How to replace a character in some specific word in a text file using python

Question

I got a task to replace "O"(capital O) by "0" in a text file by using python. But one condition is that I have to preserve the other words like Over, NATO etc. I have to replace only the words like 9OO to 900, 2OO6 to 2006 and so on. I tried a lot but yet not successful. My code is given below. Please help me any one. Thanks in advance

import re

srcpatt = 'O'
rplpatt = '0'
cre = re.compile(srcpatt)

with open('myfile.txt', 'r') as file:
    content = file.read()

wordlist = re.findall(r'(\d+O|O\d+)',str(content))
print(wordlist)

for word in wordlist:
    subcontent = cre.sub(rplpatt, word)
    newrep = re.compile(word)
    newcontent = newrep.sub(subcontent,content)

with open('myfile.txt', 'w') as file:
    file.write(newcontent)

print('"',srcpatt,'" is successfully replaced by "',rplpatt,'"')

is wordlist coming out OK? If that much of it works we can focus on the rest. — Brian
– Brian, Commented Jun 13, 2013 at 21:21
'wordlist' just for seeing the output @Brian. Thanks for edit. — user2483882
– user2483882, Commented Jun 13, 2013 at 21:40
As you'll see - regex are easy to get wrong. "And so on" is no substitute for having a bunch more testcases. — John La Rooy
– John La Rooy, Commented Jun 13, 2013 at 21:45

U2EF1 · Accepted Answer · 2013-06-13 21:46:22Z

1

re.sub can take in a replacement function, so we can pare this down pretty nicely:

import re
with open('myfile.txt', 'r') as file:
    content = file.read()
with open('myfile.txt', 'w') as file:
    file.write(re.sub(r'\d+[\dO]+|[\dO]+\d+', lambda m: m.group().replace('O', '0'), content))

edited Jun 13, 2013 at 21:46

answered Jun 13, 2013 at 21:22

U2EF1

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3 Comments

Lanaru Over a year ago

This will not work for the string "9OO". You could change \d+O+\d+ to \d+O+\d*.

user2483882 Over a year ago

yea thanks a lot! But what about the string like O4:O5 to 04:05 ?

U2EF1 Over a year ago

Edited the regex, should be fine now.

perreal · Accepted Answer · 2013-06-13 21:24:17Z

0

import re

srcpatt = 'O'
rplpatt = '0'
cre = re.compile(srcpatt)
reg = r'\b(\d*)O(O*\d*)\b'

with open('input', 'r') as f:
    for line in f:
        while re.match(reg,line): line=re.sub(reg, r'\g<1>0\2', line)
        print line

print('"',srcpatt,'" is successfully replaced by "',rplpatt,'"')

answered Jun 13, 2013 at 21:24

perreal

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Comments

John La Rooy · Accepted Answer · 2013-06-13 21:41:08Z

0

You can probably get away with matching just a leading digit followed by O. This won't handle OO7, but it will work nicely with 8080 for example. Which none of the answers here matching the trailing digits will. If you want to do that you need to use a lookahead match.

re.sub(r'(\d)(O+)', lambda m: m.groups()[0] + '0'*len(m.groups()[1]), content)

edited Jun 13, 2013 at 21:41

answered Jun 13, 2013 at 21:21

John La Rooy

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