Something like
sentence.replace(*, "newword")
(which doesn't work, btw)
Let's say
sentence = "hello world"
return sentence.replace(*, "newworld")
should return "newword newword"
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42piratas
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What would `sentence.replace(*, "newword")` return? – vaultah Apr 28 '16 at 16:50
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let's say `sentence = "hello world"`, then it should return `sentence = "newword newword"` – 42piratas Apr 28 '16 at 16:51
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1Try `sentence = ' '.join(['newword'] * len(sentence.split()))` – vaultah Apr 28 '16 at 16:53
3 Answers
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Since you aren't going to be replacing a specific word, str.replace() won't really support any kind of pattern-matching.
However, you could use the re.sub() function that will allow you to pass in a Regular Expression that would match everything and replace it :
import re
# Replace each series of non-space characters [^\s]+ with "newword"
sentence = re.sub('[^\s]+','newword',sentence)
Example
You can find a complete interactive example of this here and demonstrated below :
Rion Williams
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What you're looking for is a word replace. So instead of string.replace which replaces characters, you want something that will replace all words.
>>> sentence = "hello world this is my sentence"
>>> " ".join(["newword"] * len(sentence.split()))
'newword newword newword newword newword newword'
In the above case we're spiting the sentence into a list of it's words and simple making another list of words "newword" of the same length. Finally we're joining the new words together with the " " character in between them
Slick
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If you care about speed, just crafting the string manually seems to be twice as fast:
In [8]: import re
In [9]: sentence = "hello world this is my sentence"
In [10]: nonspace = re.compile('[^\s]+')
In [11]: %timeit re.sub(nonspace, 'newword', sentence)
100000 loops, best of 3: 6.28 µs per loop
In [12]: %timeit ' '.join('newword' for _ in xrange(len(sentence.split())))
100000 loops, best of 3: 2.52 µs per loop
In [13]: sentence *= 40 # Make the sentence longer
In [14]: %timeit re.sub(nonspace, 'newword', sentence)
10000 loops, best of 3: 70.6 µs per loop
In [15]: %timeit ' '.join('newword' for _ in xrange(len(sentence.split())))
10000 loops, best of 3: 30.2 µs per loop
And join is actually faster when you hand it a list, so ' '.join(['newword' for _ in xrange(len(sentence.split()))]) should result in some performance improvements (it caches the result in my informal %timeit tests, so I didn't include it)
