Question

Given two unordered arrays of same lengths a and b:

a = [7,3,5,7,5,7]
b = [0.2,0.1,0.3,0.1,0.1,0.2]

I’d like to group by the elements in a:

aResult = [7,3,5]

summing over the elements in b (Example used to summarize a probability density function):

bResult = [0.2 + 0.1 + 0.2, 0.1, 0.3 + 0.1] = [0.5, 0.1, 0.4]

Alternatively, random a and b in python:

import numpy as np
a = np.random.randint(1,10,10000)
b = np.array([1./len(a)]*len(a))

I have two approaches, which for sure are far from the lower performance boundary.
Approach 1 (at least nice and short): Time: 0.769315958023

def approach_2(a,b):
    bResult = [sum(b[i == a]) for i in np.unique(a)]
    aResult = np.unique(a)

Approach 2 (numpy.groupby, horribly slow) Time: 4.65299129486

def approach_2(a,b): 
    tmp = [(a[i],b[i]) for i in range(len(a))]
    tmp2 = np.array(tmp, dtype = [('a', float),('b', float)])
    tmp2 = np.sort(tmp2, order='a') 

    bResult = []
    aResult = []
    for key, group in groupby(tmp2, lambda x: x[0]):
        aResult.append(key)
        bResult.append(sum([i[1] for i in group]))

Update: Approach3, by Pablo. Time: 1.0265750885

def approach_Pablo(a,b):    

    pdf = defaultdict(int); 
    for x,y in zip(a,b):
        pdf[x] += y  

Update: Approach 4, by Unutbu. Time: 0.184849023819 [WINNER SO FAR, but a as integer only]

def unique_Unutbu(a,b):

    x=np.bincount(a,weights=b)
    aResult = np.unique(a)
    bResult = x[aResult]

Maybe someone finds a smarter solution to this problem than me 🙂

Answer 1

If a is composed of ints < 2**31-1 (that is, if a has values that can fit in dtype int32), then you could use np.bincount with weights:

import numpy as np
a = [7,3,5,7,5,7]
b = [0.2,0.1,0.3,0.1,0.1,0.2]

x=np.bincount(a,weights=b)
print(x)
# [ 0.   0.   0.   0.1  0.   0.4  0.   0.5]

print(x[[7,3,5]])
# [ 0.5  0.1  0.4]

np.unique(a) returns [3 5 7], so the result appears in a different order:

print(x[np.unique(a)])
# [ 0.1  0.4  0.5]

One potential problem with using np.bincount is that it returns an array whose length is equal to the maximum value in a. If a contains even one element with value near 2**31-1, then bincount would have to allocate an array of size 8*(2**31-1) bytes (or 16GiB).

So np.bincount might be the fastest solution for arrays a which have big length, but not big values. For arrays a which have small length (and big or small values), using a collections.defaultdict would probably be faster.

Edit: See J.F. Sebastian’s solution for a way around the integer-values-only restriction and big-values problem.