How to select certain columns of a matrix only when the values in certain rows showing the the double of the value in a particular row of a column?

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Kim Arnold
Kim Arnold 2020-8-14
编辑： Shae Morgan 2020-8-14
Hi everybody!
I have a Matrix with 5 rows and 2300 columns M(5,2300).
0.2 0.3 0.4 2 ...
3 0.5 1.9 2.5 ...
2 0.7 0.2 3 ...
1 1 1 1 ...
0.5 0.1 1.5 3.1 ...
looks like above. The numbers you see are ratios because all the values of a particular column were divided by the value in the 4th row (which is now 1). I intend to just keep the columns for future analysis when at least one value of the other rows in a column (row1:3 and row5 values) are >=2, meaning that they should be at least the double of the value in the 4th row. As an output from the columns above it should give me an index like [1,0,0,1].
Could you probably suggest a nice solution? I'm thankful for every help.

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采纳的回答

KSSV
KSSV 2020-8-14
Let A be your 5*n matrix.
% GEt ratio
R = A(1:3,:)./A(5,:) ;
idx = R>=2 ; % get indices where ratio is >= 2
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更多回答（2 个）

Bruno Luong
Bruno Luong 2020-8-14
编辑：Bruno Luong 2020-8-14
Mkeep = M(:,any(M>=2,1))
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Kim Arnold
Kim Arnold 2020-8-14
Dear Bruno, yes the site was not loading anymore.
Yeah with your code i get the columns with their values but I just wanted to have it with idx 1,0 ...
anyways thanks as well for your help!

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Shae Morgan
Shae Morgan 2020-8-14
编辑：Shae Morgan 2020-8-14
do this in two steps: 1) find the indices where the values > 2 and then 2) sum the columns - if all rows 1-3 and 5 are less than 2, then this sum will be 0. If one of them has a value >=2, then the sum will be >0.
idx= M>=2;
out= sum(idx)>0
out will give you an index of 0's and 1's for each column. if 1, then it's something you should analyze later (i.e., has a value 2x row 4)
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Kim Arnold
Kim Arnold 2020-8-14
Dear Shae,
thanks, is what KSSV also recommended to do. Just was looking for like one line of code :).
Thanks as well for your input.
Shae Morgan
Shae Morgan 2020-8-14
编辑：Shae Morgan 2020-8-14
single line answer by combining the variables :) Thanks for the consideration! Should work for KSSV's answer too:
out= sum(M>=2)>0 %my implementation, single line
id = find(sum(M>=2)) ; %KSSV's solution, single line

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