I think this does what you are asking to do. The visualization for checking isn't great, but it helps you see how the windowing works. Hopefully I understood your question right and this is what you are trying to do. Anytime there is a 1 or 2 in the time series (rather than a 0) the window steps forward some fraction of the full window length (here 50%).
To examine how to do this, start with a sample time series:
import matplotlib.pylab as plt
import numpy as np
N = 5000 # time series length
# create some sort of data set to work with
x = np.zeros(N)
# add a few 1s and 2s to the list (though really they are the same for the windowing)
y = np.random.random(N)
x[y < 0.01] = 1
x[y < 0.005] = 2
# assign a window length
M = 50 # window length
overlap = 0.5 # assume 50% overlap
M_overlap = int(M * (1-overlap))
My approach is to sum the window of interest over your time series. If the sum ==0, there is no overlap between windows and if it is >0 then there is overlap. The question, then, becomes how should we calculate these sums efficiently? I compare two approaches. The first is simply to walk through the time series and the second is to use convolution (which is much faster). For the first one, I also explore different ways of assessing window size after summation.
Summation (slow version)
def window_sum1():
# start of windows in list windows
windows = [0,]
while windows[-1] + M < N:
check = sum(x[windows[-1]:windows[-1]+M]) == 0
windows.append(windows[-1] + M_overlap + (M - M_overlap) * check)
if windows[-1] + M > N:
windows.pop()
break
# plotting stuff for checking
return(windows)
Niter = 10**4
print(timeit.timeit(window_sum1, number = Niter))
# 29.201083058
So this approach went through 10,000 time series of length 5000 in about 30 seconds. But the line windows.append(windows[-1] + M_overlap + (M - M_overlap) * check) can be streamlined in an if statement.
Summation (fast version, 33% faster than slow version)
def window_sum2():
# start of windows in list windows
windows = [0,]
while windows[-1] + M < N:
check = sum(x[windows[-1]:windows[-1]+M]) == 0
if check:
windows.append(windows[-1] + M)
else:
windows.append(windows[-1] + M_overlap)
if windows[-1] + M > N:
windows.pop()
break
# plotting stuff for checking
return(windows)
print(timeit.timeit(window_sum2, number = Niter))
# 20.456240447000003
We see a 1/3 reduction in time with the if statement.
Convolution (85% faster than fast summation)
We can use signal processing to get a lot faster, by convolving the time series with the window of interest using numpy.convolve. (Disclaimer: I got the idea from the accepted answer to this question.) Of course, it also makes sense to adopt the faster window size assessment from above.
def window_conv():
a = np.convolve(x,np.ones(M,dtype=int),'valid')
windows = [0,]
while windows[-1] + M < N:
if a[windows[-1]]:
windows.append(windows[-1] + M_overlap)
else:
windows.append(windows[-1] + M)
if windows[-1] + M > N:
windows.pop()
break
return(windows)
print(timeit.timeit(window_conv, number = Niter))
#3.3695770570000008
Sliding window
The last thing I will add is that, as shown in one of the comments of this question, as of numpy 1.20 there is a function called sliding_window_view. I still have numpy 1.19 running and was not able to test it to see if it's faster than convolution.
