I'm trying to sum +1 to some specific cells of a numpy array but I can't find any way without slow loops:
coords = np.array([[1,2],[1,2],[1,2],[0,0]])
X = np.zeros((3,3))
for i,j in coords:
X[i,j] +=1
Resulting in:
X = [[ 1. 0. 0.]
[ 0. 0. 3.]
[ 0. 0. 0.]]
X[coords[:,0],coords[:,1] += 1 returns
X = [[ 1. 0. 0.]
[ 0. 0. 1.]
[ 0. 0. 0.]]
Any help?
numpy.at is exactly for those situations.
In [1]: np.add.at(X,tuple(coords.T),1)
In [2]: X
Out[2]:
array([[ 1., 0., 0.],
[ 0., 0., 3.],
[ 0., 0., 0.]])
for loop in numba is 80 times faster :/ stackoverflow.com/q/57561500/125507
You can use np.bincount, like so -
out_shape = (3,3) # Input param
# Get linear indices corresponding to coords with the output array shape.
# These form the IDs for accumulation in the next step.
ids = np.ravel_multi_index(coords.T,out_shape)
# Use bincount to get 1-weighted accumulations. Since bincount assumes 1D
# array, we need to do reshaping before and after for desired output.
out = np.bincount(ids,minlength=np.prod(out_shape)).reshape(out_shape)
If you are trying to assign values other than 1s, you can use the additional input argument to feed in weights to np.bincount.
Sample run -
In [2]: coords
Out[2]:
array([[1, 2],
[1, 2],
[1, 2],
[0, 0]])
In [3]: out_shape = (3,3) # Input param
...: ids = np.ravel_multi_index(coords.T,out_shape)
...: out = np.bincount(ids,minlength=np.prod(out_shape)).reshape(out_shape)
...:
In [4]: out
Out[4]:
array([[1, 0, 0],
[0, 0, 3],
[0, 0, 0]], dtype=int64)
Another option is np.histogramdd:
bins = [np.arange(d + 1) for d in X.shape]
out, edges = np.histogramdd(coords, bins)
print(out)
# [[ 1. 0. 0.]
# [ 0. 0. 3.]
# [ 0. 0. 0.]]
