How to split a string within a list to create key-value pairs in Python

Question

I have a list that looks like this:

[ \'abc=lalalla\', \'appa=kdkdkdkd\', \'kkakaka=oeoeoeo\']

And I want to split this list by \'=\' so th

Answer 1

print dict([s.split("=") for s in my_list])

like this

>>> my_list = [ 'abc=lalalla', 'appa=kdkdkdkd', 'kkakaka=oeoeoeo']
>>> print dict(s.split("=") for s in my_list) #thanks gribbler
{'kkakaka': 'oeoeoeo', 'abc': 'lalalla', 'appa': 'kdkdkdkd'}

Answer 2

Example by using map function

a = ["abc=lalalla", "appa=kdkdkdkd", "kkakaka=oeoeoeo"]
d = dict(map(lambda s: s.split('='), a))

Answer 3

In addition, make sure you limit the splits to 1, in case the right-hand side contains an '='.

d = dict(s.split('=',1) for s in a)

Answer 4

You can feed a map object directly to dict. For built-in functions without arguments, map should show similar or better performance. You will see a drop-off in performance when introducing arguments:

from functools import partial

L = ['abc=lalalla', 'appa=kdkdkdkd', 'kkakaka=oeoeoeo']
L2 = ['abc lalalla', 'appa kdkdkdkd', 'kkakaka oeoeoeo']

n = 100000
L = L*n
L2 = L2*n

%timeit dict(map(partial(str.split, sep='='), L))  # 234 ms per loop
%timeit dict(s.split('=') for s in L)              # 164 ms per loop

%timeit dict(map(str.split, L2))                   # 141 ms per loop
%timeit dict(s.split() for s in L2)                # 144 ms per loop

Answer 5

a = [ 'abc=lalalla', 'appa=kdkdkdkd', 'kkakaka=oeoeoeo']
d = dict(s.split('=') for s in a)
print d


Output:
{'kkakaka': 'oeoeoeo', 'abc': 'lalalla', 'appa': 'kdkdkdkd'}

http://codepad.org/bZ8lGuHE