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Fix(guide): Reformat article

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theodoreyoong
2018-10-21 19:26:29 +01:00
committed by Heather Kusmierz
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commit 5e8d071a18
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guide/english/python/itertools/index.md
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@ -5,10 +5,11 @@ title: Itertools
Itertools is a python module of functions that return generators, objects that only function when iterated over.
Some examples of itertool functions include but not limited to: chain(), imap(), product(), and compress().
### Iterators terminating on the shortest input sequence:
### chain()
#### chain()
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.
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.
```py
import itertools
@ -18,10 +19,9 @@ list(itertools.chain([1, 2], [3, 4]))
# [1, 2, 3, 4]
```
#### islice()
### islice()
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.
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.
```py
import itertools
@ -31,9 +31,9 @@ list(itertools.islice(count(), 5))
# [0,1, 2, 3, 4]
```
### izip()
#### izip()
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.
`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.
```py
import itertools
@ -43,19 +43,31 @@ list(izip([1, 2, 3], ['a', 'b', 'c']))
# [(1, 'a'),(2, 'b'),(3, 'c')]
```
Combinatoric iterators:
### Combinatoric iterators
Iterator Arguments Results
product() p, q, … [repeat=1] cartesian product, equivalent to a nested for-loop
permutations() p[, r] r-length tuples, all possible orderings, no repeated elements
combinations() p, r r-length tuples, in sorted order, no repeated elements
combinations_with_replacement() p, r r-length tuples, in sorted order, with repeated elements
product('ABCD', repeat=2) AA AB AC AD BA BB BC BD CA CB CC CD DA DB DC DD
permutations('ABCD', 2) AB AC AD BA BC BD CA CB CD DA DB DC
#### product()
combinations('ABCD', 2) AB AC AD BC BD CD
`product()` returns a Cartesian product, equivalent to a nested for-loop. In comparison, the usual `zip()` function, which returns the convolution.
combinations_with_replacement('ABCD', 2) AA AB AC AD BB BC BD CC CD DD
```py
from itertools import product
list(product([1,2,3],[3,4]))
# Output
# [(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)]
Source:https://docs.python.org/3/library/itertools.html
A = [[1,2,3],[3,4,5]]
list(product(*A))
# Output
# [(1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5), (3, 3), (3, 4), (3, 5)]
B = [[1,2,3],[3,4,5],[7,8]]
list(product(*B))
# Output
# [(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)]
```
Source: https://docs.python.org/3/library/itertools.html
https://www.hackerrank.com/challenges/itertools-product/problem