Numpy sum elements in array based on its value

Question

I have unsorted array of indexes:

i = np.array([1,5,2,6,4,3,6,7,4,3,2])

I also have an array of values of the same length:

v = np.array([2,5,2,3,4,1,2,1,6,4,2])

I have array with zeros of desired values:

d = np.zeros(10)

Now I want to add to elements in d values of v based on it's index in i.

If I do it in plain python I would do it like this:

for index,value in enumerate(v):
    idx = i[index]
    d[idx] += v[index]

It is ugly and inefficient. How can I change it?

Answer 1

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.

Answer 2

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