Swapping columns in a numpy array?

Question

``` from numpy import * def swap_columns(my_array, col1, col2): temp = my_array[:,col1] my_array[:,col1] = my_array[:,col2] my_array[:,col2] = temp ``` Then ``` swap_columns(data, 0, 1) ``` Doesn't work. However, calling the code directly ``` temp = my_array[:,0] my_array[:,0] = my_array[:,1] my_array[:,1] = temp ``` Does. Why is this happening and how can I fix it? The Error says "IndexError: 0-d arrays can only use a single () or a list of newaxes (and a single ...) as an index", which implies the arguments aren't ints? I already tried converting the cols to int but that didn't solve it.

Accepted Answer

There are two issues here. The first is that the `data` you pass to your function apparently isn't a two-dimensional NumPy array -- at least this is what the error message says.

The second issue is that the code does not do what you expect:

```
my_array = numpy.arange(9).reshape(3, 3)
# array([[0, 1, 2],
#        [3, 4, 5],
#        [6, 7, 8]])
temp = my_array[:, 0]
my_array[:, 0] = my_array[:, 1]
my_array[:, 1] = temp
# array([[1, 1, 2],
#        [4, 4, 5],
#        [7, 7, 8]])
```

The problem is that Numpy basic slicing does not create copies of the actual data, but rather a view to the same data. To make this work, you either have to copy explicitly

```
temp = numpy.copy(my_array[:, 0])
my_array[:, 0] = my_array[:, 1]
my_array[:, 1] = temp
```

or use advanced slicing

```
my_array[:, [0, 1]] = my_array[:, [1, 0]]
```

Swapping columns in a numpy array?

Solution

Problem