Replace s substring from array of string - Python

Question

How to remove a substring from the following string 'AA.01.001 Hello' with regular expression using Python.

I tried the following but, not worked.

array= ['AB.01.001 Hello','BA.10.004','CD.10.015 Good bye']
 regex = re.compile(r'[A-Z]{2,3}\\.[0-9]{2}\\.[0-9]{3}')
filtered = filter(lambda i: not regex.search(i), array)

Edit:

Exprected output : [`Hello`,'Good bye']

I think the pattern works right? Just add \s* after it to account for the whitespaces and use a single backslash. regex101.com/r/Jkht46/1 See Python demo ideone.com/rZfGLP — The fourth bird
– The fourth bird, Commented Jul 7, 2019 at 13:31
can you add some more examples to it, will the string to replace is always in this format 'AA.01.001 — Code Maniac
– Code Maniac, Commented Jul 7, 2019 at 13:31
I think the issue is just the double backslashes. You are using a raw string. So you only need one backslash. — Neil
– Neil, Commented Jul 7, 2019 at 13:36

Wiktor Stribiżew · Accepted Answer · 2019-07-07 13:56:10Z

1

You may use re.sub:

import re
array= ['AB.01.001 Hello','BA.10.004','CD.10.015 Good bye']
regex = re.compile(r'[A-Z]{2,3}\.[0-9]{2}\.[0-9]{3}\s*')
filtered = filter(None, map(lambda i: regex.sub('', i), array))
print(list(filtered))
# => ['Hello', 'Good bye']

See the Python demo.

Note I used only one \ to escape the literal dots (as you are using a raw string literal to define the regex) and added \s* to also remove any 0 or more whitespaces after your pattern.

Details

map(lambda i: regex.sub('', i), array) - iterates over the array items and removes the matches with re.sub
filter(None, ...) removes the empty items resulting from the substitution when the pattern matches the whole string.

answered Jul 7, 2019 at 13:56

Wiktor Stribiżew

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