Numpy sum elements in array based on its value

np.add.at(d, i, v)

You'd think d[i] += v would work, but if you try to do multiple additions to the same cell that way, one of them overrides the others. The ufunc.at method avoids those problems.

We can use np.bincount which is supposedly pretty efficient for such accumulative weighted counting, so here's one with that -

counts = np.bincount(i,v)
d[:counts.size] = counts

Alternatively, using minlength input argument and for a generic case when d could be any array and we want to add into it -

d += np.bincount(i,v,minlength=d.size).astype(d.dtype, copy=False)

Runtime tests

This section compares np.add.at based approach listed in the other post with the np.bincount based one listed earlier in this post.

In [61]: def bincount_based(d,i,v):
    ...:     counts = np.bincount(i,v)
    ...:     d[:counts.size] = counts
    ...: 
    ...: def add_at_based(d,i,v):
    ...:     np.add.at(d, i, v)
    ...:     

In [62]: # Inputs (random numbers)
    ...: N = 10000
    ...: i = np.random.randint(0,1000,(N))
    ...: v = np.random.randint(0,1000,(N))
    ...: 
    ...: # Setup output arrays for two approaches
    ...: M = 12000
    ...: d1 = np.zeros(M)
    ...: d2 = np.zeros(M)
    ...: 

In [63]: bincount_based(d1,i,v) # Run approaches
    ...: add_at_based(d2,i,v)
    ...: 

In [64]: np.allclose(d1,d2)  # Verify outputs
Out[64]: True

In [67]: # Setup output arrays for two approaches again for timing
    ...: M = 12000
    ...: d1 = np.zeros(M)
    ...: d2 = np.zeros(M)
    ...: 

In [68]: %timeit add_at_based(d2,i,v)
1000 loops, best of 3: 1.83 ms per loop

In [69]: %timeit bincount_based(d1,i,v)
10000 loops, best of 3: 52.7 µs per loop