difference between 2 values in a vector

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A-Rod
A-Rod am 10 Jun. 2024
Kommentiert: Star Strider am 10 Jun. 2024
Hello Community.
Asking for your help again to find a solution.....
I have a vector with several data points, I'd like to get the difference between each value that is not equal to CERO.
I'm using "diff" command however this includes the difference betwee values that are equal to CERO, here is an example.
x = [0.2 0.0 0.0 0.0 0.25 0.0 0.0 0.0 0.3 0.0 0.0 0.4 0.0 0.0 0.0 0.1 0.0 0.0 0.0 0.35 ]
I tried this : b = diff(x)
b = -0.20 0.0 0.0 0.25 -0.25 0.0 0.0 0.30 -0.30 0.0 0.40 -0.40 0.0 0.0 0.10 -0.10 0.0 0.0 0.35
it gives me the difference between each data point
What I'm trying to get is the difference between each point that is not equal to cero like this:
b = 0.0 0.0 0.0 0.0 0.05 0.0 0.0 0.0 0.05 0.0 0.0 0.1 0.0 0.0 0.0 -0.3 0.0 0.0 0.0 0.25
I don't want to eliminate the ceros in between if I do so it will change the length of the vector and I need to keep same lenght to compare Vs other signals.
the number of ceros between each data value is not always the same, could be 3 ceros in between but it could be 2, 4, 6.....
as always I thank you in advance, your feedback will be higly appreciated.

Akzeptierte Antwort

Star Strider
Star Strider am 10 Jun. 2024
Try this —
x = [0.2 0.0 0.0 0.0 0.25 0.0 0.0 0.0 0.3 0.0 0.0 0.4 0.0 0.0 0.0 0.1 0.0 0.0 0.0 0.35 ]
x = 1x20
0.2000 0 0 0 0.2500 0 0 0 0.3000 0 0 0.4000 0 0 0 0.1000 0 0 0 0.3500
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Lv = x ~= 0
Lv = 1x20 logical array
1 0 0 0 1 0 0 0 1 0 0 1 0 0 0 1 0 0 0 1
idx = find(Lv);
dx = diff(x(Lv))
dx = 1x5
0.0500 0.0500 0.1000 -0.3000 0.2500
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b = zeros(size(x));
b(idx(2:end)) = dx
b = 1x20
0 0 0 0 0.0500 0 0 0 0.0500 0 0 0.1000 0 0 0 -0.3000 0 0 0 0.2500
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<mw-icon class=""></mw-icon>
b(1:10)
ans = 1x10
0 0 0 0 0.0500 0 0 0 0.0500 0
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
b(11:20)
ans = 1x10
0 0.1000 0 0 0 -0.3000 0 0 0 0.2500
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<mw-icon class=""></mw-icon>
.
  2 Kommentare
A-Rod
A-Rod am 10 Jun. 2024
it worked, thanks a lot for taking time to response and for sharing solution
Star Strider
Star Strider am 10 Jun. 2024
As always, my pleasure!

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Weitere Antworten (1)

Voss
Voss am 10 Jun. 2024
x = [0.2 0.0 0.0 0.0 0.25 0.0 0.0 0.0 0.3 0.0 0.0 0.4 0.0 0.0 0.0 0.1 0.0 0.0 0.0 0.35 ];
disp(x)
Columns 1 through 18 0.2000 0 0 0 0.2500 0 0 0 0.3000 0 0 0.4000 0 0 0 0.1000 0 0 Columns 19 through 20 0 0.3500
b = zeros(size(x));
idx = find(x ~= 0);
b(idx(2:end)) = diff(x(idx));
disp(b)
Columns 1 through 18 0 0 0 0 0.0500 0 0 0 0.0500 0 0 0.1000 0 0 0 -0.3000 0 0 Columns 19 through 20 0 0.2500
  2 Kommentare
A-Rod
A-Rod am 10 Jun. 2024
another solution, thank you for sharing it helps a lot to understand more how matlab works
Voss
Voss am 10 Jun. 2024
You're welcome!

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