Elegantly iterating over string replacement

Question

Say I have a string that looks like

'one, two, three,'

What is a pythonic way of iterating through replacing ',' with '.' one at a time? Ideally the return value of the function would look like:

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

Reasoning for selected answer, thanks for contributing!

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

@IgnacioVazquez-Abrams: not sure how that helps, this is a small function of a much larger project that currently takes up 15 lines that I feel could be reduced to just a couple (hence the request for a pythonic answer). I could post two working versions of this if you are interested. — gzzo, May 22 '13 at 07:20

tobias_k · Accepted Answer · 2013-05-22T09:32:42.517

5

One-liner, s being '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.']

Alternatively, replace the outermost [ ] by ( ) to have a generator object instead.

That is, of course, only for single-character replacements. For more generally replacing substrings with other strings you should use one of the other solutions, e.g., using regular expressions.

edited May 22 '13 at 09:32

answered May 22 '13 at 07:27

tobias_k

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iterator is general concept: any object whose class has a next method and an __iter__ method that does return self. http://stackoverflow.com/questions/2776829/difference-between-python-generators-vs-iterators – luke14free May 22 '13 at 07:31
You can easily change the square brackets in the list comp `[...]`, to `(...)` and it becomes a generator expression – jamylak May 22 '13 at 07:32
1

i really don't like the part where you enumerate and test all characters in the string. a regex or a find seems more appropriate. – njzk2 May 22 '13 at 07:45
@njzk2 Wouldn't regex or find essentially do the same thing, internally? – tobias_k May 22 '13 at 08:10
@tobias_k: the advantage of the regex answer is that it works with arbitrary search strings: for example, turn `A...B...C...etc` into `[A*B...C..., A...B*C... etc]`. – georg May 22 '13 at 08:57
yes, but that would be internally. The code would appear more functionnal and display less low-lewel plumbing. – njzk2 May 22 '13 at 09:00
@thg435 Of course, thanks for pointing this out! For single-character replacements, I still find this to be the most readable way, though. ;-) – tobias_k May 22 '13 at 09:35

Thomas Jung · Answer 2 · 2013-05-22T07:35:48.783

4

You can use enumerate:

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

or regular expressions:

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

edited May 22 '13 at 07:35

answered May 22 '13 at 07:30

Thomas Jung

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njzk2 · Answer 3 · 2013-05-22T08:01:53.283

4

A simple as I can find :

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)]

edited May 22 '13 at 08:01

answered May 22 '13 at 07:44

njzk2

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Good stuff, although I'd use `i.end()` in the second one to make it work for arbitrary strings. – georg May 22 '13 at 07:57

score 2 · Answer 4 · edited May 22 '13 at 07:34

2

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.']

edited May 22 '13 at 07:34

Inbar Rose

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answered May 22 '13 at 07:18

lucasg

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Elegantly iterating over string replacement

4 Answers4