I feel like this is a really simple question but I can't find the solution.
Given a boolean array of true/false values, I need the output of all the indices with the value "false". I have a way to do this for true:
test = [ True False True True]
test1 = np.where(test)[0]
This returns [0,2,3], in other words the corresponding index for each true value. Now I need to just get the same thing for the false, where the output would be [1]. Anyone know how?
Use np.where(~test) instead of np.where(test).
With enumerate:
>>> test = [True, False, True, True]
>>> [i for i, b in enumerate(test) if b]
[0, 2, 3]
>>> [i for i, b in enumerate(test) if not b]
[1]
itertools.groupby to do both in one shot, but then the type conversion gets a bit messier
numpy, so this definitely shouldn't be usedUsing explicite np.array().
import numpy as np
test = [True, False, True, True]
a = np.array(test)
test1 = np.where(a==False)[0] #np.where(test)[0]
print(test1)
