Khaled Khafagy - 2 months ago 8x

Python Question

I have a problem during substitution with data in array:

say,

`a = [1, 0, 0]`

b = [0, 0, 0]

c = [0, 0]

X = numpy.zeros((3, 3, 2))

and I have Matrix

`Y`

Now; I want to equal these elements of X by Y directly;

`X[tuple(numpy.where(a==0)[0]),`

tuple(numpy.where(b==0)[0]),

tuple(numpy.where(c==0)[0])] = Y

I got the error

`shape mismatch: objects cannot be broadcast to a single shape`

Answer

You could use `np.ix_`

to construct index arrays appropriate for indexing `X`

:

```
import numpy as np
np.random.seed(2016)
a=np.array([1, 0, 0])
b=np.array([0, 0, 0])
c=np.array([0, 0])
X = np.zeros((3,3,2))
Y = np.random.randint(1, 10, size=(2,3,2))
idx = np.ix_(a==0, b==0, c==0)
X[idx] = Y
print(X)
```

yields

```
array([[[ 0., 0.],
[ 0., 0.],
[ 0., 0.]],
[[ 9., 8.],
[ 3., 7.],
[ 4., 5.]],
[[ 2., 2.],
[ 3., 3.],
[ 9., 9.]]])
```

Alternatively, you could construct a boolean mask

```
mask = (a==0)[:,None,None] & (b==0)[None,:,None] & (c==0)[None,None,:]
X[mask] = Y
```

Indexing `(a=0)`

as in `(a==0)[:,None,None]`

adds new axes to the 1D boolean array `(a=0)`

. `(a==0)[:,None,None]`

has shape (3,1,1). Similarly, `(b==0)[None,:,None]`

has shape (1,3,1), and `(c==0)[None,None,:]`

has shape (1,1,2).

When combined with `&`

(bitwise-and), the three arrays are broadcasted to one common shape, (3,3,2). Thus, `X`

gets indexed by one boolean array of shape (3,3,2) in

```
X[mask] = Y
```

Source (Stackoverflow)

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