numpy - Python - Find K max values in each row of one matrix and compare to binary matrix -

i need determine if position (index) of k largest values in matrix in same position binary indicator matrix, b.

import numpy np = np.matrix([[.8,.2,.6,.4],[.9,.3,.8,.6],[.2,.6,.8,.4],[.3,.3,.1,.8]]) b = np.matrix([[1,0,0,1],[1,0,1,1],[1,1,1,0],[1,0,0,1]]) print "a:\n", print "b:\n", b  d = argsort(a) d[:,2:] # return whether these indices in 'b' 

returns:

a: [[ 0.8  0.2  0.6  0.4]  [ 0.9  0.3  0.8  0.6]  [ 0.2  0.6  0.8  0.4]  [ 0.3  0.3  0.1  0.8]] b: [[1 0 0 1]  [1 0 1 1]  [1 1 1 0]  [1 0 0 1]]  matrix([[2, 0],         [2, 0],         [1, 2],         [1, 3]]) 

i compare indices returned last result and, if b has ones in positions, return count. example, final desired result be:

1 2 2 1 

in other words, in first row of a, top-2 values correspond 1 of ones in b, etc.

any ideas how efficiently? maybe argsort wrong approach here. thanks.

when take argsort minimum 0 maximum 3, can reverse doing [::-1] maximum 0 , minimum 3:

s = np.argsort(a, axis=1)[:,::-1]    #array([[0, 2, 3, 1], #       [0, 2, 3, 1], #       [2, 1, 3, 0], #       [3, 1, 0, 2]]) 

now can use np.take 0s maximums , 1s second-maximums are:

s2 = s + (np.arange(s.shape[0])*s.shape[1])[:,none] s = np.take(s.flatten(),s2) #array([[0, 3, 1, 2], #       [0, 3, 1, 2], #       [3, 1, 0, 2], #       [2, 1, 3, 0]]) 

in b, 0 values should replaced np.nan 0==np.nan gives false:

b = np.float_(b) b[b==0] = np.nan #array([[  1.,  nan,  nan,   1.], #       [  1.,  nan,   1.,   1.], #       [  1.,   1.,   1.,  nan], #       [  1.,  nan,  nan,   1.]]) 

and following comparison give desired result:

print np.logical_or(s==b-1, s==b).sum(axis=1) #[[1] # [2] # [2] # [1]] 

the general case, compare n biggest values of a against binary b:

def check_a_b(a,b,n=2):     b = np.float_(b)     b[b==0] = np.nan     s = np.argsort(a, axis=1)[:,::-1]     s2 = s + (np.arange(s.shape[0])*s.shape[1])[:,none]     s = np.take(s.flatten(),s2)     ans = s==(b-1)     in range(n-1):         ans = np.logical_or( ans, s==b+i )     return ans.sum(axis=1) 

this pair-wise comparisons in logical_or.