lambda with floats
monkeys paw
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
Patrick Maupin
Not sure why you are returning a lambda (which is just a function that
does not have a name) from an outer function.
A function that does this multiplication would simply be:
def s(n):
return 208.0 * 208.0 * n
Regards,
Pat
Patrick Maupin
I realized I didn't show the use. A bit different than what you were
doing:
>>> def s(n):
... return 208.0 * 208.0 * n
...
>>> s(1)
43264.0
>>> s(0.5)
21632.0
>>> s(3)
129792.0
I'm not sure exactly what you mean by "acre meter" though; this
returns the number of square feet in 'n' acres.
Regards,
Pat
Patrick Maupin
I should stop making a habit of responding to myself, BUT. This isn't
quite an acre in square feet. I just saw the 43xxx and assumed it
was, and then realized it couldn't be, because it wasn't divisible by
10. (I used to measure land with my grandfather with a 66 foot long
chain, and learned at an early age that an acre was 1 chain by 10
chains, or 66 * 66 * 10 = 43560 sqft.)
That's an exact number, and 208 is a poor approximation of its square
root.
Regards,
Pat
Peter Pearson
The expression "lambda x: (208 * 208) * n" is independent of x.
Is that what you intended?
--
To email me, substitute nowhere->spamcop, invalid->net.
monkeys paw
> 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
Patrick Maupin
Yes, but then what is the 'n' for. When you do that, you are not
using it, and it is still confusing.
Regards,
Pat
monkeys paw
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.
monkeys paw
You are absolutely right Pat, so here is the correct equate which also
utilizes my original question of using floats in a lambda, perfectly...
g = lambda x: 208.71 * 208.71 * x
g(1)
43559.864100000006
but truly the easiest to remember is based on your chain:
g = lambda x: 660 * 66 * x
g(1)
43560
Now back to python...
Steven D'Aprano
> 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.
Perhaps you feel that lambda is a special kind of object. It isn't. It's
just a short-cut for creating an anonymous function object.
f = lambda x: x+1
is almost exactly the same as:
def function(x):
return x+1
f = function
del function
The only advantages of lambda are:
(1) you can write a simple function as a one-liner; and
(2) it's an expression, so you can embed it in another expression:
list_of_functions = [math.sin, lambda x: 2*x-1, lambda x, y=1: x**y]
for func in list_of_functions:
plot(func)
The disadvantage of lambda is that you can only include a single
expression as the body of the function.
You will generally find lambdas used as callback functions, and almost
nowhere else.
--
Steven
Bas
> I should stop making a habit of responding to myself, BUT. This isn't
> quite an acre in square feet. I just saw the 43xxx and assumed it
> was, and then realized it couldn't be, because it wasn't divisible by
> 10. (I used to measure land with my grandfather with a 66 foot long
> chain, and learned at an early age that an acre was 1 chain by 10
> chains, or 66 * 66 * 10 = 43560 sqft.)
> That's an exact number, and 208 is a poor approximation of its square
> root.
There is no need to remember those numbers for the imperially
challenged people:
In [1]: import scipy.constants as c
In [2]: def acre2sqft(a):
...: return a * c.acre / (c.foot * c.foot)
...:
In [3]: acre2sqft(1)
Out[3]: 43560.0
Cheers,
Bas
monkeys paw
> On Apr 7, 6:15 am, Patrick Maupin<pmau...@gmail.com> wrote:
>> I should stop making a habit of responding to myself, BUT. This isn't
>> quite an acre in square feet. I just saw the 43xxx and assumed it
>> was, and then realized it couldn't be, because it wasn't divisible by
>> 10. (I used to measure land with my grandfather with a 66 foot long
>> chain, and learned at an early age that an acre was 1 chain by 10
>> chains, or 66 * 66 * 10 = 43560 sqft.)
>> That's an exact number, and 208 is a poor approximation of its square
>> root.
>
> There is no need to remember those numbers for the imperially
> challenged people:
>
> In [1]: import scipy.constants as c
scipy.constants ??
doesn't work for me.
Patrick Maupin
Basically, he's saying that, instead of remembering the very simple
"66" and "10" values, you can download and install a multi-megabyte
gzipped tar file for the scipy project. ;-)
(Of course, you get a few nice functions thrown in for free along with
your constants, but downloading scipy for its constants is like
choosing a sports car for its cupholders.)
monkeys paw
yea, the 66 foot chain story is a good one, i cant forget that.
Appreciate the help, i just looked up the SciPY Project download,
it is 40 MB. I'm gonna check it out none the less...
Albert van der Horst
Patrick Maupin <pma...@gmail.com> wrote:
>On Apr 6, 11:10=A0pm, Patrick Maupin <pmau...@gmail.com> wrote:
>> On Apr 6, 11:04=A0pm, Patrick Maupin <pmau...@gmail.com> wrote:
>>
>>
>>
>> > On Apr 6, 10:16=A0pm, 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)
>>
>> > > ... =A0 =A0 return lambda x: (208 * 208) * n
>> > > ...
>> > > =A0>>> f =3D s(1)
>> > > =A0>>> f(1)
>> > > 43264
>> > > =A0>>> 208 * 208
>> > > 43264
>> > > =A0>>> f(.25)
>> > > 43264
>>
>> > Not sure why you are returning a lambda (which is just a function that
>> > does not have a name) from an outer function.
>>
>> > A function that does this multiplication would simply be:
>>
>> > def s(n):
>>
>> > Regards,
>> > Pat
>>
>> I realized I didn't show the use. =A0A bit different than what you were
>> doing:
>>
>> >>> def s(n):
>>
>> ... =A0 =A0 return 208.0 * 208.0 * n
>> ...>>> s(1)
>> 43264.0
>> >>> s(0.5)
>> 21632.0
>> >>> s(3)
>>
>> 129792.0
>>
>> I'm not sure exactly what you mean by "acre meter" though; this
>> returns the number of square feet in 'n' acres.
>>
>> Regards,
>> Pat
>
>I should stop making a habit of responding to myself, BUT. This isn't
>quite an acre in square feet. I just saw the 43xxx and assumed it
>was, and then realized it couldn't be, because it wasn't divisible by
>10. (I used to measure land with my grandfather with a 66 foot long
>chain, and learned at an early age that an acre was 1 chain by 10
>That's an exact number, and 208 is a poor approximation of its square
>root.
208 = 2^6 * 13^2
66 = 2 * 3 * 11
Now I understand why the state of Wisconsin wants to drop out of
the USA and join the EU ;-)
>
>Regards,
>Pat
Groetjes Albert
--
--
Albert van der Horst, UTRECHT,THE NETHERLANDS
Economic growth -- being exponential -- ultimately falters.
albert@spe&ar&c.xs4all.nl &=n http://home.hccnet.nl/a.w.m.van.der.horst