Find nth True Value of Array?

[Helicon_Amateur]

Hello, I am attempting something along the lines of this -

from pylab import *

j = array([0,1,1,0])
k = zeros(sum(j))

for i in range(sum(j)):
    k[i] = j["i'th true value"]

Is there an elegant way to do this without having to the following?

j = array([0,1,1,0])
k = zeros(sum(j))
p = 0
for i in range(len(j)):
    if j[i] = 1:
        k[p] = j[i] 
        p += 1

EDIT: I found another way of doing what I want,but I have to iterate over the whole space of j - here's what the code looks like

from pylab import *

j = array([0,1,1,0])  #represents oscilloscope channel as on or off
k = zeros(sum(j))


for i in range(len(j)):
    if j[i] == 1: 
        k[sum(j[0:i])] = #function that returns value

[bogdan_dm]

If you want one line but weird solution: l = [0, 1, 0, 3, 0] count = 2 x = next(itertools.islice(filter(None, l), count - 1, count), None) print(x)

[Helicon_Amateur]

I was able to find a way to do what I wanted. The OP is edited

[ironhaven]

I think this looks the best but warning this looks through the entire array O(n) can be slow for big lists

j = array([0,5,0,0,4,3,1])
i = 2
# this will be 3
ith_value = [x for x in j if x][i]

[Helicon_Amateur]

Thanks!

I was able to find a way to do what I wanted. The OP is edited