ctorizing
return newfunc(x,y) #< calling the vectorized func
def test_test():
test_instance=test()
x=numpy.arange(1,10)
y=numpy.arange(11,20)
print test_instance.operate('add',x,y)
if __name__=='__main__':
test_test()
==
On Fri, Apr 2, 2010 at 12:34 PM, Shailendra wrote:
> Hi All,
>
>>>> x=arange(10)
>>>> indices=[(1,),(2,)]
>>>> x[indices]
> Traceback (most recent call last):
> File "", line 1, in
> IndexError: unsupported iterator index
>
>
Well, this is just a toy problem. argmax represent a method which will
give me a index in x[cond] . And for the case of multiple value my
requirement is fine with getting any "max" index.
Thanks,
Shailendra
On Fri, Apr 2, 2010 at 3:00 PM, eat wrote:
> Shailendra gmail.com> w
[2, 3],
[4, 5],
[6, 7],
[8, 9]])
>>> indices=[(1,0),(0,1)]
>>> x[indices]
array([2, 1])
Should not the first one also work? How to convert [(1,),(2,)] to
[1,2] efficiently.
Thanks,
Shailendra
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I forgot to mention that i wanted this to work for general shape. So i
modified it little bit
>>> x = array([[1,2,3,4,5], [6,7,8,7,6], [1,2,3,4,5]])
>>> cond = (x > 5)
>>> loc= where(cond)
>>> arg_max=argmax(x[cond])
>>> x[tuple([e[arg_max] for e in
ax=argmax(x[cond])
>>> max
2
>>> x[cond][max]
8
Now , I want to get the index of this element in x. How to acheive
this. In real situation x will be huge.
Thanks,
Shailendra
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ro=numpy.nonzero(A)
>>> non_zero
(array([0, 0, 1, 2]), array([0, 1, 0, 2]))
>>> A[non_zero]
array([1, 1, 1, 1])
>>> A[non_zero]=X
>>> A
array([[ 2, 9, 0],
[10, 0, 0],
[ 0, 0, 3]])
Thanks,
Shailendra
On Fri, Apr 2, 2010 at 10:35 AM, gerardob w
Thanks everyone for replies/suggestion. It is simple to avoid this
problem. But my point that given the behavior of python this behavior
seems inconsistent. There could other method provided which could
evaluate bool value depending on values stored in the array.
Thanks,
Shailendra
On Fri, Apr 2
they have.
Also, please suggest the best way to differentiate between an empty
array and non-empty array( irrespective to what is inside array).
Thanks,
Shailendra
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t has either 1 "match"(closest point) or none
Also, the size of the cordinates1 and cordinates2 are quite large and
"outer" should not be used. I can think of only C style code to
achieve this. Can any one suggest pythonic way of doing this?
Thanks,
Shailendra
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