First row of numpy.ones is still populated after referencing another matrix

Question

I have a matrix \'A\' whose values are shown below. After creating a matrix \'B\' of ones using numpy.ones and assigning the values from \'A\' to \'B\' by indexing \'i\' rows an

Answer 1

Here first what I think you are trying to do, with minimal corrections, comments to your code:

import numpy as np

A = np.matrix([[8,8,8,7,7,6,8,2],
               [8,8,7,7,7,6,6,7],
               [1,8,8,7,7,6,6,6],
               [1,1,8,7,7,6,7,7],
               [1,1,1,1,8,7,7,6],
               [1,1,2,1,8,7,7,6],
               [2,2,2,1,1,8,7,7],
               [2,1,2,1,1,8,8,7]])

B = np.ones((8,8),dtype=np.int)

for i in np.arange(1,9): #  i= 1...8
    for j in np.arange(1,9):  # j= 1..8, but A[8,j] and A[j,8] do not exist,
                              # if you insist on 1-based indeces, numpy still expects 0... n-1, 
                              # so you'll have to subtract 1 from each index to use them
        B[i-1,j-1] = A[i-1,j-1]


C = np.zeros((6,6),dtype=np.int)
D = np.matrix([[1,1,2,3,3,2,2,1],
               [1,2,1,2,3,3,3,2],
               [1,1,2,1,1,2,2,3],
               [2,2,3,2,2,2,1,3],
               [1,2,2,3,2,3,1,3],
               [1,2,3,3,2,3,2,3],
               [1,2,2,3,2,3,1,2],
               [2,2,3,2,2,3,2,2]])


for k in np.arange(2,8):       # k = 2..7
    for l in np.arange(2,8):   # l = 2..7   ; matrix B has indeces 0..7, so if you want inner points, you'll need 1..6
            b = B[k-1,l-1]   # so this is correct, gives you the inner matrix
            if b == 8:           # here b is  a value in the matrix , not the index, careful not to mix those
                # Matrix C is smaller than Matrix B  ; yes C has indeces from 0..5 for k and l
                # so to address C you'll need to subtract 2 from the k,l that you defined in the for loop
                C[k-2,l-2] = C[k-2,l-2] + 1*D[k-1,l-1]

print C

output:

[[2 0 0 0 0 0]
 [1 2 0 0 0 0]
 [0 3 0 0 0 0]
 [0 0 0 2 0 0]
 [0 0 0 2 0 0]
 [0 0 0 0 3 0]]

But there are more elegant ways to do it. In particular look up slicing, ( numpy conditional array arithmetic, possibly scipy threshold.All of the below should be much faster than Python loops too (numpy loops are written in C).

B=np.copy(A)  #if you need a copy of A, this is the way

# one quick way to make a matrix that's 1 whereever A==8, and is smaller
from scipy import stats
B1=stats.threshold(A, threshmin=8, threshmax=8, newval=0)/8   # make a matrix with ones where there is an 8
B1=B1[1:-1,1:-1]
print B1

#another quick way  to make a matrix that's 1 whereever A==8
B2 = np.zeros((8,8),dtype=np.int)
B2[A==8]=1
B2=B2[1:-1,1:-1]
print B2

# the following would obviously work with either B1 or B2 (which are the same)

print np.multiply(B2,D[1:-1,1:-1])

Output:

[[1 0 0 0 0 0]
 [1 1 0 0 0 0]
 [0 1 0 0 0 0]
 [0 0 0 1 0 0]
 [0 0 0 1 0 0]
 [0 0 0 0 1 0]]
[[1 0 0 0 0 0]
 [1 1 0 0 0 0]
 [0 1 0 0 0 0]
 [0 0 0 1 0 0]
 [0 0 0 1 0 0]
 [0 0 0 0 1 0]]
[[2 0 0 0 0 0]
 [1 2 0 0 0 0]
 [0 3 0 0 0 0]
 [0 0 0 2 0 0]
 [0 0 0 2 0 0]
 [0 0 0 0 3 0]]