Removing rows from a column depending on another column

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Christopher
Christopher 2013년 9월 8일
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

채택된 답변

Image Analyst
Image Analyst 2013년 9월 8일
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

추가 답변 (2개)

the cyclist
the cyclist 2013년 9월 8일
Cf_new = Cf(Cp<=1);
  댓글 수: 4
Christopher
Christopher 2013년 9월 8일
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 2013년 9월 8일
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 2013년 9월 8일
편집: Andrei Bobrov 2013년 9월 8일
out = Cf(1:find(Cp>=.99,1,'first'));

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