CODEWITHSUNDEEP
pythonregexpython-3.x
user15051990
user15051990

Reputation: 1895

Remove part of the string using Regex in python

I have text file which contains the information in below format.

2018/03/21-17:08:48.638553  508     7FF4A8F3D704     snononsonfvnosnovoosr
2018/03/21-17:08:48.985053 346K     7FE9D2D51706     ahelooa afoaona woom
2018/03/21-17:08:50.486601 1.5M     7FE9D3D41706     qojfcmqcacaeia
2018/03/21-17:08:50.980519  16K     7FE9BD1AF707     user: number is 93823004
2018/03/21-17:08:50.981908 1389     7FE9BDC2B707     user 7fb31ecfa700
2018/03/21-17:08:51.066967    0     7FE9BDC91700     Exit Status = 0x0
2018/03/21-17:08:51.066968    1     7FE9BDC91700     std:ZMD:

Expected Result

I want to remove part of the string till 3rd space (that is 7FF4A8F3D704). Result should look like

snononsonfvnosnovoosr
ahelooa afoaona woom
qojfcmqcacaeia
user: number is 93823004
user 7fb31ecfa700
Exit Status = 0x0
std:ZMD:

Solution

I can remove "2018/03/21-17:08:48.638553" with the below code. But I am trying to replace the whole part with ''.

import re
Regex_list = [r'\d{4}/\d{2}/\d{2}-\d{2}:\d{2}:\d{2}.\d{6}']
for p in Regex_list:
    text = re.sub(p, ' ', file)

Upvotes: 0

Views: 1276

Answers (3)

Chris Charley
Chris Charley

Reputation: 6573

Another approach that uses re.split() (and limits the split to 3 splits). This assumes that there are no spaces in the first three fields.

It splits on 1 or more spaces.

for data in L.splitlines():
    print(re.split(r'\s+', data, 3)[-1])

Output:

snononsonfvnosnovoosr
ahelooa afoaona woom
qojfcmqcacaeia
user: number is 93823004
user 7fb31ecfa700
Exit Status = 0x0
std:ZMD:

Upvotes: 0

SpghttCd
SpghttCd

Reputation: 10860

If this is the exact structure of your text file, why don't you simply cut off the first n uninteresting characters?

for line in txt.splitlines():
    print(line[53:])


#snononsonfvnosnovoosr
#ahelooa afoaona woom                                      
#qojfcmqcacaeia                                             
#user: number is 93823004                                    
#user 7fb31ecfa700                                      
#Exit Status = 0x0                                           
#std:ZMD:

Upvotes: 1

CertainPerformance
CertainPerformance

Reputation: 370659

Because it looks like the first 3 column values won't ever have any spaces in them, match \S+\s+ to get a column value and its associated space-padding to the right, and repeat it 3 times:

output = re.sub(r'(?m)^(?:\S+\s+){3}', '', input)

https://regex101.com/r/YHXTJs/1

Upvotes: 0

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