mapping functions and lambda

[Suresh Jeevanandam]

Given a string
s = 'a=1,b=2'

I want to create a dictionary {'a': '1', 'b': '2'}

I did,

dict(map(lambda k: k.split('='), s.split(',')))

Is it possible to get rid of the lambda here, without having to define
another function just for this.

Is this the easiest/straight-forward way to do this?

regards,
Suresh

[Suresh Jeevanandam]

Re: mapping functions and lambda

I got it:
dict([k.split('=') for k in s.split(',')])

regards,
Suresh
Suresh Jeevanandam wrote:[color=blue]
> Given a string
> s = 'a=1,b=2'
>
> I want to create a dictionary {'a': '1', 'b': '2'}
>
> I did,
>
> dict(map(lambda k: k.split('='), s.split(',')))
>
> Is it possible to get rid of the lambda here, without having to define
> another function just for this.
>
> Is this the easiest/straight-forward way to do this?
>
> regards,
> Suresh[/color]

[Steve Holden]

Re: mapping functions and lambda

Suresh Jeevanandam wrote:[color=blue]
> I got it:
> dict([k.split('=') for k in s.split(',')])
>
> Suresh Jeevanandam wrote:
>[color=green]
>>Given a string
>>s = 'a=1,b=2'
>>
>>I want to create a dictionary {'a': '1', 'b': '2'}
>>
>>I did,
>>
>>dict(map(lamb da k: k.split('='), s.split(',')))
>>
>>Is it possible to get rid of the lambda here, without having to define
>>another function just for this.
>>
>>Is this the easiest/straight-forward way to do this?
>>[/color][/color]

In Python 2.4 you don't even need to construct the list, you can just
use a generator expression instead:

dict(k.split('= ') for k in s.split(','))

regards
Steve
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