CODEWITHSUNDEEP
pythonregexparsing
Anon
Anon

Reputation: 875

Parsing string in Python

Here's a string that I'm trying to parse in python

    s1="One : Two : Three : Four  Value  : Five  Value  : Six  Value : Seven  Value : Eight  Value :"

Can someone tell me a re function I can use to parse the above string so that s1 becomes as follows without any ':'

One

Two

Three

Four Value

Five Value 

Six Value

Seven Value

Eight Value

I've tried making use of strip, lstrip and rstrip after spliting the string by using the following code but I don't get the format I need

    res1=s1.split(' : ')

UPDATE: Thanks a lot for your answers but the output I'm getting looks like this whether I use

1->

    for index in s1:
      print index

or....

2->

    pprint(s1)

OUTPUT:

O

n

e

:

T

w

o

:

T

h

r

e

e

:

F

o

u

r

V

a

l

u

e

:

F

i

v

e

V

a

l

u

e

:

S

i

x

V

a

l

u

e

:

S

e

v

e

n

V

a

l

u

e

:

E

i

g

h

t

V

a

l

u

e

:

Upvotes: 0

Views: 391

Answers (5)

Omnifarious
Omnifarious

Reputation: 56128

 import re
 re.split(r'\s*:\s*', s1)

And, slightly more efficiently, if you have to do a lot of splitting...

 import re
 split_re = re.compile(r'\s*:\s*')
 split_re.split(s1)

And this will also work. It would be interesting to do a speed test.

 [a.strip() for a in s1.split(':')]

These will all get you an array containing each word. If you want a string containing multiple lines with a blank line between each word, you can use '\n\n'.join(foo) for each of them to get that string. But this also works:

 import re
 split_re = re.compile(r'\s*:\s*')
 res1 = split_re.subn('\n\n', s1)[0]

Testing shows though that:

 res1 = '\n\n'.join(a.strip() for a in s1.split(':'))

is actually the fastest, and it's certainly the prettiest. And if you want to avoid the blank line at the end from the final ':' that doesn't have anything after it:

 res1 = '\n\n'.join(a.strip() for a in s1.split(':')).strip()

Upvotes: 0

pupizoid
pupizoid

Reputation: 311

The easiest way is: res1 = ' '.join(s1.split(':')) if you want a single row string, else you should try: res1 = '\n'.join(s1.split(':'))

Upvotes: 0

dmg
dmg

Reputation: 7726

A list comprehension approach (for diversity reasons and cause it's the only answer that doesn't leave a blank item at the end). Either of these:

filter(lambda x: x != '', [item.strip() for item in s1.split(':')])
[item.strip() for item in s1.split(':') if item.strip() != '']
[item for item in map(lambda x: x.strip(), s1.split(':')) if item != '']

Upvotes: 1

Cyrille
Cyrille

Reputation: 14593

1/ If you want a string:

From your split() :

res1 = '\n'.join(res1)

Other solution:

res1 = s1.replace(' : ', '\n')

2/ If you want a list:

res1 = [item.strip() for item in s1.split(':')]

[...] will return a list containing your strings. Check "List Comprehensions" in http://docs.python.org/2/tutorial/datastructures.html for more informations

Upvotes: 0

eumiro
eumiro

Reputation: 213125

'\n'.join(a.strip() for a in s1.split(':'))

returns

One
Two
Three
Four  Value
Five  Value
Six  Value
Seven  Value
Eight  Value

If you need extra empty lines:

'\n\n'.join(a.strip() for a in s1.split(':'))

Upvotes: 6

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