Easy way to replace placeholders in string with the list of values?

Question

I have a long string that contains placeholders, that should be replaced with some data.

strOut = "text text {{ }} text text {{ }}"

with this pattern

pattern = r'\{{(.+?)\}}'

It's easy for me to do something like this

pattern.sub(self.mymethod, strOut)

where mymethod will be called for substitution. It actually works great. However, this is the problem now. I need to replace all placeholders in string with the values from a list. For example, this is the string again:

strOut = "text text {{ }} text {{ }} text"

It will always have undetermined number of placeholders. If I also have a list of let's say 2 values

myList = [2, 3]

I need a way to inject these values into placeholders, and end up with this

"text text 2 text 3 text"

Number of values in a list and number of placeholder will always be the same, I just don't know in advance how many there will be.

score 6 · Answer 1 · edited May 23 '17 at 11:53

6

my_csv = '%(first)s,%(last)s'
print my_csv % dict(last='Doe', first='John')  # John,Doe

I like this for the following reasons :

the ordering of values is no longer needed
multiple placeholders can be replaced in one shot.

inspired from https://stackoverflow.com/a/11023271/2147228

edited May 23 '17 at 11:53

Community

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answered Feb 03 '16 at 18:04

Kalyanaraman Santhanam

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Jon Clements · Answer 2 · 2012-12-14T01:01:02.500

3

Sample code:

>>> strOut = "text text {{ }} text text {{ }}"
>>> pattern = r'\{{(.+?)\}}'
>>> import re
>>> re.findall(pattern, strOut)
[' ', ' ']
>>> items = iter(str(el) for el in [1, 2])
# as pointed out by @eyquem, this can also be written as:
# items = imap(str, [1, 2])

>>> re.sub(pattern, lambda L: next(items), strOut)
'text text 1 text text 2'

This will break if there's not enough parameters to meet the placeholders, but just ignore the remainder otherwise.

And if you're using 2.7+, you're laughing if you can get away with {} as a place holder, as you can then just use 'text text {} text text {}'.format(*[1,2]) for instance

And for "fun"

(ab)using the lambda, you can do:

re.sub(pattern, lambda i, j=(str(el) for el in [1, 2]): next(j), strOut)

edited Dec 14 '12 at 01:01

answered Dec 14 '12 at 00:41

Jon Clements

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@Zed Out of curiousity - any reason you can't use the builtin `str.format` and use `{}` as a placeholder? – Jon Clements Dec 14 '12 at 00:51
@Jon Clements Isn't ``iter(str(el) for el in [1, 2])`` the fommowing: ``imap(str,[1, 2])`` ? – eyquem Dec 14 '12 at 00:53
@eyquem d0h! Yes it is, I just had iter([1, 2]) and then realised it wouldn't work, so just changed it to a generator – Jon Clements Dec 14 '12 at 00:56

score 1 · Answer 3 · answered Dec 14 '12 at 00:54

If there are no appearances of '{}' (as opposed to '{{ }}') in your strOut text, the following approach works. For strOut and myList as shown, the string it produces is 'text text 2 text 3 text'.

strOut = "text text {{ }} text {{ }} text"
myList = [2, 3]
strOut.replace('{{ }}','{}').format(*myList)

score 1 · Answer 4 · answered Dec 24 '17 at 11:07

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You can use format function and pass list with argument unpacking (*)

>>> myList = [2, 3]
>>> strOut = 'text text {} text {} text'
>>> print(strOut.format(*myList))
text text 2 text 3 text

answered Dec 24 '17 at 11:07

Vlad Bezden

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score 0 · Answer 5 · answered Dec 14 '12 at 00:43

You could try something like this (it uses the count parameter of re.sub):

In [1]: import re

In [2]: strOut = "text text {{ }} text text {{ }}"

In [3]: pattern = r'\{{(.+?)\}}'

In [4]: reduce(lambda x, y: re.sub(pattern, str(y), x, 1), [2, 3], strOut)
Out[4]: 'text text 2 text text 3'

Easy way to replace placeholders in string with the list of values?

5 Answers5