how to assign values to numpy array given the row indices and starting column indices?

Question

a = np.array([2,3,1,4])
b = np.array([2,3,7,1])
c = np.zeros((4, 10))

I wanna assign value 1 to some elements in c . a and b define the positions of such elements. a is the starting column indices of value 1 in each row. And b represents how many consecutive 1 there are in the row. The output I am expecting is:

array([[ 0.,  0.,  1.,  1.,  0.,  0.,  0.,  0.,  0.,  0.],
       [ 0.,  0.,  0.,  1.,  1.,  1.,  0.,  0.,  0.,  0.],
       [ 0.,  1.,  1.,  1.,  1.,  1.,  1.,  1.,  0.,  0.],
       [ 0.,  0.,  0.,  0.,  1.,  0.,  0.,  0.,  0.,  0.]])

I can use a simple for loop as below:

for i in range(c.shape[0]):
    for k in range(a[i], a[i]+b[i]):
        c[i,k]=1

But it would be slow for large arrays, is there any faster numpy indexing to do this? Thanks.

Answer 1

You can cast it into a 1D problem

def convert_inds(a,b,array_shape):
    
    nrows,ncols = array_shape
    to_take = np.zeros(sum(b))
    count = 0
    for ind,item in enumerate(b):
        start_ind = ind*ncols+a[ind]
        to_take[count:count+item] = np.arange(start_ind,start_ind+item)
        count += item
        
    return to_take.astype(np.int)

to_take = convert_inds(a,b,c.shape)

c.ravel()[to_take] = 1

In the code above, convert_inds will convert a and b to

array([ 2,  3, 13, 14, 15, 21, 22, 23, 24, 25, 26, 27, 34])

which are indices of 1 s in the flattened c . By doing this, you only need to iterate through b in the function convert_inds .

Answer 2

I implemented next solution without any Python loops, just pure NumPy code. Maybe it is not that simple as python-loop solution, but definitely will be much faster especially for large data.

Try it online!

import numpy as np

def set_val_2d(a, val, starts, lens):
    begs = starts + np.arange(a.shape[0]) * a.shape[1]
    ends = begs + lens
    clens = lens.cumsum()
    ix = np.ones((clens[-1],), dtype = np.int64)
    ix[0] = begs[0]
    ix[clens[:-1]] = begs[1:] - ends[:-1] + 1
    ix = ix.cumsum()
    a.ravel()[ix] = val

a = np.array([2,3,1,4])
b = np.array([2,3,7,1])
c = np.zeros((4, 10))

set_val_2d(c, 1, a, b)
print(c)

Output:

[[0. 0. 1. 1. 0. 0. 0. 0. 0. 0.]
 [0. 0. 0. 1. 1. 1. 0. 0. 0. 0.]
 [0. 1. 1. 1. 1. 1. 1. 1. 0. 0.]
 [0. 0. 0. 0. 1. 0. 0. 0. 0. 0.]]

Answer 3

If you choose a fancy indexing based approach, the most difficult part is finding indexes of axis 1. This is very similar to:

>>> np.repeat(a, b)
array([2, 2, 3, 3, 3, 1, 1, 1, 1, 1, 1, 1, 4])

except that each groups of indexes should be incrementing. This fix can be done with this function:

def accumulative_count(counts, initial):
    counter = np.ones(np.sum(counts), dtype=int)
    marker_idx = np.r_[0, np.cumsum(counts)[:-1]]
    subtract_vals = np.r_[1, counts[:-1]]
    initial_vals = np.r_[initial[0], np.diff(initial)]
    counter[marker_idx] = counter[marker_idx] - subtract_vals + initial_vals
    return np.cumsum(counter)

>>> accumulative_count(counts, initial)
array([2, 3, 3, 4, 5, 1, 2, 3, 4, 5, 6, 7, 4], dtype=int32)

After all, you're capable to finish it:

c[np.repeat(np.arange(len(c)), b), accumulative_count(b, a)] = 1

c:

array([[0., 0., 1., 1., 0., 0., 0., 0., 0., 0.],
       [0., 0., 0., 1., 1., 1., 0., 0., 0., 0.],
       [0., 1., 1., 1., 1., 1., 1., 1., 0., 0.],
       [0., 0., 0., 0., 1., 0., 0., 0., 0., 0.]])