2016年7月11日月曜日

開発環境

Think Python (Allen B. Downey (著)、 O'Reilly Media)のChapter 10.(Lists)のExercises 10-1, 2, 3, 4, 5, 6, 7.(No. 2339)を取り組んでみる。

Exercises 10-1、 2、 3、 4、 5、 6、 7.(No. 2339)

コード(Emacs)

#!/usr/bin/env python3
# -*- coding: utf-8 -*-


def nested_sum(nested_list):
    result = 0
    for nums in nested_list:
        result += sum(nums)
    return result


def cumsum(nums):
    result = []
    for i, num in enumerate(nums):
        if i == 0:
            result.append(num)
        else:
            result.append(result[i - 1] + num)
    return result


def middle(items):
    t = items[1:-1]
    return t


def chop(items):
    items.pop(0)
    items.pop()


def is_sorted(items):
    return sorted(items) == items


def is_anagram(s1, s2):
    return sorted(s1) == sorted(s2)


def has_duplicates(l):
    return len(l) != len(set(l))

if __name__ == '__main__':
    print('10-1.')
    t = [[1, 2], [3], [4, 5, 6]]
    print(nested_sum(t))

    print('10-2.')
    t = [1, 2, 3]
    print(cumsum(t))

    print('10-3.')
    t = [1, 2, 3, 4]
    print(middle(t))
    print(t)

    print('10-4.')
    t = [1, 2, 3, 4]
    print(chop(t))
    print(t)

    print('10-5.')
    print(is_sorted([1, 2, 2]))
    print(is_sorted(['b', 'a']))

    print('10-6.')
    print(is_anagram('cinema', 'iceman'))
    print(is_anagram('cinemaa', 'icemana'))
    print(is_anagram('cinemaa', 'iceman'))

    print('10-7.')
    print(has_duplicates([1, 2, 1, 3, 4, 5, ]))
    print(has_duplicates([1, 2, 3, 4, 5]))

入出力結果(Terminal, IPython)

$ ./sample1.py 
10-1.
21
10-2.
[1, 3, 6]
10-3.
[2, 3]
[1, 2, 3, 4]
10-4.
None
[2, 3]
10-5.
True
False
10-6.
True
True
False
10-7.
True
False
$

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