views:

958

answers:

7

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?

+14  A: 

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}
S.Lott
While that's cool, I think I prefer the version in the original question from a long-term code maintenance point of view. The code in the original question is clear and obvious, the above requires multiple tens of seconds to grok.
Bryan Oakley
@Bryan Oakley: I agree. The question ("is there a way to eliminate the for loop using list comprehension") has an answer. However, the answer may not be what they **really** wanted.
S.Lott
Yes that is what I was looking for. Thank you.@Bryan: I'm starting to think that maybe keeping that for loop is probably a good idea. Nevertheless, at least now I know how to do this in one line.
Pavel
+1  A: 
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}
nosklo
Nice.........15.
S.Lott
A: 

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!

Agos
you wanted to keep this simple? not to be mean or pick on anyone but where is this simple?
Casey
+3  A: 

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}
Vicki Laidler
A: 

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

OP wanted the second element of the list cast to an integer.
hughdbrown
+2  A: 

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()))
A: 

I would do this:

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