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How to split data matrix conditionally?

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Andi
Andi 2022 年 4 月 14 日
コメント済み: Andi 2022 年 4 月 14 日
I have a data matrix of 59 columns and variabale number of rows, Required to extract a new matrix such that it include only those values that are in a specific bound.
As a result, we left with variable number of observations in each coloumn. How can i get new matrix, in such a condition:
An examplry random data set with my approach as below, but did not get required results.
p = rand(10, 10)
for i=1:10
q = p((p(:,ii) > .2) & (p(:,ii) < .4) , :)
end

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Chunru
Chunru 2022 年 4 月 14 日
編集済み: Chunru 2022 年 4 月 14 日
You will not get an matrix for the output since the number of entries satisfying the condition for each column will be different.
Your output can be combined as a cell array instead.
n = 50;
p = rand(n, n);
for i=1:n
pi = p(:, i);
q{i} = pi(pi > .2 & pi < .4);
end
q
q = 1×50 cell array
{13×1 double} {13×1 double} {9×1 double} {12×1 double} {11×1 double} {6×1 double} {9×1 double} {14×1 double} {8×1 double} {13×1 double} {10×1 double} {9×1 double} {9×1 double} {10×1 double} {8×1 double} {10×1 double} {17×1 double} {10×1 double} {14×1 double} {11×1 double} {9×1 double} {7×1 double} {5×1 double} {9×1 double} {8×1 double} {12×1 double} {12×1 double} {14×1 double} {9×1 double} {7×1 double} {5×1 double} {13×1 double} {9×1 double} {10×1 double} {10×1 double} {15×1 double} {9×1 double} {13×1 double} {12×1 double} {9×1 double} {11×1 double} {13×1 double} {10×1 double} {6×1 double} {12×1 double} {11×1 double} {12×1 double} {10×1 double} {11×1 double} {10×1 double}
%% counting base on q (actually you can do that on p instead)
count = cellfun(@numel, q)
count = 1×50
13 13 9 12 11 6 9 14 8 13 10 9 9 10 8 10 17 10 14 11 9 7 5 9 8 12 12 14 9 7
% number of cells with >=10 elements
nc = sum(count>=10)
nc = 31
  12 件のコメント
Andi
Andi 2022 年 4 月 14 日
For each point in data set, I have an upper and lower limit, then by using that upper and lower limits i need to search for observations in ecah coloumn of dataset 2. So technically each column should have some value or just no value. There is no other choice to give answer like NaN or etc.
Andi
Andi 2022 年 4 月 14 日
we did mistake here that why we get NaN
e{ii, kk}=a(a(:, ii)>L_lim(:,kk) & a(:,ii)<U_lim(:, kk), ii);
Thank you for help.

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