How can I add a 1D array to a segment of a 2D array?

Question

I have a 2D NumPy array filled with zeroes (placeholder values). I would like to add a 1D array filled with ones and zeroes to a part of it. eg.

2D array:

array([[0, 0, 0, 0, 0],
       [0, 0, 0, 0, 0],
       [0, 0, 0, 0, 0]])

1D array:
```
array([1, 0, 1])
```

Desired end product: I want the array starting in position [2, 1]

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

Or an insertion in any other position it could reasonably fit in. I have tried to do it with boolean masks but have not had any luck creating one in the correct shape. I have also tried flattening the 2D array, but couldn't figure out how to replace the values in the correct space.

next time at least provide us with the correct format for the arrays (i.e. [[11,2],....] , ) , please — pippo1980
– pippo1980, Commented Aug 10, 2021 at 15:27
Can't believe I tried boolean masks but forgot about fancy indexing. Thanks! — Calvin Adams
– Calvin Adams, Commented Aug 10, 2021 at 15:29

Ivan · Accepted Answer · 2021-08-10 15:35:19Z

1

You can indeed flatten the array and create a sequence of positions where you will insert your 1D array segment:

>>> pos = [1, 2]
>>> start = x.shape[1]*pos[0] + pos[1]
>>> seq = start + np.arange(len(segment))

>>> seq
array([7, 8, 9])

Then, you can either index the flattened array:

>>> x_f = x.flatten()
>>> x_f[seq] = segment
>>> x_f.reshape(x.shape)
array([[0, 0, 0, 0, 0],
       [0, 0, 1, 0, 1],
       [0, 0, 0, 0, 0]])

Alternatively, you can np.ravel_multi_index to get seq and apply np.unravel_index on it.

>>> seq = np.arange(len(segment)) + np.ravel_multi_index(pos, x.shape)
array([7, 8, 9])

>>> indices = np.unravel_index(seq, x.shape)
(array([1, 1, 1]), array([2, 3, 4]))

>>> x[indices] = segment
>>> x
array([[0, 0, 0, 0, 0],
       [0, 0, 1, 0, 1],
       [0, 0, 0, 0, 0]])

edited Aug 10, 2021 at 15:35

answered Aug 10, 2021 at 15:19

Ivan

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