Replacing values in numpy array

Question

I two numpy arrays, both M by N. X contains random values. Y contains true/false. Array A contains indices for rows in X that need replacement, with the value -1. I want to only replace values where Y is true.

Here is some code to do that:

M=30
N=40
X = np.zeros((M,N))  # random values, but 0s work too
Y = np.where(np.random.rand(M,N) > .5, True, False)
A=np.array([ 7,  8, 10, 13]), # in my setting, it's (1,4), not (4,)
for i in A[0]:
    X[i][Y[A][i]==True]=-1

However, what I actually want is only replace some of the entries. List B contains how many need to be replaced for each index in A. It's already ordered so A[0][0] corresponds to B[0], etc. Also, it's true that if A[i] = k, then the corresponding row in Y has at least k trues.

B = [1,2,1,1]

Then for each index i (in loop),

X[i][Y[A][i]==True][0:B[i]] = -1

This doesn't work. Any ideas on a fix?

Saullo G. P. Castro · Accepted Answer · 2014-05-08 20:05:32Z

2

It is not clear what you want to do, here is my understanding:

import numpy as np
m,n = 30,40
x = np.zeros((m,n))
y = np.random.rand(m,n) > 0.5    #no need for where here
a = np.array([7,8,10,13])
x[a] = np.where(y[a],-1,x[a])    #need where here

edited May 8, 2014 at 20:05

Saullo G. P. Castro

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answered Sep 8, 2013 at 4:15

Developer

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1 Comment

Kevin Over a year ago

Right, but I actually don't want them all replaced -- I need the first B replaced. If B[0] = 2, instead of all y[7] being changed to -1, I only want the first y[7][0:2] replaced. If B[1] = 3, then change y[8][0:3] = -1.

Kevin · Accepted Answer · 2013-09-08 13:51:09Z

1

Unfortunately, I don't have an elegant answer; however, this works:

M=30
N=40
X = np.zeros((M,N))  # random values, but 0s work too
Y = np.where(np.random.rand(M,N) > .5, True, False)
A=np.array([ 7,  8, 10, 13]), # in my setting, it's (1,4), not (4,)
B = [1,2,1,1]

# position in row where X should equal - 1, i.e. X[7,a0], X[8,a1], etc
a0=np.where(Y[7]==True)[0][0]
a1=np.where(Y[8]==True)[0][0]
a2=np.where(Y[8]==True)[0][1]
a3=np.where(Y[10]==True)[0][0]
a4=np.where(Y[13]==True)[0][0]

# For each row (i) indexed by A, take only B[i] entries where Y[i]==True.  Assume these indices in X = -1
for i in range(len(A[0])):
    X[A[0][i]][(Y[A][i]==True).nonzero()[0][0:B[i]]]=-1

np.sum(X) # should be -5
X[7,a0]+X[8,a1]+X[8,a2]+X[10,a3]+X[13,a4] # should be -5

answered Sep 8, 2013 at 13:51

Kevin

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1 Comment

Kevin Over a year ago

This code would look a bit better if A was (k,) instead of (1,k). Any ideas for speed improvement?

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