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How to extract colon separated values from the same line?

I am using python regular expressions. I want all colon separated values in a line.

e.g.

input = 'a:b c:d e:f'

expected_output = [('a','b'), ('c', 'd'), ('e', 'f')]


But when I do

>>> re.findall('(.*)s?:s?(.*)','a:b c:d')


I get

[('a:b c', 'd')]


I have also tried

>>> re.findall('(.*)s?:s?(.*)[s$]','a:b c:d')
[('a', 'b')]


4 Answers
4

The following code works for me:

inpt = 'a:b c:d e:f'
re.findall('(S+):(S+)',inpt)


Output:

[('a', 'b'), ('c', 'd'), ('e', 'f')]


(S+)s*:s*(S+)

Use split instead of regex, also avoid giving variable name like keywords
:

inpt = 'a:b c:d e:f'
k= [tuple(i.split(':')) for i in inpt.split()]
print(k)

# [('a', 'b'), ('c', 'd'), ('e', 'f')]


The easiest way using list comprehension and split :

list comprehension

split

[tuple(ele.split(':')) for ele in input.split(' ')]


#driver values :

IN : input = 'a:b c:d e:f'
OUT : [('a', 'b'), ('c', 'd'), ('e', 'f')]


You may use

list(map(lambda x: tuple(x.split(':')), input.split()))


where

input.split() is

input.split()

>>> input.split()
['a:b', 'c:d', 'e:f']


lambda x: tuple(x.split(':')) is function to convert string to tuple 'a:b' => (a, b)

lambda x: tuple(x.split(':'))

'a:b' => (a, b)

map applies above function to all list elements and returns a map object (in Python 3) and this is converted to list using list

map

list

Result

>>> list(map(lambda x: tuple(x.split(':')), input.split()))
[('a', 'b'), ('c', 'd'), ('e', 'f')]


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