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タプルの中の配列の連続した合計値を算出したい

退会済みユーザー

退会済みユーザー

総合スコア0

Python

Pythonは、コードの読みやすさが特徴的なプログラミング言語の1つです。 強い型付け、動的型付けに対応しており、後方互換性がないバージョン2系とバージョン3系が使用されています。 商用製品の開発にも無料で使用でき、OSだけでなく仮想環境にも対応。Unicodeによる文字列操作をサポートしているため、日本語処理も標準で可能です。

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投稿2017/07/06 02:27

タプルの中の配列の連続した合計値を算出したいです。

total_ans =[ ( sum( [ dict( zip( a, b ) )[v] for i, v in enumerate(x) if i > a.index( v ) ] ), list(x) ) for x in permutations( a ) ] print(total_ans)

と書いて実行したら

[(0, [2, 5, 7, 3, 6]), (8, [2, 5, 7, 6, 3]), (6, [2, 5, 3, 7, 6]), (6, [2, 5, 3, 6, 7]), (14, [2, 5, 6, 7, 3]), (6, [2, 5, 6, 3, 7]), (3, [2, 7, 5, 3, 6]), (11, [2, 7, 5, 6, 3]), (3, [2, 7, 3, 5, 6]), (3, [2, 7, 3, 6, 5]), (11, [2, 7, 6, 5, 3]), (3, [2, 7, 6, 3, 5]), (9, [2, 3, 5, 7, 6]), (9, [2, 3, 5, 6, 7]), (3, [2, 3, 7, 5, 6]), (3, [2, 3, 7, 6, 5]), (9, [2, 3, 6, 5, 7]), (9, [2, 3, 6, 7, 5]), (17, [2, 6, 5, 7, 3]), (9, [2, 6, 5, 3, 7]), (11, [2, 6, 7, 5, 3]), (3, [2, 6, 7, 3, 5]), (9, [2, 6, 3, 5, 7]), (9, [2, 6, 3, 7, 5]), (1, [5, 2, 7, 3, 6]), (9, [5, 2, 7, 6, 3]), (7, [5, 2, 3, 7, 6]), (7, [5, 2, 3, 6, 7]), (15, [5, 2, 6, 7, 3]), (7, [5, 2, 6, 3, 7]), (1, [5, 7, 2, 3, 6]), (9, [5, 7, 2, 6, 3]), (1, [5, 7, 3, 2, 6]), (1, [5, 7, 3, 6, 2]), (9, [5, 7, 6, 2, 3]), (1, [5, 7, 6, 3, 2]), (7, [5, 3, 2, 7, 6]), (7, [5, 3, 2, 6, 7]), (1, [5, 3, 7, 2, 6]), (1, [5, 3, 7, 6, 2]), (7, [5, 3, 6, 2, 7]), (7, [5, 3, 6, 7, 2]), (15, [5, 6, 2, 7, 3]), (7, [5, 6, 2, 3, 7]), (9, [5, 6, 7, 2, 3]), (1, [5, 6, 7, 3, 2]), (7, [5, 6, 3, 2, 7]), (7, [5, 6, 3, 7, 2]), (4, [7, 2, 5, 3, 6]), (12, [7, 2, 5, 6, 3]), (4, [7, 2, 3, 5, 6]), (4, [7, 2, 3, 6, 5]), (12, [7, 2, 6, 5, 3]), (4, [7, 2, 6, 3, 5]), (1, [7, 5, 2, 3, 6]), (9, [7, 5, 2, 6, 3]), (1, [7, 5, 3, 2, 6]), (1, [7, 5, 3, 6, 2]), (9, [7, 5, 6, 2, 3]), (1, [7, 5, 6, 3, 2]), (4, [7, 3, 2, 5, 6]), (4, [7, 3, 2, 6, 5]), (4, [7, 3, 5, 2, 6]), (4, [7, 3, 5, 6, 2]), (4, [7, 3, 6, 2, 5]), (4, [7, 3, 6, 5, 2]), (12, [7, 6, 2, 5, 3]), (4, [7, 6, 2, 3, 5]), (12, [7, 6, 5, 2, 3]), (4, [7, 6, 5, 3, 2]), (4, [7, 6, 3, 2, 5]), (4, [7, 6, 3, 5, 2]), (10, [3, 2, 5, 7, 6]), (10, [3, 2, 5, 6, 7]), (4, [3, 2, 7, 5, 6]), (4, [3, 2, 7, 6, 5]), (10, [3, 2, 6, 5, 7]), (10, [3, 2, 6, 7, 5]), (7, [3, 5, 2, 7, 6]), (7, [3, 5, 2, 6, 7]), (1, [3, 5, 7, 2, 6]), (1, [3, 5, 7, 6, 2]), (7, [3, 5, 6, 2, 7]), (7, [3, 5, 6, 7, 2]), (4, [3, 7, 2, 5, 6]), (4, [3, 7, 2, 6, 5]), (4, [3, 7, 5, 2, 6]), (4, [3, 7, 5, 6, 2]), (4, [3, 7, 6, 2, 5]), (4, [3, 7, 6, 5, 2]), (10, [3, 6, 2, 5, 7]), (10, [3, 6, 2, 7, 5]), (10, [3, 6, 5, 2, 7]), (10, [3, 6, 5, 7, 2]), (4, [3, 6, 7, 2, 5]), (4, [3, 6, 7, 5, 2]), (18, [6, 2, 5, 7, 3]), (10, [6, 2, 5, 3, 7]), (12, [6, 2, 7, 5, 3]), (4, [6, 2, 7, 3, 5]), (10, [6, 2, 3, 5, 7]), (10, [6, 2, 3, 7, 5]), (15, [6, 5, 2, 7, 3]), (7, [6, 5, 2, 3, 7]), (9, [6, 5, 7, 2, 3]), (1, [6, 5, 7, 3, 2]), (7, [6, 5, 3, 2, 7]), (7, [6, 5, 3, 7, 2]), (12, [6, 7, 2, 5, 3]), (4, [6, 7, 2, 3, 5]), (12, [6, 7, 5, 2, 3]), (4, [6, 7, 5, 3, 2]), (4, [6, 7, 3, 2, 5]), (4, [6, 7, 3, 5, 2]), (10, [6, 3, 2, 5, 7]), (10, [6, 3, 2, 7, 5]), (10, [6, 3, 5, 2, 7]), (10, [6, 3, 5, 7, 2]), (4, [6, 3, 7, 2, 5]), (4, [6, 3, 7, 5, 2])]

