Substitute commas after a certain amount of pipes
Question:
I have the following string
s = 'AAAnA|A33, 3|BB,C|CC,C|CC555|AVENUE ,STREET ,POTATO ,JOSPH'
s = 'AAAnA|A33, 3|BB,C|CC,C|STREET ,POTATO ,JOSPH'
What I want to do is take the values after the "last pipeline". And substitute all the commas for ‘|’.
Important infos, there is a chance of having empty spaces and commas before handed, yes the pipeline varies the amount. (Just noticed now)
My earlier attempt:
print(re.sub(r'[|]{5}',"|",s))
Answers:
You can try this code without regex
s = 'AAAA|A333|BBC|CCC|CC555|AVENUE ,STREET ,POTATO ,JOSPH'
s.split('|')[5].replace(',', '|')
Split the string at the | characters. Do the comma replacements in the 6th element of that list, then join them back together.
fields = s.split('|')
fields[5] = fields[5].replace('|', ',')
s = '|'.join(fields)
You may use this re.sub with a lambda:
import re
s = 'AAAA,LTD|A333|BBC|CCC|CC555|AVENUE ,STREET ,POTATO ,JOSPH'
print (re.sub(r'^((?:[^|]*|){5})(.*)', lambda m: m[1] + m[2].replace(',', '|'), s))
Output:
AAAA,LTD|A333|BBC|CCC|CC555|AVENUE |STREET |POTATO |JOSPH
RegEx Breakup:
^: Start(: Start capture group #1
(?:: Start non-capture group
[^|]*: Match 0 or more of any char that is not ||: Match a |
){5}: End non-capture group. Repeat this group 5 times
): End capture group #1(.*): Match and capture remaining text in capture group #2- In lambda code we replace
, with | in 2nd capture group only
One alternative -I assume you want the first part to remain as it is. This will work for any number of commas or white spaces.
Example strings –
s1= r'AAAnA|A33, 3|BB,C|CC,C|CC555|AVENUE ,STREET , POTATO ,JOSPH'
s2=r'AAAnA|A33, 3|BB,C|CC,C|CC555|AVENUE ,STREET ,,,, POTATO ,JOSPH'
s3=r'AAAnA|A33, 3|BB,C|CC,C|CC555|AVENUE ,STREET , POTATO ,JOSPH'
Code :
m=re.sub('[ ]{0,}[,]{1,}[ ]{0,}',r'|',re.search(r'[^|]+$',s)[0])
o=re.search('(.*)[|]',s)[0]
print(o+m)
Output:
AAAnA|A33, 3|BB,C|CC,C|CC555|AVENUE|STREET|POTATO|JOSPH
I have the following string
s = 'AAAnA|A33, 3|BB,C|CC,C|CC555|AVENUE ,STREET ,POTATO ,JOSPH'
s = 'AAAnA|A33, 3|BB,C|CC,C|STREET ,POTATO ,JOSPH'
What I want to do is take the values after the "last pipeline". And substitute all the commas for ‘|’.
Important infos, there is a chance of having empty spaces and commas before handed, yes the pipeline varies the amount. (Just noticed now)
My earlier attempt:
print(re.sub(r'[|]{5}',"|",s))
You can try this code without regex
s = 'AAAA|A333|BBC|CCC|CC555|AVENUE ,STREET ,POTATO ,JOSPH'
s.split('|')[5].replace(',', '|')
Split the string at the | characters. Do the comma replacements in the 6th element of that list, then join them back together.
fields = s.split('|')
fields[5] = fields[5].replace('|', ',')
s = '|'.join(fields)
You may use this re.sub with a lambda:
import re
s = 'AAAA,LTD|A333|BBC|CCC|CC555|AVENUE ,STREET ,POTATO ,JOSPH'
print (re.sub(r'^((?:[^|]*|){5})(.*)', lambda m: m[1] + m[2].replace(',', '|'), s))
Output:
AAAA,LTD|A333|BBC|CCC|CC555|AVENUE |STREET |POTATO |JOSPH
RegEx Breakup:
^: Start(: Start capture group #1(?:: Start non-capture group[^|]*: Match 0 or more of any char that is not||: Match a|
){5}: End non-capture group. Repeat this group 5 times
): End capture group #1(.*): Match and capture remaining text in capture group #2- In lambda code we replace
,with|in 2nd capture group only
One alternative -I assume you want the first part to remain as it is. This will work for any number of commas or white spaces.
Example strings –
s1= r'AAAnA|A33, 3|BB,C|CC,C|CC555|AVENUE ,STREET , POTATO ,JOSPH'
s2=r'AAAnA|A33, 3|BB,C|CC,C|CC555|AVENUE ,STREET ,,,, POTATO ,JOSPH'
s3=r'AAAnA|A33, 3|BB,C|CC,C|CC555|AVENUE ,STREET , POTATO ,JOSPH'
Code :
m=re.sub('[ ]{0,}[,]{1,}[ ]{0,}',r'|',re.search(r'[^|]+$',s)[0])
o=re.search('(.*)[|]',s)[0]
print(o+m)
Output:
AAAnA|A33, 3|BB,C|CC,C|CC555|AVENUE|STREET|POTATO|JOSPH