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问题:
I have a Numpy 3 axis array whose elements are 3 dimensional. I'd like to average them and return the same shape of the array. The normal average function removes the 3 dimensions and replace it with the average (as expected):
a = np.array([[[0.1, 0.2, 0.3], [0.2, 0.3, 0.4]],
[[0.4, 0.4, 0.4], [0.7, 0.6, 0.8]]], np.float32)
b = np.average(a, axis=2)
# b = [[0.2, 0.3],
# [0.4, 0.7]]
Result required:
# b = [[[0.2, 0.2, 0.2], [0.3, 0.3, 0.3]],
# [[0.4, 0.4, 0.4], [0.7, 0.7, 0.7]]]
Can you do this elegantly or do I just have to iterate over the array in Python (which will be a lot slower compared to a powerful Numpy function).
Can you set the Dtype argument, for the np.mean function, to a 1D array perhaps?
Thanks.
回答1:
>>> import numpy as np
>>> a = np.array([[[0.1, 0.2, 0.3], [0.2, 0.3, 0.4]],
... [[0.4, 0.4, 0.4], [0.7, 0.6, 0.8]]], np.float32)
>>> b = np.average(a, axis=2)
>>> b
array([[ 0.2 , 0.29999998],
[ 0.40000001, 0.69999999]], dtype=float32)
>>> c = np.dstack((b, b, b))
>>> c
array([[[ 0.2 , 0.2 , 0.2 ],
[ 0.29999998, 0.29999998, 0.29999998]],
[[ 0.40000001, 0.40000001, 0.40000001],
[ 0.69999999, 0.69999999, 0.69999999]]], dtype=float32)
回答2:
Ok, CAUTION I don't have my masters in numpyology yet, but just playing around, I came up with:
>>> np.average(a,axis=-1).repeat(a.shape[-1]).reshape(a.shape)
array([[[ 0.2 , 0.2 , 0.2 ],
[ 0.29999998, 0.29999998, 0.29999998]],
[[ 0.40000001, 0.40000001, 0.40000001],
[ 0.69999999, 0.69999999, 0.69999999]]], dtype=float32)
回答3:
Have you considered using broadcasting? Here is more info about broadcasting if you're new to the concept.
Here is an example using broadcast_arrays
, keep in mind that the b
produced here by broadcast_arrays
should be treated as read only, you should make a copy if you want to write to it:
>>> b = np.average(a, axis=2)[:, :, np.newaxis]
>>> b, _ = np.broadcast_arrays(b, a)
>>> b
array([[[ 0.2 , 0.2 , 0.2 ],
[ 0.29999998, 0.29999998, 0.29999998]],
[[ 0.40000001, 0.40000001, 0.40000001],
[ 0.69999999, 0.69999999, 0.69999999]]], dtype=float32)
回答4:
Here is a method that avoids making copies:
a = a.T
a[:] = a.mean(axis=0)
a = a.T
Or if you don't want to overwrite a
:
b = np.empty_like(a)
b = b.T
b[:] = a.mean(axis=-1).T
b = b.T
回答5:
This is for an arbitrary axis:
array
is the ndimentional array
and axis
is the axis to average
np.repeat( np.expand_dims( np.mean( array, axis ), axis ), array.shape[axis], axis )