我们可以计算线性索引，然后累加到使用 np.add.at .因此，以xpos和ypos作为数组，这是一个实现-

We could compute the linear indices, then accumulate into zeros-initialized output array with np.add.at. Thus, with xpos and ypos as arrays, here's one implementation -

m,n = xpos.max()+1, ypos.max()+1
out = np.zeros((m,n),dtype=int)
np.add.at(out.ravel(), xpos*n+ypos, 1)

样品运行-

In [95]: # 1d arrays holding x and y indices
    ...: xpos    =   np.array([0,0,1,2,1,2,1,0,0,0,0,1,1,1,2,2,3])
    ...: ypos    =   np.array([3,2,1,1,3,0,1,0,0,1,2,1,2,3,3,2,0])
    ...: 

In [96]: cnt_a   =   np.zeros((4,4))

In [97]: # This method works, but is very slow for a large array
    ...: for i in range(0,len(xpos)):
    ...:     cnt_a[xpos[i],ypos[i]] = cnt_a[xpos[i],ypos[i]] + 1
    ...:     

In [98]: m,n = xpos.max()+1, ypos.max()+1
    ...: out = np.zeros((m,n),dtype=int)
    ...: np.add.at(out.ravel(), xpos*n+ypos, 1)
    ...: 

In [99]: cnt_a
Out[99]: 
array([[ 2.,  1.,  2.,  1.],
       [ 0.,  3.,  1.,  2.],
       [ 1.,  1.,  1.,  1.],
       [ 1.,  0.,  0.,  0.]])

In [100]: out
Out[100]: 
array([[2, 1, 2, 1],
       [0, 3, 1, 2],
       [1, 1, 1, 1],
       [1, 0, 0, 0]])