On Tue, Dec 6, 2011 at 7:55 PM, Olivier Delalleau <sh...@keba.be> wrote:
> It may not be the most efficient way to do this, but you can do:
> mask = b > a
> a[mask] = b[mask]
>
> -=- Olivier
>
> 2011/12/6 questions anon <questions.a...@gmail.com>
>>
>> I would like to produce an array with the maximum values out of many
>> (10000s) of arrays.
>> I need to loop through many multidimentional arrays and if a value is
>> larger (in the same place as the previous array) then I would like that
>> value to replace it.
>>
>> e.g.
>> a=[1,1,2,2
>> 11,2,2
>> 1,1,2,2]
>> b=[1,1,3,2
>> 2,1,0,0
>> 1,1,2,0]
>>
>> where b>a replace with value in b, so the new a should be :
>>
>> a=[1,1,3,2]
>> 2,1,2,2
>> 1,1,2,2]
>>
>> and then keep looping through many arrays and replace whenever value is
>> larger.
>>
>> I have tried numpy.putmask but that results in
>> TypeError: putmask() argument 1 must be numpy.ndarray, not list
>> Any other ideas? Thanks
if I understand correctly it's a minimum.reduce
numpy
>>> a = np.concatenate((np.arange(5)[::-1], np.arange(5)))*np.ones((4,3,1))
>>> np.minimum.reduce(a, axis=2)
array([[ 0., 0., 0.],
[ 0., 0., 0.],
[ 0., 0., 0.],
[ 0., 0., 0.]])
>>> a.T.shape
(10, 3, 4)
python with iterable
>>> reduce(np.maximum, a.T)
array([[ 4., 4., 4., 4.],
[ 4., 4., 4., 4.],
[ 4., 4., 4., 4.]])
>>> reduce(np.minimum, a.T)
array([[ 0., 0., 0., 0.],
[ 0., 0., 0., 0.],
[ 0., 0., 0., 0.]])
Josef
>>
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>
>
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