numpy swap multiple elements in an array

Question

I have a numpy array containing a random spread of 1s and 0s. I want to replace all the 1s with a 0 and all the zeros with a 1.

arr[arr == 0] = 2
arr[arr == 1] = 0
arr[arr == 2] = 1

I'm currently having to use a temporary value (2 in this case) in order to avoid all the 0s becoming 1s and then subsequently making the entire array full of 0s. Is there a more elegant/efficient way to do this?

Answer 1

You can calculate and store your Boolean indexers before overwriting any values:

ones = a == 1
zeros = a == 0

a[ones] = 0
a[zeros] = 1

The solution also works if you have values other than 0 and 1.

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If you don't need an in place solution, you can use np.where:

a = np.where(ones, 0, np.where(zeros, 1, a))

Answer 2

Here's a solution that's very specific to your problem, but should also be very fast. Given the array:

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

You can subtract 1 from all the values and multiply by negative 1:

>>> (a-1)*-1
array([[0, 1, 1, 0],
       [0, 0, 0, 1]])