简短（有用）的python代码段 [英] Short (and useful) python snippets

问题描述

本着现有什么是您最有用的C / C ++代码段的精神-线程：

你们有（经常）使用的简短的，单功能的Python代码段，并想与StackOverlow社区共享吗？请保持条目较小（也许在25个
行以下？），并且每个帖子仅给出一个示例。

Do you guys have short, monofunctional Python snippets that you use (often) and would like to share with the StackOverlow Community? Please keep the entries small (under 25 lines maybe?) and give only one example per post.

我将以一小段开始不时计算python项目中的sloc（代码的源代码行）：

I'll start of with a short snippet i use from time to time to count sloc (source lines of code) in python projects:

# prints recursive count of lines of python source code from current directory
# includes an ignore_list. also prints total sloc

import os
cur_path = os.getcwd()
ignore_set = set(["__init__.py", "count_sourcelines.py"])

loclist = []

for pydir, _, pyfiles in os.walk(cur_path):
    for pyfile in pyfiles:
        if pyfile.endswith(".py") and pyfile not in ignore_set:
            totalpath = os.path.join(pydir, pyfile)
            loclist.append( ( len(open(totalpath, "r").read().splitlines()),
                               totalpath.split(cur_path)[1]) )

for linenumbercount, filename in loclist: 
    print "%05d lines in %s" % (linenumbercount, filename)

print "\nTotal: %s lines (%s)" %(sum([x[0] for x in loclist]), cur_path)

推荐答案

初始化2D列表

虽然可以安全地初始化列表：

While this can be done safely to initialize a list:

lst = [0] * 3

相同的技巧不适用于2D列表（列表o f个列表）：

The same trick won’t work for a 2D list (list of lists):

>>> lst_2d = [[0] * 3] * 3
>>> lst_2d
[[0, 0, 0], [0, 0, 0], [0, 0, 0]]
>>> lst_2d[0][0] = 5
>>> lst_2d
[[5, 0, 0], [5, 0, 0], [5, 0, 0]]

运算符*复制其操作数，并且用[]构造的重复列表指向同一列表。正确的方法是：

The operator * duplicates its operands, and duplicated lists constructed with [] point to the same list. The correct way to do this is:

>>> lst_2d = [[0] * 3 for i in xrange(3)]
>>> lst_2d
[[0, 0, 0], [0, 0, 0], [0, 0, 0]]
>>> lst_2d[0][0] = 5
>>> lst_2d
[[5, 0, 0], [0, 0, 0], [0, 0, 0]]

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