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

Python starts counting at 0, so your code should work find if you replace np.arange(1,9) with np.arange(9)

In [11]: np.arange(1,9)
Out[11]: array([1, 2, 3, 4, 5, 6, 7, 8])

In [12]: np.arange(9)
Out[12]: array([0, 1, 2, 3, 4, 5, 6, 7, 8])

Answer 2

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]]

Answer 3

For one thing, as others have mentioned, NumPy uses 0-based indexing. But even once you fix that, this is not what you want to use:

for i in np.arange(9):
    for j in np.arange(9):
        B[i:j] = A[i:j]

The : indicates slicing, so i:j means "all items from the i-th, up to the j-th, excluding the last one." So your code is copying every row over several times, which is not a very efficient way of doing things.

You probable wanted to use ,:

for i in np.arange(8): # Notice the range only goes up to 8
    for j in np.arange(8): # ditto
        B[i, j] = A[i, j]

This will work, but is also pretty wasteful performancewise when using NumPy. A much faster approach is to simply ask for:

B[:] = A

Answer 4

1. As stated above: python indices start at 0.

2. In order to iterate over some (say matrix) indices, you should use the builtin function 'range' and not 'numpy.arange'. The arange returns an ndarray, while range returns a generator in a recent python version.

3. The syntax 'B[i:j]' does not refer to the element at row i and column j in an array B. It rather means: all rows of B starting at row i and going up to (but not including) row j (if B has so many rows, otherwise it returns until includingly the last row). The element at position i, j is in fact 'B[i,j]'.

The indexing syntax of python / numpy is quite powerful and performant.

Answer 5

A cleaner way, in my opinion, of writing the C loop is:

for k in range(1,7):
    for l in range(1,7):
        if B[k,l]==8:
            C[k-1, l-1] += D[k,l]

That inner block of B (and D) can be selected with slices, B[1:7, 1:7] or B[1:-1, 1:-1].

A and D are defined as np.matrix. Since we aren't doing matrix multiplications here (no dot products), that can create problems. For example I was puzzled why

In [27]: (B[1:-1,1:-1]==8)*D[1:-1,1:-1]
Out[27]: 
matrix([[2, 1, 2, 3, 3, 3],
        [3, 3, 3, 4, 5, 5],
        [1, 2, 1, 1, 2, 2],
        [2, 2, 3, 2, 3, 1],
        [2, 2, 3, 2, 3, 1],
        [2, 3, 3, 2, 3, 2]])

What I expected (and matches the loop C) is:

In [28]: (B[1:-1,1:-1]==8)*D.A[1:-1,1:-1]
Out[28]: 
array([[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]])

B = A.copy() still leaves B as matrix. B=A.A returns an np.ndarray. (as does np.copy(A))

D.A is the array equivalent of D. B[1:-1,1:-1]==8 is boolean, but when used in the multiplication context it is effectively 0s and 1s.

But if we want to stick with np.matrix then I'd suggest using the element by element multiply function:

In [46]: np.multiply((A[1:-1,1:-1]==8), D[1:-1,1:-1])
Out[46]: 
matrix([[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]])

or just multiply the full matrixes, and select the inner block after:

In [47]: np.multiply((A==8), D)[1:-1, 1:-1]
Out[47]: 
matrix([[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]])