### Question :

## Original question

I am getting a very odd error message when I try to assign some of the elements of an array. I am using a combination of a slice and a set of indices. See the following simple example.

```
import scipy as sp
a = sp.zeros((3, 4, 5))
b = sp.ones((4, 5))
I = sp.array([0, 1, 3])
b[:, I] = a[0, :, I]
```

This code raises the following `ValueError`

:

ValueError: shape mismatch: value array of shape (3,4) could not be broadcast to indexing result of shape (3,4)

—

### Follow up

Be careful when using a combination of a slice and seq. of integers. As pointed out on github:

```
x = rand(3, 5, 7)
print(x[0, :, [0,1]].shape)
# (2, 5)
print(x[0][:, [0,1]].shape)
# (5, 2)
```

This is how numpy is designed to work, but it is nevertheless a bit confusing that x[0][:, I] is not the same as x[0, :, I]. Since this is the behavior I want I choose to use x[0][:, I] in my code.

##
Answer #1:

Looks like there are some errors in copying your code to question.

But I suspect there’s a known problem with indexing:

```
In [73]: a=np.zeros((2,3,4)); b=np.ones((3,4)); I=np.array([0,1])
```

Make `I`

2 elements. Indexing `b`

gives the expected (3,2) shape. 3 rows from the slice, 2 columns from `I`

indexing

```
In [74]: b[:,I].shape
Out[74]: (3, 2)
```

But with 3d `a`

we get the transpose.

```
In [75]: a[0,:,I].shape
Out[75]: (2, 3)
```

and assignment would produce an error

```
In [76]: b[:,I]=a[0,:,I]
...
ValueError: array is not broadcastable to correct shape
```

It’s putting the 2 element dimension defined by `I`

first, and the 3 element from `:`

second. It’s a case of mixed advanced indexing that has been discussed earlier – and there’s a bug issue as well. (I’ll have to look those up).

You are probably using a newer `numpy`

(or `scipy`

) and getting a different error message.

It’s documented that indexing with two arrays or lists, and slice in the middle, puts the slice at the end, e.g.

```
In [86]: a[[[0],[0],[1],[1]],:,[0,1]].shape
Out[86]: (4, 2, 3)
```

The same thing is happening with `a[0,:,[0,1]]`

. But there’s a good argument that it shouldn’t be this way.

As to a fix, you could transpose a value, or change the indexing

```
In [88]: b[:,I]=a[0:1,:,I]
In [90]: b[:,I]=a[0,:,I].T
In [91]: b
Out[91]:
array([[ 0., 0., 1., 1.],
[ 0., 0., 1., 1.],
[ 0., 0., 1., 1.]])
In [92]: b[:,I]=a[0][:,I]
```

##
Answer #2:

First of all it looks like you’re missing a comma on the line 6:

```
I = sp.array([0,1,4])
```

Secondly, I would expect the value 4 in the array I to raise an IndexError, since both a and b have a max dimension of 4. I suspect you might want:

```
I = sp.array([0,1,3])
```

Making these changes run the program for me, and I got b as:

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

Which I suspect is what you want.

##
Answer #3:

Here I get this error with indices `[0,1,4]`

:

```
IndexError: index 4 is out of bounds for axis 2 with size 4
```

Its suggesting the value `4`

is being used as an index, while the SIZE 4 implies the max index would be 3.

EDIT: now that you changed it to `[0, 1, 3]`

, it’s running fine here.

EDIT: with your current code, I get the same error, but when I print the arrays themselves, they have a transverse shape:

```
print b[:, I]
print a[0, :, I]
[[ 1. 1. 1.]
[ 1. 1. 1.]
[ 1. 1. 1.]
[ 1. 1. 1.]]
[[ 0. 0. 0. 0.]
[ 0. 0. 0. 0.]
[ 0. 0. 0. 0.]]
```