Mix lambda and list comprehension?

Re: Mix lambda and list comprehension?

Try this:
[color=blue][color=green][color=darkred]
>>> [lambda x, y=y:x+y for y in range(10)][4](2)[/color][/color][/color]
6


It is important to bind y in a closure at the time
the lambda is defined. Otherwise, y remains unbound
until you invoke the function call. At that time, the most
recent value of y is the last value in the range loop (namely, 9).


Raymond Hettinger



"Peter Barth" <peter.barth@ t-online.de> wrote in message
news:6f5b3f88.0 307142302.1a153 [email protected] gle.com...[color=blue]
> Hi,
> trying to mix lambda expresions and list comprehension
> doesn't seem to work.
> ---[color=green][color=darkred]
> >>> [lambda x:x+y for y in range(10)][9](2)[/color][/color]
> 11[color=green][color=darkred]
> >>> [lambda x:x+y for y in range(10)][4](2)[/color][/color]
> 11
> ---
> I expected the second expression to return 6.
> What did I do wrong? Any hints?
> Thanks
> - Peter[/color]

Re: Mix lambda and list comprehension?

[email protected] (Peter Barth) wrote in message news:<6f5b3f88. 0307142302.1a15 [email protected] ogle.com>...[color=blue]
> Hi,
> trying to mix lambda expresions and list comprehension
> doesn't seem to work.
> ---[color=green][color=darkred]
> >>> [lambda x:x+y for y in range(10)][9](2)[/color][/color]
> 11[color=green][color=darkred]
> >>> [lambda x:x+y for y in range(10)][4](2)[/color][/color]
> 11
> ---
> I expected the second expression to return 6.
> What did I do wrong? Any hints?
> Thanks
> - Peter[/color]

It is a scope issue. The last value for y is used for all
the created lambdas. All lambdas users are bitten by that,
soon or later. The solution is to make y local to the
lambda function, with the optional argument trick:
[color=blue][color=green][color=darkred]
>>> [lambda x,y=y:x+y for y in range(10)][4](2)[/color][/color][/color]
6

Michele

Re: Mix lambda and list comprehension?

Thanks a lot, works fine.
However, the solution does not really feel "pythonesqu e".
Is it considered a usability bug or fine as is?
- Peter

"Raymond Hettinger" <vze4rx4y@veriz on.net> wrote in message news:<hMNQa.113 4$I4.740@nwrdny 01.gnilink.net> ...[color=blue]
> Try this:
>[color=green][color=darkred]
> >>> [lambda x, y=y:x+y for y in range(10)][4](2)[/color][/color]
> 6
>
>
> It is important to bind y in a closure at the time
> the lambda is defined. Otherwise, y remains unbound
> until you invoke the function call. At that time, the most
> recent value of y is the last value in the range loop (namely, 9).
>
>
> Raymond Hettinger
>
>
>
> "Peter Barth" <peter.barth@ t-online.de> wrote in message
> news:6f5b3f88.0 307142302.1a153 [email protected] gle.com...[color=green]
> > Hi,
> > trying to mix lambda expresions and list comprehension
> > doesn't seem to work.
> > ---[color=darkred]
> > >>> [lambda x:x+y for y in range(10)][9](2)[/color][/color]
> 11[color=green][color=darkred]
> > >>> [lambda x:x+y for y in range(10)][4](2)[/color]
> > 11
> > ---
> > I expected the second expression to return 6.
> > What did I do wrong? Any hints?
> > Thanks
> > - Peter[/color][/color]

Re: Mix lambda and list comprehension?

[email protected] (Bengt Richter) wrote in message news:<bf3pjm$gs [email protected]. 122>...[color=blue]
> On 15 Jul 2003 05:41:02 -0700, [email protected] (Michele Simionato) wrote:
>[color=green]
> >peter.barth@ t-online.de (Peter Barth) wrote in message news:<6f5b3f88. 0307142302.1a15 [email protected] ogle.com>...[color=darkred]
> >> Hi,
> >> trying to mix lambda expresions and list comprehension
> >> doesn't seem to work.
> >> ---
> >> >>> [lambda x:x+y for y in range(10)][9](2)[/color][/color]
> 11[color=green][color=darkred]
> >> >>> [lambda x:x+y for y in range(10)][4](2)
> >> 11
> >> ---
> >> I expected the second expression to return 6.
> >> What did I do wrong? Any hints?
> >> Thanks
> >> - Peter[/color]
> >
> >It is a scope issue. The last value for y is used for all
> >the created lambdas. All lambdas users are bitten by that,
> >soon or later. The solution is to make y local to the
> >lambda function, with the optional argument trick:
> >[color=darkred]
> >>>> [lambda x,y=y:x+y for y in range(10)][4](2)[/color]
> >6[/color]
>
> or you could capture y as constants in the lambdas ;-)
>[color=green][color=darkred]
> >>> [eval('lambda x:x+%s'%y) for y in range(10)][4](2)[/color][/color]
> 6
>
>
> Regards,
> Bengt Richter[/color]

Thanks to God, there is a smile in your post!!

;)

Michele