.org/3/library/itertools.html)
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- [Hackerrank](https://www.hackerrank.com/challenges/itertools-product/problem)
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