#itertools

Using Itertools, we go through the combinations() method. However, before digging more into the subject, it is critical to understand how it is used. Let’s have a look at it first. When conducting various computations, we constantly encounter combinations or permutations. Even though humans can calculate numbers, dealing with high quantities may be difficult at times. Consider what may happen if we had technologies that could make this decision for people.

itertools Package

The Itertools package perfectly meets our expectations. However, its scope goes beyond that. It provides extra methods that help with the other preset operations. However, this package is divided into three types: infinite iterators, combinatorial iterators, and terminating iterators.

We’d just talk about the combinations() method since this module is too vital to go through in detail right now. While we’re here, let’s take a look at how to install it, integrate it, and see what combinations are available.

Advanced python part 2

In part 1, we introduced advanced string, bytes manipulation in Python. This article covers some advanced python knowledge with built-in functions and other useful tools for sequence iteration, data transformation.

Useful built-in functions

Use any() to return true if any of the sequence values are true

Use all() to return true only if all values are true

Quickly find the minimum/maximum…

How to: All permutations of a Windows license key

All permutations of a Windows license key

I need to apply for a Windows 8 upgrade for my laptop, for which I need the Windows 7 license key on the underside of the laptop.

Because Microsoft decided in their infinite wisdom to create license labels that wear off, and I cannot read my license key clearly, it means I can’t register my laptop for the windows upgrade offer using an automated process.

以前こんなエントリを書いたけど、こういったコードは自分で書かずに itertools を使うのが Python の定石っぽい。

1. 繰り返しを許さない組み合わせ

itertools.combinations() を使う。

#!/usr/bin/env python # -*- coding: utf-8 -*- import itertools if __name__ == '__main__': l = [1, 2, 3, 4, 5] # 組み合わせ (1) # 繰り返しを許さない: 1,1 はない # 順序が違っても同じと見なす: 1,2 と 2,1 は同じ for element in itertools.combinations(l, 2): print(element) # 以下のコードと意味的に等価 ''' for i, iv in enumerate(l): for j, jv in enumerate(l[i + 1:], start=(i + 1)): print(iv, jv) '''

実行結果は以下の通り。

(1, 2) (1, 3) (1, 4) (1, 5) (2, 3) (2, 4) (2, 5) (3, 4) (3, 5) (4, 5)

2. 繰り返しを許す組み合わせ

itertools.combinations_with_replacement() を使う。

#!/usr/bin/env python # -*- coding: utf-8 -*- import itertools if __name__ == '__main__': l = [1, 2, 3, 4, 5] # 組み合わせ (2) # 繰り返しを許す: 1,1 がある # 順序が違っても同じと見なす: 1,2 と 2,1 は同じ for element in itertools.combinations_with_replacement(l, 2): print(element) # 以下のコードと意味的に等価 ''' for i, iv in enumerate(l): for j, jv in enumerate(l[i:], start=(i + 1)): print(iv, jv) '''

実行結果は以下の通り。

(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (2, 2) (2, 3) (2, 4) (2, 5) (3, 3) (3, 4) (3, 5) (4, 4) (4, 5) (5, 5)

3. 順列

itertools.permutations() を使う。

#!/usr/bin/env python # -*- coding: utf-8 -*- import itertools if __name__ == '__main__': l = [1, 2, 3, 4, 5] # 順列 # 繰り返しを許さない: 1,1 はない # 順序が違えば別と見なす: 1,2 と 2,1 は別 for element in itertools.permutations(l, 2): print(element) # 以下のコードと意味的に等価 ''' for i in l: for j in l: if i != j: print(i, j) '''

実行結果は以下の通り。

(1, 2) (1, 3) (1, 4) (1, 5) (2, 1) (2, 3) (2, 4) (2, 5) (3, 1) (3, 2) (3, 4) (3, 5) (4, 1) (4, 2) (4, 3) (4, 5) (5, 1) (5, 2) (5, 3) (5, 4)

4. デカルト積

単純にループをネストしてシーケンスを走査する場合にも itertools.product() を使った方が見通しが良くなる。

#!/usr/bin/env python # -*- coding: utf-8 -*- import itertools if __name__ == '__main__': l = [1, 2, 3, 4, 5] # デカルト積 # 繰り返しを許す: 1,1 がある # 順序が違えば別と見なす: 1,2 と 2,1 は別 for element in itertools.product(l, repeat=2): print(element) # 以下のコードと意味的に等価 ''' for i in l: for j in l: print(i, j) '''

実行結果は以下の通り。

(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (5, 1) (5, 2) (5, 3) (5, 4) (5, 5)

So I have a data set like this,

data = [('a',1),('a',2),('a',3),('b',1),('b',2),('a',4)]

Let's say I want to group them by consecutive occurences of the first element of each tuple. So the output I'm expecting is something like this,

a 1,2,3 b 1,2 a 4

Take a deep breath, and don't be surprised, I'm going to do this in just two lines.

>>> for k, v in itertools.groupby(data, key = lambda x : x[0]): >>> print k, [_[1] for _ in list(v)] a [1, 2, 3] b [1, 2] a [4]