Menu
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.
845
mean() Arithmetic mean is the sum of elements along an axis divided by the number of elements. The numpy. mean() function returns the arithmetic mean of elements in the array.
np. mean always computes an arithmetic mean, and has some additional options for input and output (e.g. what datatypes to use, where to place the result). np. average can compute a weighted average if the weights parameter is supplied.
NumPy arrays have an attribute called shape that returns a tuple with each index having the number of corresponding elements.
WeldNumpy is a Weld-enabled library that provides a subclass of NumPy's ndarray module, called weldarray, which supports automatic parallelization, lazy evaluation, and various other optimizations for data science workloads.
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)
If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With
