Accumulate constant value in Numpy Array

Question

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?

Answer 1

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.]])

Answer 2

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)