lambda with floats

[Patrick Maupin]

On Apr 8, 6:06 pm, monkeys paw <mon... at joemoney.net> wrote:
> On 4/7/2010 1:08 PM, Peter Pearson wrote:
>
>
>
> > On Tue, 06 Apr 2010 23:16:18 -0400, monkeys paw<mon... at joemoney.net>  wrote:
> >> I have the following acre meter which works for integers,
> >> how do i convert this to float? I tried
>
> >> return float ((208.0 * 208.0) * n)
>
> >>>>> def s(n):
> >> ...    return lambda x: (208 * 208) * n
> >> ...
> >>>>> f = s(1)
> >>>>> f(1)
> >> 43264
> >>>>> 208 * 208
> >> 43264
> >>>>> f(.25)
> >> 43264
>
> > The expression "lambda x: (208 * 208) * n" is independent of x.
> > Is that what you intended?
>
> Seems i should have done this:
> g = lambda x: 208.0 * 208.0 * x
> g(1)
> 43264.0

Yes, but then what is the 'n' for.  When you do that, you are not
using it, and it is still confusing.

Regards,
Pat