I'm trying find the index of the first element bigger than a threshold, like this:
index = 0
while timeStamps[index] < self.stopCount and index < len(timeStamps):
index += 1
Can this be done in a one-liner? I found:
index = next((x for x in timeStamps if x <= self.stopCount), 0)
I'm not sure what this expression does and it seems to return 0 always... Could somebody point out the error and explain the expression?
Another option is to use np.argmax (see this post for details). So your code would become something like
(timeStamps > self.stopCount).argmax()
the caveat is that if the condition is never satisfied the argmax will return 0.
I would do it this way:
import numpy as np
threshold = 20
sample_array = np.array([10,11,12,13,21,200,1,2])
idx = np.array([np.where(sample_array > threshold)]).min()
print(idx)
#4
this one liner will work
sample_array = np.array([10,11,12,13,21,200,1,2])
# oneliner
print(sum(np.cumsum(arr>threshold)==0))
np.cumsum(sample_array>threshold)==0) will have value 0 until element is bigger than threshold
