CODEWITHSUNDEEP
pythonarraysnumpymatrixindices
user9998992
user9998992

Reputation: 124

Summing selected elements of a matrix in Python

I have a [n x n] matrix containing values belonging to different groups, and a [1 x n] vector defining to which group each element belongs. (n usually ~1E4, in this example n=4)

I want to calculate a matrix obtained by summing together all the elements belonging to the same group.

I use np.where() to calculate the indices where the elements of each group are located. When I use the calculated indices, I do not obtain the expected elements, because I select pairs of positions instead of ranges (I'm used to Matlab, where I can simply select M(idx1,idx2) ).

import numpy as np

n=4
M = np.random.rand(n,n)
print(M)

# This vector defines to which group each element belong
belongToGroup = np.array([0, 1, 0, 2])

nGroups=np.max(belongToGroup);

# Calculate a matrix obtained by summing elements belonging to the same group
M_sum = np.zeros((nGroups+1,nGroups+1))
for g1 in range(nGroups+1):
    idxG1 = np.where(belongToGroup==g1)
    for g2 in range(nGroups+1):
        idxG2 = np.where(belongToGroup==g2)
        print('g1 = ' + str(g1))
        print('g2 = ' + str(g2))
        print(idxG1[0])
        print(idxG2[0])
        print(M[idxG1[0],idxG2[0]])
        print(np.sum(M[idxG1[0],idxG2[0]]))
        M_sum[g1,g2]=np.sum(M[idxG1[0],idxG2[0]])

print('')
print('Example of the problem:')
print('Elements I would like to sum to obtain M_sum[0,0]')
print(M[0:2,0:2])
print('Elements that are summed instead')
print(M[[0,1],[0,1]])

Example of the problem: In the example above, the element M_sum[0,0] should be the sum of M[0,0], M[0,1], M[1,0], and M[1,1] Instead, it is calculated as the sum of M[0,0] and M[1,1]

Upvotes: 1

Views: 1350

Answers (2)

Paul Panzer
Paul Panzer

Reputation: 53029

You can use np.ix_ to obtain matlab-ish behavior:

A = np.arange(9).reshape(3, 3)
A[[1,2],[0,2]]
# array([3, 8])
A[np.ix_([1,2],[0,2])]
# array([[3, 5],
#        [6, 8]])

Under the hood, np.ix_ does what @hpaulj describes in detail:

np.ix_([1,2],[0,2])
# (array([[1],
#        [2]]), array([[0, 2]]))

You can apply this to your specific problem as follows:

M = np.random.randint(0, 10, (n, n))
M
# array([[6, 2, 7, 1],
#        [6, 7, 9, 5],
#        [9, 4, 3, 2],
#        [3, 1, 7, 9]])
idx = np.array([0, 1, 0, 2])

ng = idx.max() + 1
out = np.zeros((ng, ng), M.dtype)
np.add.at(out, np.ix_(idx, idx), M)
out
# array([[25,  6,  3],
#        [15,  7,  5],
#        [10,  1,  9]])

As an aside: There is a faster but less obvious solution that relies on flat indexing:

np.bincount(np.ravel_multi_index(np.ix_(idx, idx), (ng, ng)).ravel(), M.ravel(), ng*ng).reshape(ng, ng)
# array([[25.,  6.,  3.],
#        [15.,  7.,  5.],
#        [10.,  1.,  9.]])

Upvotes: 1

hpaulj
hpaulj

Reputation: 231385

In MATLAB, indexing with 2 lists (actually matrices) picks a block. numpy on the other hand tries to broadcast the indexing arrays against each other, and returns selected points. Its behavior is close to what sub2ind does in MATLAB.

In [971]: arr = np.arange(16).reshape(4,4)                                      
In [972]: arr                                                                   
Out[972]: 
array([[ 0,  1,  2,  3],
       [ 4,  5,  6,  7],
       [ 8,  9, 10, 11],
       [12, 13, 14, 15]])
In [973]: i1, i2 = np.array([0,2,3]), np.array([1,2,0])

Indexing with 2 1d arrays of the same size:

In [974]: arr[i1,i2]
Out[974]: array([ 1, 10, 12])

This, in effect returns [arr[0,1], arr[2,2], arr[3,0]], one element for each point of matching indices.

But if I turn one index into a 'column vector', it selects from rows, while i2 selects from columns.

In [975]: arr[i1[:,None], i2]                                                   
Out[975]: 
array([[ 1,  2,  0],
       [ 9, 10,  8],
       [13, 14, 12]])

MATLAB makes the block indexing easy, while individual access is harder. In numpy the block access is a bit harder, though the underlying mechanics is the same.

With your example i1[0] and i2[0] can be arrays like:

array([0, 2]), array([3])
(2,) (1,)

The shape (1,) array can broadcast with the (2,) or with a (2,1) array just as well. Your code would fail if instead is[0] were np.array([0,1,2]), a (3,) array which can't pair with the (2,) array. But with a (2,1) it produces a (2,3) block.

Upvotes: 2

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