3

I'm using python2 and I want to get rid of these empty strings in the output of the following python regular expression: 

import re
x = "010101000110100001100001"
print re.split("([0-1]{8})", x)

and the output is this :

['', '01010100', '', '01101000', '', '01100001', '']

I just want to get this output:

['01010100', '01101000', '01100001']
Cœur
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Ahmed Ramadan
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6 Answers6

4

Regex probably isn't what you want to use in this case. It seems that you want to just split the string into groups of n (8) characters.

I poached an answer from this question.

def split_every(n, s):
    return [ s[i:i+n] for i in xrange(0, len(s), n) ]

split_every(8, "010101000110100001100001")
Out[2]: ['01010100', '01101000', '01100001']
Community
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Tim
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  • I actually made another code similar to this and I didn't get to know the Dynamic looping stuff yet, but it's really awesome :D – Ahmed Ramadan Dec 15 '16 at 04:33
2

One possible way:

print filter(None, re.split("([0-1]{8})", x))
Nurjan
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1
import re
x = "010101000110100001100001"
l = re.split("([0-1]{8})", x)
l2 = [i for i in l if i]

out:

['01010100', '01101000', '01100001']
宏杰李
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1

This is exactly what is split for. It is split string using regular expression as separator.

If you need to find all matches try use findall instead:

import re
x = "010101000110100001100001"
print(re.findall("([0-1]{8})", x))
neverwalkaloner
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1
print([a for a in re.split("([0-1]{8})", x) if a != ''])
G. Bahaa
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0

Following your regex approach, you can simply use a filter to get your desired output.

import re
x = "010101000110100001100001"
unfiltered_list = re.split("([0-1]{8})", x)
print filter(None, unfiltered_list)

If you run this, you should get:

['01010100', '01101000', '01100001']
gom1
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