python, numpy boolean array: negation in where statement

Question

with:

import numpy as np
array = get_array()

I need to do the following thing:

for i in range(len(array)):
    if random.uniform(0, 1) < prob:
        array[i] = not array[i]

with array being a numpy.array.

I wish I could do something similar to:

array = np.where(np.random.rand(len(array)) < prob, not array, array)

but I obtain the following result (referring to 'not array'):

The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()

Why can I take the value of array but not its negation?

Currently I solved with:

array = np.where(np.random.rand(len(array)) < prob, - array + 1, array)

but it looks really clumsy to me.

Thank you for your help

p.s.: I don't care if the statement modifies array or not. I just need the result of the operation.

just another question: I want to do this change for 2 reasons: readability and efficiency. Is there a real performance improvement with it? Thank you again

Answer 1

I suggest using

array ^= numpy.random.rand(len(array)) < prob

This is probably the most efficient way of getting the desired result. It will modify the array in place, using "xor" to invert the entries which the random condition evaluates to True for.

Why can I take the value of array but not its negation?

You can't take the truth value of the array either:

>>> bool(array)
ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()

The not operator implicitly tries to convert its operand to bool, and then returns the opposite truth value. It is not possible to overload not to perform any other behaviour. To negate a NumPy array of bools, you can use

~array

or

numpy.logical_not(array)

or

numpy.invert(array)

though.

Answer 2

putmask is very efficient if you want to replace selected elements:

import numpy as np

np.putmask(array, numpy.random.rand(array.shape) < prob, np.logical_not(array))