と出力されました。
この

(0, [2, 5, 7, 3, 6]), (8, [2, 5, 7, 6, 3]), (6, [2, 5, 3, 7, 6]), (6, [2, 5, 3, 6, 7]), (14, [2, 5, 6, 7, 3])・・・

と続くタプル(配列?)の[2, 5, 7, 3, 6]、[2, 5, 7, 6, 3]、[2, 5, 3, 7, 6]、[2, 5, 3, 6, 7]、[2, 5, 6, 7, 3]の連続した合計値を算出したいです。
ただし、”合計値”と書いていますがただの合計値ではなく
例えばa=[2, 5, 7, 3, 6] とすると
a[0]+(a[0]+a[1])+(a[0]+a[1]+a[2])+(a[0]+a[1]+a[2]+a[3])+(a[0]+a[1]+a[2]+a[3]+a[4])=63 のような配列の要素の連続の合計値を算出したいです。

なので理想のアウトプットは

(0, [63]),・・・

のような形です。

total_ans =[ ( sum( [ dict( zip( a, b ) )[v] for i, v in enumerate(x) if i > a.index( v ) ] ), list(x) ) for x in permutations( a ) ] x2= sorted(total_ans.items(), key=lambda x: x[1]) print(x2)

までコードを実装したのですが
連続した合計値を算出する方法の実装がわかりません。
どのように書けますか?

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回答1

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ベストアンサー

いま欲しいのは [2, 5, 7, 3, 6]63 をする方法です。

In [1]: x = [2, 5, 7, 3, 6] In [2]: list(range(1, len(x) + 1)) Out[2]: [1, 2, 3, 4, 5] In [3]: [x[:i] for i in range(1, len(x) + 1)] Out[3]: [[2], [2, 5], [2, 5, 7], [2, 5, 7, 3], [2, 5, 7, 3, 6]] In [4]: [sum(x[:i]) for i in range(1, len(x) + 1)] Out[4]: [2, 7, 14, 17, 23] In [5]: sum(sum(x[:i]) for i in range(1, len(x) + 1)) Out[5]: 63

という具合に欲しいものに向かって変形していって、欲しい式を得ます。

def sum_of_sum_of_sub_sequence(x): return sum(sum(x[:i]) for i in range(1, len(x) + 1))

