On 4/8/2010 7:19 PM, Patrick Maupin wrote:
On Apr 8, 6:06 pm, monkeys paw<mon...@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...@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
I was going from example and looking for something useful from
the lambda feature. I come from C -> Perl -> Python (recent). I
don't find lambda very useful yet.
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