如何在列表中拆分字符串以在 Python 中创建键值对
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How to split a string within a list to create key-value pairs in Python
提问by Vor
I have a list that looks like this:
我有一个看起来像这样的列表：
[ 'abc=lalalla', 'appa=kdkdkdkd', 'kkakaka=oeoeoeo']
And I want to split this list by '=' so that everything on the left side will become keys and on the right, values.
我想用 '=' 分割这个列表，这样左侧的所有内容都将成为键，而右侧的所有内容都将成为值。
{
'abc':'lalalla',
'appa':'kdkdkdkd',
'kkakaka':'oeoeo'
}
采纳答案by Demian Brecht
a = [ 'abc=lalalla', 'appa=kdkdkdkd', 'kkakaka=oeoeoeo']
d = dict(s.split('=') for s in a)
print d
Output:
{'kkakaka': 'oeoeoeo', 'abc': 'lalalla', 'appa': 'kdkdkdkd'}
回答by Joran Beasley
print dict([s.split("=") for s in my_list])
like this
像这样
>>> 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'}
回答by user117529
In addition, make sure you limit the splits to 1, in case the right-hand side contains an '='.
此外，确保将拆分限制为 1，以防右侧包含“=”。
d = dict(s.split('=',1) for s in a)
回答by jpp
You can feed a mapobject directly to dict. For built-in functions without arguments, mapshould show similar or better performance. You will see a drop-off in performance when introducing arguments:
您可以将map对象直接提供给dict. 对于没有参数的内置函数，map应该表现出相似或更好的性能。引入参数时，您会看到性能下降：
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
