Elements multiplication in NumPy array with all other elements-CodePudding

If I have a NumPy array [a b c d], is there a way to get the sum of the multiplication of each element with all other elements excluding itself and without duplicates?

For example, the sum for this array would be: ab ac ad bc bd cd.

Using for loops is an option, but it wouldn't be very efficient when you have very large arrays. So I was wondering if there is a more efficient way in NumPy.

CodePudding user response：

Using math, you can refactor the sum of product into a product of sums.

a*b   a*c   a*d   b*c   b*d   c*d

(a b c)*d   (a b)*c   a*b

This gives you the following:

out = (a[1:]*np.cumsum(a[:-1])).sum()

Example:

a = np.array([3, 6, 4, 9, 10])

np.triu(a*a[:,None], k=1).sum()
# 391

(a[1:]*np.cumsum(a[:-1])).sum()
# 391

CodePudding user response：

Here's a test of @mozwy's idea of using the upper triangle.

A sample array:

In [11]: x = np.array([1,2,3,4]); a,b,c,d = x

the sum of products:

In [12]: a*b a*c a*d b*c b*d c*d
Out[12]: 35

The upper tri:

In [13]: A = np.triu(np.ones((4,4),int),k=1)    
In [14]: A
Out[14]: 
array([[0, 1, 1, 1],
       [0, 0, 1, 1],
       [0, 0, 0, 1],
       [0, 0, 0, 0]])

And matrix product:

In [15]: x@A@x
Out[15]: 35

Another approach, starting with the full outer product:

In [17]: outer = x[:,None]*x    
In [18]: outer
Out[18]: 
array([[ 1,  2,  3,  4],
       [ 2,  4,  6,  8],
       [ 3,  6,  9, 12],
       [ 4,  8, 12, 16]])

Remove the trace, and you are left with twice the desired result:

In [19]: np.trace(outer)          # or x@x
Out[19]: 30

In [20]: (np.sum(outer)-np.trace(outer))/2
Out[20]: 35.0

equivalently using einsum:

In [25]: (np.einsum('i,j->',x,x)-np.einsum('i,i',x,x))/2
Out[25]: 35.0

With numpy it is easy, and relatively efficient, to do the full sum of outer products; "vectorization" means doing the whole-array calculations in compiled code, even if means some "extra" work and memory use.