Reputation: 511
I have a 5 dim array (comes from binning operations) and would like to have it normed (sum == 1 for the last dimension).
I thought I found the answer here but it says:
ValueError: Found array with dim 5. the normalize function expected <= 2.
I achieve the result with 5 nested loops, like:
for en in range(en_bin.nb):
for zd in range(zd_bin.nb):
for az in range(az_bin.nb):
for oa in range(oa_bin.nb):
# reduce fifth dimension (en reco) for normalization
b = np.sum(a[en][zd][az][oa])
for er in range(er_bin.nb):
a[en][zd][az][oa][er] /= b
but I want to vectorise operations.
For example:
In [18]: a.shape
Out[18]: (3, 1, 1, 2, 4)
In [20]: b.shape
Out[20]: (3, 1, 1, 2)
In [22]: a
Out[22]:
array([[[[[ 0.90290316, 0.00953237, 0.57925688, 0.65402645],
[ 0.68826638, 0.04982717, 0.30458093, 0.0025204 ]]]],
[[[[ 0.7973917 , 0.93050739, 0.79963614, 0.75142376],
[ 0.50401287, 0.81916812, 0.23491561, 0.77206141]]]],
[[[[ 0.44507296, 0.06625994, 0.6196917 , 0.6808444 ],
[ 0.8199077 , 0.02179789, 0.24627425, 0.43382448]]]]])
In [23]: b
Out[23]:
array([[[[ 2.14571886, 1.04519487]]],
[[[ 3.27895899, 2.33015801]]],
[[[ 1.81186899, 1.52180432]]]])
Upvotes: 3
Views: 3095
Reputation: 221574
Sum along the last axis by listing axis=-1
with numpy.sum
, keeping dimensions and then simply divide by the array itself, thus bringing in NumPy broadcasting
-
a/a.sum(axis=-1,keepdims=True)
This should be applicable for ndarrays of generic number of dimensions.
Alternatively, we could sum
with axis-reduction and then add a new axis with None/np.newaxis
to match up with the input array shape and then divide -
a/(a.sum(axis=-1)[...,None])
Upvotes: 4