Python: replace an exact matching substring with variable

Question

I have a list of strings like 'cdbbdbda', 'fgfghjkbd', 'cdbbd' etc. I have also a variable fed from another list of strings. What I need is to replace a substring in the first list's strings, say b by z, only if it is preceeded by a substring from the variable list, all the other occurrences being intouched. What I have:

a = ['cdbbdbda', 'fgfghjkbd', 'cdbbd']
c = ['d', 'f', 'l']

What I do:

for i in a:
    for j in c:
        if j+'b' in i:
            i = re.sub('b', 'z', i)

What I need:

'cdzbdzda'
'fgfghjkbd'
'cdzbd'

What I get:

'cdzzdzda'
'fgfghjkbd'
'cdzzd'

all instances of 'b' are replaced. I'm new in it, any help is very welcome. Looking for answer at Stackoverflow I have found many solutions with regex based on word boundaries or with re either with str.replace based on count, but I can't use it as the lenght of the string and number of occurrences of 'b' can vary.

Nick Russo · Accepted Answer · 2018-05-02 18:48:10Z

1

I think if you include j in the find and replace, you'll get what you want.

>>> for i in a:
...     for j in c:
...         i = re.sub(j+'b', j+'z', i)
...     print i
... 
cdzbdzda
fgfghjkbd
cdzbd
>>>

I added print i because your loop doesn't make in-place changes, so without that output, it's not possible to see what replacements were made.

answered May 2, 2018 at 18:48

Nick Russo

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Comments

Alper · Accepted Answer · 2018-05-02 19:01:51Z

1

You should simply use regular expressions with a positive lookbehind assertion.

Like this:

import re

for i in a:
  for j in c:
    i = re.sub('(?<=' + j + ')b', 'z', i)

The base case is:

re.sub('(?<=d)b', 'z', 'cdbbdbda')

answered May 2, 2018 at 19:01

Alper

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Comments

Ajax1234 · Accepted Answer · 2018-05-02 19:15:58Z

0

You can use a list comprehension:

import re
a = ['cdbbdbda', 'fgfghjkbd', 'cdbbd']
c = ['d', 'f', 'l']
new_a = [re.sub('|'.join('(?<={})b'.format(i) for i in c), 'z', b) for b in a]

Output:

['cdzbdzda', 'fgfghjkbd', 'cdzbd']

answered May 2, 2018 at 19:15

Ajax1234

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