使用行索引为 numpy 数组赋值

Question

Suppose I have two arrays, a=np.array([0,0,1,1,1,2]), b=np.array([1,2,4,2,6,5]) . Elements in a mean the row indices of where b should be assigned. And if there are multiple elements in the same row, the values should be assigned in order. So the result is a 2D array c :

c = np.zeros((3, 4))
counts = {k:0 for k in range(3)}
for i in range(a.shape[0]):
    c[a[i], counts[a[i]]]=b[i]
    counts[a[i]]+=1
print(c)

Is there a way to use some fancy indexing method in numpy to get such results faster (without a for loop) in case these arrays are big.

Answer 1

I had to run your code to actually see what it produced. There are limits to what I can 'run' in my head.

In [230]: c                                                                                            
Out[230]: 
array([[1., 2., 0., 0.],
       [4., 2., 6., 0.],
       [5., 0., 0., 0.]])
In [231]: counts                                                                                       
Out[231]: {0: 2, 1: 3, 2: 1}

Omitting this information may be delaying possible answers. 'vectorization' requires thinking in whole-array terms, which is easiest if I can visualize the result, and look for a pattern.

This looks like a padding problem.

In [260]: u, c = np.unique(a, return_counts=True)                                                      
In [261]: u                                                                                            
Out[261]: array([0, 1, 2])
In [262]: c                                                                                            
Out[262]: array([2, 3, 1])      # cf with counts

Load data with rows of different sizes into Numpy array

Working from previous padding questions, I can construct a mask:

In [263]: mask = np.arange(4)<c[:,None]                                                                
In [264]: mask                                                                                         
Out[264]: 
array([[ True,  True, False, False],
       [ True,  True,  True, False],
       [ True, False, False, False]])

and use that to assign the b values to c :

In [265]: c = np.zeros((3,4),int)                                                                      
In [266]: c[mask] = b                                                                                  
In [267]: c                                                                                            
Out[267]: 
array([[1, 2, 0, 0],
       [4, 2, 6, 0],
       [5, 0, 0, 0]])

Since a is already sorted we might get the counts faster than with unique . Also it will have problems if a doesn't have any values for some row(s).