Replacing elements of a logical index vector

“Idx” is a logical index vector when a certain “power on” condition is true (e.g. idx=Pwr_on>threshold) Say the result is 5 "zeros" and 5 "ones" per cycle i.e. idx=[0;0;0;0;0;1;1;1;1;1;0;0;0;0;0;1;1;1;1;1;0;0;0;0;0.........] Every time idx is true; I need to replace the first element and last element with “0” within the index. So effectively idx=[0;0;0;0;0;0;1;1;1;0;0;0;0;0;0;0;1;1;1;0;0;0;0;0;0.........] Any suggestions

[Adam]

How are you defining "first element and last element...within the index" for a general case? It is clear in this example, but is this as complicated as your data gets? i.e. there are always n 0's followed by n 1's and then this just cycles every 2n elements?

Answer 1

[Image Analyst]

This will do it. It's explicit so you can see what's going on. Of course, you could collapse it all down into a single line of code (though that may be hard to follow).

idx=[0;0;0;0;0;1;1;1;1;1;0;0;0;0;0;1;1;1;1;1;0;0;0;0;0]
di = [0; diff(idx)]
leadingEdges = find(di>0)
trailingEdges = find(di<0)-1
idx(leadingEdges) = 0;
idx(trailingEdges) = 0

Answer 2

[Kashmir Singh]

There's always zero's followed by one's. However the power duty cycle varies i.e. it's not always 5 zero's followed by 5 one's. This can be varied by increasing/decreasing the duty cycle.

Answer 3

[Jos (10584)]

X = logical([0 1 1 0 1 1 1 0 0 1 1 1 1 0 0 0 1 1 1 1 1])
Y = fliplr(conv(fliplr(conv(double(X),[1 1],'same')),[1 1],'same'))>3