CODEWITHSUNDEEP
pythonregexpython-re
KAIPA UDAY KUMAR REDDY
KAIPA UDAY KUMAR REDDY

Reputation: 1

Python re module, length of occurrence of a word

I have regex like

m=re.search('z+',"i want to zzz zzzzzzz this is end")

how to find the length of consicutive z ignoring spaces 10(3z+7z)

"example str want to zz z this is end" ->3(2+1)

I am new here, please comment if anything is missing

Upvotes: 0

Views: 78

Answers (3)

Wiktor Stribiżew
Wiktor Stribiżew

Reputation: 627334

You can use

import re
matches = re.findall("z+(?:\s+z+)*", "i want to zzz zzzzzz this is end")
z_count = [(s, s.count('z')) for s in matches]
wht_count = [(s, len(s) - s.count('z')) for s in matches]
print( f'Z count: {z_count}\nWhitespace count: {wht_count}' )
# => 
##    Z count: [('zzz zzzzzz', 9)]
##    Whitespace count: [('zzz zzzzzz', 1)]

See the online Python demo.

Please note that z+(?:\s+z+)* regex won't capture any leading/trailing whitespace, so the number of spaces will be correct.

The regex matches

  • z+ - one or more z letters
  • (?:\s+z+)* - zero or more repetitions of one or more whitespace and one or more letters z.

See the regex demo.

Upvotes: 0

Tim Roberts
Tim Roberts

Reputation: 54767

You will need to use a slightly different regular expression, like z[z ]* to return a string of z's and spaces, and then use s.count() to count how many z's there were:

matches = re.findall("z[z ]*", "i want to zzz zzzzzz this is end")
result = [s.count('z') for s in matches]
print( result )

Output:

[9]

Upvotes: 1

wasif
wasif

Reputation: 15498

You need to use findall method instead, it will return all matching occurrences. Then you need to map the Len function to return length of the matched substrings of z, and later you can get both it and it's sum:

m=re.findall('z+',"i want to zzz zzzzzzz this is end")
m=list(map(len,m))
print(sum(m),m)

Upvotes: 0

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