CODEWITHSUNDEEP
pythonregexpython-3.x
Bob
Bob

Reputation: 289

get rid of a specific part of an element in a set

I have the set 

set1={'*klj?', 'bl:VOLTe?', 'abkjld:Sure:STe?', 'JKLJS?', 'TRered[:AMide]?', 'DKJ[:dkja]?'}

I want to have the set look like 

set1={'*klj?', 'bl:VOLTe?', 'abkjld:Sure:STe?', 'JKLJS?', 'TRered?','DKJ?'}

where I want to get rid of the [:AMide] and [:dkja] inside the set.

I was trying to use regex

What I have so far is 

set2={}
    for element in set:
        x=re.sub("([\(\[]).*?([\)\]])", "", str(element))
        set2.add(x)

This gets rid of the [] and what is inside but doesn't properly recreate the set, ie set2.add(x) doesn't work

Upvotes: 0

Views: 75

Answers (4)

Samuel Nde
Samuel Nde

Reputation: 2743

Here is another option 

res = {re.sub('(:AMide)|(:dkja)', '', s) for s in set1}
{re.sub(']|\[', '', t) for t in res}

The output is:

>>>>  {'*klj?', 'DKJ?', 'JKLJS?', 'TRered?', 'abkjld:Sure:STe?', 'bl:VOLTe?'}

Upvotes: 0

Ajax1234
Ajax1234

Reputation: 71461

You can try this:

import re
set1={'*klj?', 'bl:VOLTe?', 'abkjld:Sure:STe?', 'JKLJS?', 'TRered[:AMide]?', 'DKJ[:dkja]?'}
new_set = {re.sub('\[\:[a-zA-Z]+\]', '', i) for i in set1}

Output:

{'*klj?', 'abkjld:Sure:STe?', 'DKJ?', 'JKLJS?', 'TRered?', 'bl:VOLTe?'}

Upvotes: 0

Kasravnd
Kasravnd

Reputation: 107337

You don't need such a complicated regex for this task. Just use two replaces with a set-comprehension: 

In [10]: {i.replace('[:AMide]', '').replace('[:dkja]', '') for i in set1}
Out[10]: {'*klj?', 'DKJ?', 'JKLJS?', 'TRered?', 'abkjld:Sure:STe?', 'bl:VOLTe?'}

After all, if you want to remove everything between brackets I think you could simply use a negated character class as following:

In [11]: import re

In [12]: {re.sub(r'\[[^]]+\]', r'', i) for i in set1}
Out[12]: {'*klj?', 'DKJ?', 'JKLJS?', 'TRered?', 'abkjld:Sure:STe?', 'bl:VOLTe?'}

Upvotes: 1

Mad Physicist
Mad Physicist

Reputation: 114440

Strings are immutable. You can not replace a string in-place. The proper way to modify your set is either to remove the offending elements and put in the correct versions, or to create an entirely new set. The latter approach is a one-liner:

set1 = set(re.sub("([\(\[]).*?([\)\]])", "\g<1>\g<2>", str(element)) for element in set1)

Upvotes: 1

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