python - 优雅地迭代字符串替换

Question

假设我有一个看起来像的字符串

'one, two, three,'

什么是通过用“。”替换“，”进行迭代的pythonic方式 一次一个？理想情况下，函数的返回值如下所示：

['one. two, three,' , 'one, two. three,' , 'one, two, three.']

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选择答案的推理，感谢您的贡献！

import timeit

def one(s):
    b = ["%s.%s" % (s[:i], s[i+1:]) for i, c in enumerate(s) if c == ","]

def two(s):
    b = [s[:i] + "." + s[i+1:] for i in range(len(s)) if s[i] == ","]

def thr(s):
    b = [s[:i] + "." + s[i+1:] for i, c in enumerate(s) if s[i] == ","]

def fou(s):
    ss = s.split(',')
    b = [','.join(ss[:i]) + '.' + ','.join(ss[i:]) for i in range(1,len(ss))]

a = 'one,two,three,four,five,six,seven,eight,nine,ten,eleven,twelve,'

print(timeit.timeit('one(a)', 'from __main__ import one, a', number = 1000000))
print(timeit.timeit('two(a)', 'from __main__ import two, a', number = 1000000))
print(timeit.timeit('thr(a)', 'from __main__ import thr, a', number = 1000000))
print(timeit.timeit('fou(a)', 'from __main__ import fou, a', number = 1000000))

#   C:\dev\goddangit>python timin.py
#   14.3008527857
#   11.8759967856
#   13.3739626708
#   18.8536401851

Answer 1

单线，s是'one, two, three,'：

>>> [s[:i] + "." + s[i+1:] for i in range(len(s)) if s[i] == ","]
['one. two, three,', 'one, two. three,', 'one, two, three.']

或者，将最外层替换[ ]为( )生成器对象。

当然，这仅适用于单字符替换。为了更一般地用其他字符串替换子字符串，您应该使用其他解决方案之一，例如，使用正则表达式。

Answer 2

您可以使用枚举：

 ["%s.%s" % (s[:i], s[i+1:]) for i, c in enumerate(s) if c == ","]

或正则表达式：

 ["%s.%s" % (s[:m.start()], s[m.start()+1:]) for m in re.finditer(',', s)]

Answer 3

我能找到一个简单的：

import re
s = 'one, two, three,'
# delim will be used as a regex
delim = ','
[s[:i.start()] + "." + s[i.end():] for i in re.finditer(delim, s)]

Answer 4

import itertools

l_string = "one, two, three,".rstrip(',').split(',')
separators = lambda pos: [ '.' if i==pos else ',' for i,x in enumerate(l_string) ]
print [ "".join([ "".join(elem) for elem in itertools.izip( l_string, separators(pos)  ) ]) for pos in range(len(l_string)) ]

>>> ['one. two, three,', 'one, two. three,', 'one, two, three.']