Binary String To List Python

I've got a binary string like '1100011101'. I would like to parse it into a list where each chunk of 1's or 0's is a separate value in the list. Such as: '1100011101' becomes ['11'

Solution 1:

You can scrape a (minor) bit of performance out of this by using a regex instead of groupby() + join(). This just finds groups of 1 or 0:

import re

s = '1100011101'
l = re.findall(r"0+|1+", s)
# ['11', '000', '111', '0', '1']

Timings:

s = '1100011101' * 1000

%timeit l = [''.join(g) for _, g in groupby(s)]
# 1.16 ms ± 9.79 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

%timeit re.findall(r"0+|1+", s)
# 723 µs ± 5.32 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

Solution 2:

Use itertools.groupby:

from itertools import groupby

binary= "1100011101"

result= ["".join(repeat) for _, repeat in groupby(binary)]
print(result)

Output

['11', '000', '111', '0', '1']

Solution 3:

Insert a space between the 01 and 10 transitions using .replace() and then split the resulting string:

'1100011101'.replace("01","0 1").replace("10","1 0").split()

['11', '000', '111', '0', '1']

Solution 4:

with groupby

>>> from itertools import groupby as f
>>> x = str(1100011101)
>>> sol = [''.join(v) for k, v in f(x)]
>>> print(sol)
['11', '000', '111', '0', '1']

without using groupby and if you want more faster execution

def func(string):
    if not string:
        return []
    def get_data(string):
            if not string:
                return 
            count = 0
            target = string[0]
            for i in string:
                if i==target:
                    count+=1else:
                    yield target*count
                    count = 1
                    target = i
            if count>0:
                yield target*count
    return list(get_data(string))
        
    x = '1100011101'
    sol =func(x)print(sol)

output

['11', '000', '111', '0', '1']

Timings on my machine

from itertools import groupby

s = '11000111010101' * 100000

%timeit l = [''.join(g) for _, g in groupby(s)]
318 ms ± 2.75 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)


import re

s = '11000111010101' * 100000

%timeit l = re.findall(r"0+|1+", s)
216 ms ± 2.01 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)


deffunc(string):
    ifnot string:
        return []
    defget_data(string):
            ifnot string:
                return 
            count = 0
            target = string[0]
            for i in string:
                if i==target:
                    count+=1else:
                    yield target*count
                    count = 1
                    target = i
            if count>0:
                yield target*count
    returnlist(get_data(string))

s = '11000111010101' * 100000

%timeit func(s)
205 ms ± 11.5 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
####################################################################from itertools import groupby

s = '11000111010101' * 1000

%timeit l = [''.join(g) for _, g in groupby(s)]
3.28 ms ± 178 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)


import re

s = '11000111010101' * 1000

%timeit l = re.findall(r"0+|1+", s)
2.06 ms ± 57.9 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)


deffunc(string):
    ifnot string:
        return []
    defget_data(string):
            ifnot string:
                return 
            count = 0
            target = string[0]
            for i in string:
                if i==target:
                    count+=1else:
                    yield target*count
                    count = 1
                    target = i
            if count>0:
                yield target*count
    returnlist(get_data(string))

s = '11000111010101' * 1000

%timeit func(s)
1.91 ms ± 153 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)