Help with lambda

[lallous]

Yes it should be listed somewhere, now I get it. Thanks Arnaud.

--
Elias

On Feb 18, 1:47 pm, Arnaud Delobelle <arno... at googlemail.com> wrote:
> lallous <elias.bachaal... at gmail.com> writes:
> > Hello,
>
> > I am still fairly new to Python. Can someone explain to me why there
> > is a difference in f and g:
>
> > def make_power(n):
> >     return lambda x: x ** n
>
> > # Create a set of exponential functions
> > f = [lambda x: x ** n for n in xrange(2, 5)]
> > g = [make_power(n) for n in xrange(2, 5)]
>
> > print f[0](3), f[1](3)
> > print g[0](3), g[1](3)
>
> > I expect to have "f" act same like "g".
>
> > Thanks
>
> It's a FAQ!  Except I don't think it's in the official Python FAQ, but
> it ought to be.
>
> The reason (very quickly) is that each time you call make_power() you
> create a new name 'n' which is bound to a different value each time,
> whereas in f:
>
>     [lambda x: x ** n for n in xrange(2, 5)]
>
> The 'n' is the same name for each of the lambda functions and so is
> bound to the same value, which by the end of the loop is 4.  So each of
> the lambdas in the list f is the same as:
>
>     lambdsa x; x**4
>
> after the end of the list comprehension.
>
> The usual fix for this is to change f to:
>
> f = [lambda x, n=n: x ** n for n in xrange(2, 5)]
>
> I'll let you think why this works.
>
> HTH
>
> --
> Arnaud