convert string to dict using list comprehension

问题

I have came across this problem a few times and can't seem to figure out a simple solution. Say I have a string

string = "a=0 b=1 c=3"

I want to convert that into a dictionary with a, b and c being the key and 0, 1, and 3 being their respective values (converted to int). Obviously I can do this:

list = string.split()
dic = {}
for entry in list:
    key, val = entry.split('=')
    dic[key] = int(val)

But I don't really like that for loop, It seems so simple that you should be able to convert it to some sort of list comprehension expression. And that works for slightly simpler cases where the val can be a string.

dic = dict([entry.split('=') for entry in list])

However, I need to convert val to an int on the fly and doing something like this is syntactically incorrect.

dic = dict([[entry[0], int(entry[1])] for entry.split('=') in list])

So my question is: is there a way to eliminate the for loop using list comprehension? If not, is there some built in python method that will do that for me?

回答1:

Do you mean this?

>>> dict( (n,int(v)) for n,v in (a.split('=') for a in string.split() ) )
{'a': 0, 'c': 3, 'b': 1}

回答2:

How about a one-liner without list comprehension?

 foo="a=0 b=1 c=3"
 ans=eval( 'dict(%s)'%foo.replace(' ',',')) )
 print ans
{'a': 0, 'c': 3, 'b': 1}

回答3:

Try the next:

dict([x.split('=') for x in s.split()])

回答4:

I sometimes like this approach, especially when the logic for making keys and values is more complicated:

s = "a=0 b=1 c=3"

def get_key_val(x):
    a,b = x.split('=')
    return a,int(b)

ans = dict(map(get_key_val,s.split()))

回答5:

Nowadays you should probably use a dictionary comprehension, introduced in 2.7:

mydict = {key: int(value) for key, value in (a.split('=') for a in mystring.split())}

The dictionary comprehension is faster and shinier (and, in my opinion, more readable).

from timeit import timeit

setup = """mystring = "a=0 b=1 c=3\""""
code1 = """mydict = dict((n,int(v)) for n,v in (a.split('=') for a in mystring.split()))""" # S.Lott's code
code2 = """mydict = {key: int(value) for key, value in (a.split('=') for a in mystring.split())}"""

print timeit(code1, setup=setup, number=10000) # 0.115524053574
print timeit(code2, setup=setup, number=10000) # 0.105328798294

回答6:

from cgi import parse_qsl
text = "a=0 b=1 c=3"
dic = dict((k, int(v)) for k, v in parse_qsl(text.replace(' ', '&')))
print dic

prints

{'a': 0, 'c': 3, 'b': 1}

回答7:

I would do this:

def kv(e): return (e[0], int(e[1]))
d = dict([kv(e.split("=")) for e in string.split(" ")])

回答8:

I like S.Lott's solution, but I came up with another possibility.
Since you already have a string resembling somehow the way you'd write that, you can just adapt it to python syntax and then eval() it :)

import re
string = "a=0 b=1 c=3"
string2 = "{"+ re.sub('( |^)(?P<id>\w+)=(?P<val>\d+)', ' "\g<id>":\g<val>,', string) +"}"
dict = eval(string2)
print type(string), type(string2), type(dict)
print string, string2, dict

The regex here is pretty raw and won't catch all the possible python identifiers, but I wanted to keep it simple for simplicity's sake. Of course if you have control over how the input string is generated, just generate it according to python syntax and eval it away. BUT of course you should perform additional sanity checks to make sure that no code is injected there!

来源：https://stackoverflow.com/questions/1246444/convert-string-to-dict-using-list-comprehension