Add value to every "other" field ((i+j)%2==0) of numpy array

You can easily do the operation in two steps, like

import numpy as npa = np.zeros((5, 14))# Even rows, odd columnsa[::2, 1::2] += 1# Odd rows, even columnsa[1::2, ::2] += 1print a

Here's one way using NumPy broadcasting -

a[(np.arange(a.shape[0])[:,None] + np.arange(a.shape[1]))%2==0] += 1

Explanation : We basically create two arrays that are equivalent of the i-th and j-th iterators. Let's call them I and J.

I = np.arange(a.shape[0])J = np.arange(a.shape[1])

Now, to perform an operation between all possible i and j, we create extend I to 2D by pushing its elements into the first axis and thus creating singleton dimension along its second axis.

Figuratively, the effect of broadcasting could be put like this :

    I[:,None] : M , 1            J : 1 , NI[:,None] + J : M,  N

Thus, the final setting would be -

a[(I[:,None] + J)%2==0] += 1

To put it other way with the intention to avoid comparing with 0 and directly use mod-2 which would be essentially 0 or 1 -

a += (np.arange(a.shape[0])[:,None]-1 + np.arange(a.shape[1]))%2

One can also use np.ix_ to process odd and then even rows for setting, like so -

a[np.ix_(np.arange(0,a.shape[0],2),np.arange(0,a.shape[1],2))] += 1a[np.ix_(np.arange(1,a.shape[0],2),np.arange(1,a.shape[1],2))] += 1

One can build a mask for "every other" element, and apply the addition on the mask.

# Create maskm00 = np.zeros(a.shape[0], dtype=bool)m00[0::2] = Truem01 = np.zeros(a.shape[1], dtype=bool)m01[0::2] = Truem0 = np.logical_and.outer(m00, m01)m10 = np.zeros(a.shape[0], dtype=bool)m10[1::2] = Truem11 = np.zeros(a.shape[1], dtype=bool)m11[1::2] = Truem1 = np.logical_and.outer(m10, m11)m = np.logical_or(m0, m1)a[m] += 1