と関数化します。

total_ans = [(0, [2, 5, 7, 3, 6]), (8, [2, 5, 7, 6, 3]), (6, [2, 5, 3, 7, 6]), (6, [2, 5, 3, 6, 7]), (14, [2, 5, 6, 7, 3]), (6, [2, 5, 6, 3, 7]), (3, [2, 7, 5, 3, 6]), (11, [2, 7, 5, 6, 3]), (3, [2, 7, 3, 5, 6]), (3, [2, 7, 3, 6, 5]), (11, [2, 7, 6, 5, 3]), (3, [2, 7, 6, 3, 5]), (9, [2, 3, 5, 7, 6]), (9, [2, 3, 5, 6, 7]), (3, [2, 3, 7, 5, 6]), (3, [2, 3, 7, 6, 5]), (9, [2, 3, 6, 5, 7]), (9, [2, 3, 6, 7, 5]), (17, [2, 6, 5, 7, 3]), (9, [2, 6, 5, 3, 7]), (11, [2, 6, 7, 5, 3]), (3, [2, 6, 7, 3, 5]), (9, [2, 6, 3, 5, 7]), (9, [2, 6, 3, 7, 5]), (1, [5, 2, 7, 3, 6]), (9, [5, 2, 7, 6, 3]), (7, [5, 2, 3, 7, 6]), (7, [5, 2, 3, 6, 7]), (15, [5, 2, 6, 7, 3]), (7, [5, 2, 6, 3, 7]), (1, [5, 7, 2, 3, 6]), (9, [5, 7, 2, 6, 3]), (1, [5, 7, 3, 2, 6]), (1, [5, 7, 3, 6, 2]), (9, [5, 7, 6, 2, 3]), (1, [5, 7, 6, 3, 2]), (7, [5, 3, 2, 7, 6]), (7, [5, 3, 2, 6, 7]), (1, [5, 3, 7, 2, 6]), (1, [5, 3, 7, 6, 2]), (7, [5, 3, 6, 2, 7]), (7, [5, 3, 6, 7, 2]), (15, [5, 6, 2, 7, 3]), (7, [5, 6, 2, 3, 7]), (9, [5, 6, 7, 2, 3]), (1, [5, 6, 7, 3, 2]), (7, [5, 6, 3, 2, 7]), (7, [5, 6, 3, 7, 2]), (4, [7, 2, 5, 3, 6]), (12, [7, 2, 5, 6, 3]), (4, [7, 2, 3, 5, 6]), (4, [7, 2, 3, 6, 5]), (12, [7, 2, 6, 5, 3]), (4, [7, 2, 6, 3, 5]), (1, [7, 5, 2, 3, 6]), (9, [7, 5, 2, 6, 3]), (1, [7, 5, 3, 2, 6]), (1, [7, 5, 3, 6, 2]), (9, [7, 5, 6, 2, 3]), (1, [7, 5, 6, 3, 2]), (4, [7, 3, 2, 5, 6]), (4, [7, 3, 2, 6, 5]), (4, [7, 3, 5, 2, 6]), (4, [7, 3, 5, 6, 2]), (4, [7, 3, 6, 2, 5]), (4, [7, 3, 6, 5, 2]), (12, [7, 6, 2, 5, 3]), (4, [7, 6, 2, 3, 5]), (12, [7, 6, 5, 2, 3]), (4, [7, 6, 5, 3, 2]), (4, [7, 6, 3, 2, 5]), (4, [7, 6, 3, 5, 2]), (10, [3, 2, 5, 7, 6]), (10, [3, 2, 5, 6, 7]), (4, [3, 2, 7, 5, 6]), (4, [3, 2, 7, 6, 5]), (10, [3, 2, 6, 5, 7]), (10, [3, 2, 6, 7, 5]), (7, [3, 5, 2, 7, 6]), (7, [3, 5, 2, 6, 7]), (1, [3, 5, 7, 2, 6]), (1, [3, 5, 7, 6, 2]), (7, [3, 5, 6, 2, 7]), (7, [3, 5, 6, 7, 2]), (4, [3, 7, 2, 5, 6]), (4, [3, 7, 2, 6, 5]), (4, [3, 7, 5, 2, 6]), (4, [3, 7, 5, 6, 2]), (4, [3, 7, 6, 2, 5]), (4, [3, 7, 6, 5, 2]), (10, [3, 6, 2, 5, 7]), (10, [3, 6, 2, 7, 5]), (10, [3, 6, 5, 2, 7]), (10, [3, 6, 5, 7, 2]), (4, [3, 6, 7, 2, 5]), (4, [3, 6, 7, 5, 2]), (18, [6, 2, 5, 7, 3]), (10, [6, 2, 5, 3, 7]), (12, [6, 2, 7, 5, 3]), (4, [6, 2, 7, 3, 5]), (10, [6, 2, 3, 5, 7]), (10, [6, 2, 3, 7, 5]), (15, [6, 5, 2, 7, 3]), (7, [6, 5, 2, 3, 7]), (9, [6, 5, 7, 2, 3]), (1, [6, 5, 7, 3, 2]), (7, [6, 5, 3, 2, 7]), (7, [6, 5, 3, 7, 2]), (12, [6, 7, 2, 5, 3]), (4, [6, 7, 2, 3, 5]), (12, [6, 7, 5, 2, 3]), (4, [6, 7, 5, 3, 2]), (4, [6, 7, 3, 2, 5]), (4, [6, 7, 3, 5, 2]), (10, [6, 3, 2, 5, 7]), (10, [6, 3, 2, 7, 5]), (10, [6, 3, 5, 2, 7]), (10, [6, 3, 5, 7, 2]), (4, [6, 3, 7, 2, 5]), (4, [6, 3, 7, 5, 2])] [(i, [sum_of_sum_of_sub_sequence(x)]) for i, x in total_ans]

と適用します。

In [8]: [(i, [sum_of_sum_of_sub_sequence(x)]) for i, x in total_ans] Out[8]: [(0, [63]), (8, [66]), (6, [59]), (6, [58]), (14, [65]), (6, [61]), (3, [65]), (11, [68]), 【略】

を得ます。

投稿2017/07/06 02:49

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