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I'm having this string in my app which stores different percentage, I want to just extract the percentage value and convert it to int( ). My String looks something like this..

note: there will be one or more blank space(s) before the number.

result ='bFilesystem Size Used Avail Use% Mounted on\n/dev/sda4 30G 20G 8.4G 71% /\n'
percentage = getPercentage(result)   #This method should output 71 in this example
print("Percentage is: ", percentage)

Thank you in advance.

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2 Answers 2

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4

You need a positive lookahead regex:

import re

result = 'demo percentage is   18%'
percentage = re.search(r'\d+(?=%)', result)        
print("Percentage is: ", percentage.group())

# Percentage is:  18
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Using regular expression:

import re

inp_str = "demo percentage is   18%"
x = re.findall(r'\d+(?=%)', inp_str)
# x is a list of all numbers appeared in the input string
x = [int(i) for i in x]
print(x) # prints - [18]
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5 Comments

This doesn't work for inp_str = "demo 11 percentage is 18%".
Yes, I know that my solution will not handle the above scenario but the question was not asked only to extract numbers that precede the % sign. Anyway, I have updated the answer.
result ='bFilesystem Size Used Avail Use% Mounted on\n/dev/sda4 30G 20G 8.4G 71% /\n' this is my string to be precise
@ManojJahgirdar please mention your requirements precisely from next time :)
@WasiAhmad yes sorry for the inconvenience. thank you.. :)

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