How about this (though it is getting a little clunky):
plot(*transpose(map(lambda x:(x,sin(x**2)/x**3), arange(3,6, 0.01))))
On Wed, Mar 10, 2010 at 7:57 PM, Chloe Lewis <chle...@berkeley.edu> wrote:
> ...although
>
> >>plot(map(lambda x:x**2, range(5,15)))
>
> probably doesn't do exactly what you want; is the idea that
>
> >>plot(range(5,15),lambda x:x**2) #DOESN'T WORK
>
> should automatically work like
>
> >>plot(range(5,15), map(lambda x:x**2, range(5,15)))
>
> by recognizing that the second argument is a function rather than a
> list?
>
> &C
>
>
> On Mar 10, 2010, at 10:47 AM, Chloe Lewis wrote:
>
>> You'd always have to specify the domain, so
>>
>> plot(map(lambda x:x**2, range(1,10)))
>>
>> shouldn't be much longer than the minimal command.
>>
>> &C
>>
>>
>> On Mar 10, 2010, at 10:12 AM, max ulidtko wrote:
>>
>>> Hi.
>>>
>>> Is it possible to plot arbitrary lambda function with matplotlib?
>>> Say, if i have f = lambda x: x*sin(x), can i just plot it without
>>> building argument-value arrays? It would be a very convenient and
>>> useful feature.
>>>
>>>
>>> ------
>>> Sincerely,
>>> max ulidtko
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>>
>>
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>
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