Joshua Ginsberg wrote:
> Try this one:
>
d = {}
for x in [1,2,3]:
> ... d[x] = lambda *args: args[0]*x
> ...
d[1](3)
try it with:
d[x] = (lambda x=x: (lambda *args: args[0]*x))()
the outer lambda fixes the value of x and produces the inner lambda
with the fixed x value
That's a damned elegant solution -- thank you...
However in trying to simplify my problem for presentation here, I think
I oversimplified a bit too much. Try this one:
>>> d = {}
>>> for x in [1,2,3]:
... d[x] = lambda *args: args[0]*x
...
>>> d[1](3)
9
The lambda is going to have to take ar
Jp Calderone wrote:
> On Thu, 06 Oct 2005 16:18:15 -0400, Joshua Ginsberg
> <[EMAIL PROTECTED]> wrote:
>>So this part makes total sense to me:
>>
> d = {}
> for x in [1,2,3]:
>>... d[x] = lambda y: y*x
>>...
> d[1](3)
>>9
>>
>>Because x in the lambda definition isn't evaluated unt
Here's another one:
>>> d = {}
>>> for x in [1,2,3]:
... d[x] = (lambda z: lambda y: y * z) (x)
...
>>> d[1](3)
3
>>> d[2](3)
6
--
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Joshua Ginsberg wrote:
> So this part makes total sense to me:
>
> >>> d = {}
> >>> for x in [1,2,3]:
> ... d[x] = lambda y: y*x
> ...
> >>> d[1](3)
> 9
>
> Because x in the lambda definition isn't evaluated until the lambda is
> executed, at which point x is 3.
>
> Is there a way to specifica
On Thu, 06 Oct 2005 16:18:15 -0400, Joshua Ginsberg <[EMAIL PROTECTED]> wrote:
>So this part makes total sense to me:
>
d = {}
for x in [1,2,3]:
>... d[x] = lambda y: y*x
>...
d[1](3)
>9
>
>Because x in the lambda definition isn't evaluated until the lambda is
>executed, at which
So this part makes total sense to me:
>>> d = {}
>>> for x in [1,2,3]:
... d[x] = lambda y: y*x
...
>>> d[1](3)
9
Because x in the lambda definition isn't evaluated until the lambda is
executed, at which point x is 3.
Is there a way to specifically hard code into that lambda definition the
c
