Python - Searching and Sorting(list, merge)

Practical Programming
An Introduction to Computer Science
Using Python 3
(Pragmatic Programmers)
(Pragmatic Bookshelf)
Paul Gries (著) Jennifer Campbell (著)
Jason Montojo (著) Lynn Beighley (編集)

開発環境

• OS X Yosemite - Apple (OS)
• Emacs (CUI)、BBEdit - Bare Bones Software, Inc. (GUI) (Text Editor)
• Python 3.4 (プログラミング言語)

Practical Programming: An Introduction to Computer Science Using Python 3 (Pragmatic Programmers) (Paul Gries (著)、Jennifer Campbell (著)、Jason Montojo (著)、Lynn Beighley (編集)、Pragmatic Bookshelf)のChapter 13(Searching and Sorting)、13.7(Exercises) 9.を解いてみる。

13.7(Exercises) 9.

コード(BBEdit)

sample9.py

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

def merge1(l1, l2):
    '''(list, list) -> list
    >>> merge1([1, 3, 4, 6], [1, 2, 5, 7])
    [1, 1, 2, 3, 4, 5, 6, 7]
    '''
    new_l = []
    i1 = 0
    i2 = 0

    while 0 <= i1 < len(l1) and 0 <= i2 < len(l2):
        if l1[i1] <= l2[i2]:
            new_l.append(l1[i1])
            i1 += 1
        else:
            new_l.append(l2[i2])
            i2 += 1
    new_l.extend(l1[i1:])
    new_l.extend(l2[i2:])
    return new_l

def merge2(l1, l2):
    '''(list, list) -> list
    >>> merge2([1, 3, 4, 6], [1, 2, 5, 7])
    [1, 1, 2, 3, 4, 5, 6, 7]
    '''
    new_l = []
    i1 = 0
    i2 = 0

    while not (i1 == len(l1) or i2 == len(l2)):
        if l1[i1] <= l2[i2]:
            new_l.append(l1[i1])
            i1 += 1
        else:
            new_l.append(l2[i2])
            i2 += 1
    new_l.extend(l1[i1:])
    new_l.extend(l2[i2:])
    return new_l

if __name__ == '__main__':
    import doctest
    doctest.testmod()

入出力結果(Terminal, IPython)

$ ./sample9.py -v
Trying:
    merge1([1, 3, 4, 6], [1, 2, 5, 7])
Expecting:
    [1, 1, 2, 3, 4, 5, 6, 7]
ok
Trying:
    merge2([1, 3, 4, 6], [1, 2, 5, 7])
Expecting:
    [1, 1, 2, 3, 4, 5, 6, 7]
ok
1 items had no tests:
    __main__
2 items passed all tests:
   1 tests in __main__.merge1
   1 tests in __main__.merge2
2 tests in 3 items.
2 passed and 0 failed.
Test passed.
$