Cfis Yoi - 1 month ago 9

Python Question

Simple enough implementation with a for loop. Example:

`s = np.zeros((3,3))`

b = np.array(([0,1],[0,1],[2,0]))

for line in range(s.shape[0]):

s[line][b[line]] = 1

What i would like to do is to update the matrix without the for loop. Something like

`s[b] = 1`

But the s[b].shape returns a (3,2,3) matrix.

Is there any way to fix that or i have to make a for loop for every line?

Answer

```
s[np.arange(s.shape[0])[..., None], b] = 1
```

We use some tricky advanced indexing here. The `np.arange(s.shape[0])[..., None]`

produces the column vector

```
array([[0],
[1],
[2]])
```

which broadcasts against `b`

to select, for each row of `s`

, the column indices given by the elements of that row of `b`

.

Source (Stackoverflow)

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