I got this list:
words = ['how', 'much', 'is[br]', 'the', 'fish[br]', 'no', 'really']
What I would like is to replace [br] with some fantastic value similar to <br /> and thus getting a new list:
words = ['how', 'much', 'is<br />', 'the', 'fish<br />', 'no', 'really']
words = [w.replace('[br]', '<br />') for w in words]
This is called a list comprehension.
You can use, for example:
words = [word.replace('[br]','<br />') for word in words]
Besides list comprehension, you can try map()
>>> map(lambda x: str.replace(x, "[br]", "<br/>"), words)
['how', 'much', 'is<br/>', 'the', 'fish<br/>', 'no', 'really']
Or another way is enumerate() function
words = ['how', 'much', 'is[br]', 'the', 'fish[br]', 'no', 'really']
# Replacing items using enumerate()
for index, value in enumerate(words):
# if value includes [br] then replace it with <br>
if '[br]' in value:
words[index] = value.replace('[br]', '<br>')
print(words)
# Output: ['how', 'much', 'is<br>', 'the', 'fish<br>', 'no', 'really']
enumerate is faster than map.
In case you're wondering about the performance of the different approaches, here are some timings:
In [1]: words = [str(i) for i in range(10000)]
In [2]: %timeit replaced = [w.replace('1', '<1>') for w in words]
100 loops, best of 3: 2.98 ms per loop
In [3]: %timeit replaced = map(lambda x: str.replace(x, '1', '<1>'), words)
100 loops, best of 3: 5.09 ms per loop
In [4]: %timeit replaced = map(lambda x: x.replace('1', '<1>'), words)
100 loops, best of 3: 4.39 ms per loop
In [5]: import re
In [6]: r = re.compile('1')
In [7]: %timeit replaced = [r.sub('<1>', w) for w in words]
100 loops, best of 3: 6.15 ms per loop
as you can see for such simple patterns the accepted list comprehension is the fastest, but look at the following:
In [8]: %timeit replaced = [w.replace('1', '<1>').replace('324', '<324>').replace('567', '<567>') for w in words]
100 loops, best of 3: 8.25 ms per loop
In [9]: r = re.compile('(1|324|567)')
In [10]: %timeit replaced = [r.sub('<\1>', w) for w in words]
100 loops, best of 3: 7.87 ms per loop
This shows that for more complicated substitutions a pre-compiled reg-exp (as in 9-10) can be (much) faster. It really depends on your problem and the shortest part of the reg-exp.
An example with for loop (I prefer List Comprehensions).
a, b = '[br]', '<br />'
for i, v in enumerate(words):
if a in v:
words[i] = v.replace(a, b)
print(words)
# ['how', 'much', 'is<br/>', 'the', 'fish<br/>', 'no', 'really']
If performance is important, including an if-else clause improves performance (by about 5% for a list of 1mil strings, no negligible really).
replaced = [w.replace('[br]','<br />') if '[br]' in w else w for w in words]
map() implementation can be improved by calling replace via operator.methodcaller() (by about 20%) but still slower than list comprehension (as of Python 3.9).
from operator import methodcaller
list(map(methodcaller('replace', '[br]', '<br />'), words))
If it suffices to modify the strings in-place, a loop implementation may be the fastest.
for i, w in enumerate(words):
if '[br]' in w:
words[i] = w.replace('[br]', '<br />')
