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Removing rows from a column depending on another column

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Hello,
I need a little bit of help with this problem. I will explain it the best I can then give a basic example below. Basically I have two column vectors. I am trying to split one of them up depending on the other. Basically I have:
Cp=[0;0.2;0.4;0.6;0.8;1;1.2;1.4];
Cf=[0;1;2;3;4;5;6;7;8];
I am trying to split up Cf depending on when Cp hits 1. So essentially, when Cp<=1 I am trying to obtain a column for Cf with only those values. So the resultant Cf would be:
Cf_new=[0;1;2;3;4;5]
So the only values obtained for Cf are when the corresponding row of Cp is less than or equal to 1. I hope this makes sense. Thank you

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Accepted Answer

Image Analyst
Image Analyst on 8 Sep 2013
Regarding your new question about extracting values of Cf only up to and including the index of the peak in Cp:
Cp=[0.2 0.4 0.8 0.99 0.8 0.4 0.2]
Cf=[0;1;2;3;4;5;6;7;8]
peakIndex = find(diff(Cp) < 0, 1, 'first')
Cf_new = Cf(1 : peakIndex)
In the command window:
Cp = 0.2 0.4 0.8 0.99 0.8 0.4 0.2
Cf =
0
1
2
3
4
5
6
7
8
peakIndex =
4
Cf_new =
0
1
2
3

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More Answers (2)

the cyclist
the cyclist on 8 Sep 2013
Cf_new = Cf(Cp<=1);

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Christopher
Christopher on 8 Sep 2013
Yes, but the problem is all the values are less than 0.99, so it returns the entire original Cf column. Some how I need to obtain the number of rows from the beginning of Cf until it hits the peak.
Image Analyst
Image Analyst on 8 Sep 2013
That's a different question than you first asked. How are you defining "peak"? There is a findpeaks() function in the Signal Processing Toolbox. Or else you could just call diff(Cp) and look for the first negative value - it just depends on how you want to define a "Peak."

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Andrei Bobrov
Andrei Bobrov on 8 Sep 2013
Edited: Andrei Bobrov on 8 Sep 2013
out = Cf(1:find(Cp>=.99,1,'first'));

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