Search code examples
pythonreplacesequential

replace pattern with a sequential number string in python


I'm trying to achieve the following replacement in python. Replace all html tags with {n} & create a hash of [tag, {n}]
Original string -> "<h> This is a string. </H><P> This is another part. </P>"
Replaced text -> "{0} This is a string. {1}{2} This is another part. {3}"

Here's my code. I've started with replacement but I'm stuck at the replacement logic as I cannot figure out the best way to replace each occurrence in a consecutive manner i.e with {0}, {1} and so on:

import re
text = "<h> This is a string. </H><p> This is another part. </P>"

num_mat = re.findall(r"(?:<(\/*)[a-zA-Z0-9]+>)",text)
print(str(len(num_mat)))

reg = re.compile(r"(?:<(\/*)[a-zA-Z0-9]+>)",re.VERBOSE)

phctr = 0
#for phctr in num_mat:
#    phtxt = "{" + str(phctr) + "}"
phtxt = "{" + str(phctr) + "}"
newtext = re.sub(reg,phtxt,text)

print(newtext)

Can someone help with a better way of achieving this? Thank you!


Solution

  • import re
    import itertools as it
    
    text = "<h> This is a string. </H><p> This is another part. </P>"
    
    cnt = it.count()
    print re.sub(r"</?\w+>", lambda x: '{{{}}}'.format(next(cnt)), text)
    

    prints

    {0} This is a string. {1}{2} This is another part. {3}
    

    Works for simple tags only (no attributes/spaces in tags). For extended tags, you have to adapt the regexp.

    Also, not reinitializing cnt = it.count() will keep the numbering going on.

    UPDATE to get a mapping dict:

    import re
    import itertools as it
    
    text = "<h> This is a string. </H><p> This is another part. </P>"
    
    cnt = it.count()
    d = {}
    def replace(tag, d, cnt):
        if tag not in d:
            d[tag] = '{{{}}}'.format(next(cnt))
        return d[tag]
    print re.sub(r"(</?\w+>)", lambda x: replace(x.group(1), d, cnt), text)
    print d
    

    prints:

    {0} This is a string. {1}{2} This is another part. {3}
    {'</P>': '{3}', '<h>': '{0}', '<p>': '{2}', '</H>': '{1}'}