How can I rearrange this matrix into new one?
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data=[50 67.5;
50 67.9;
60 75.4
60 75.9;
70 12.3;
70 12.9];
Hello, this is a data what I want to rearrage into new one.
First column of the data is time, and second is value.
This matrix is sepearated every two rows, first is starting value and second is end value.
I want to rearrage into
- fix the time
- expand the value (starting value to end value every 0.1)
Finally, I want to make this kind of data,
data=[50 67.5;
50 67.6;
50 67.7;
50 67.8;
50 67.9;
60 75.4;
60 75.5;
60 75.6;
60 75.7;
60 75.8;
60 75.9;
70 12.3;
70 12.4;
70 12.5;
70 12.6;
70 12.7;
70 12.8;
70 12.9];
How can I make like this using loop?
Best,
HyoJae.
0 Comments
Accepted Answer
Dyuman Joshi
on 25 Jan 2023
Assuming data is homogenous (In pairs throught-out), and end value is greater than or equal to first value.
data=[50 67.5;
50 67.9;
60 75.4
60 75.9;
70 12.3;
70 12.9];
row=size(data,1)/2;
%using cell array to store generated arrays
out=cell(row,1);
for i=1:row
y=(data(2*i-1,2):0.1:data(2*i,2))';
out{i}=[repelem(data(2*i-1,1),numel(y),1) y];
end
cell2mat(out)
2 Comments
Askic V
on 25 Jan 2023
This is another approach:
data=[50 67.5;
50 67.9;
60 75.4
60 75.9;
70 12.3;
70 12.9];
step = 0.1;
time_col = unique(data(:,1));
data_new = [];
for i = 1:numel(time_col)
ind_i = find(data(:,1)==time_col(i));
sec_column = data(ind_i(1),2):step:data(ind_i(end),2);
first_column = time_col(i)*ones(size(sec_column));
data_new = [data_new; [first_column', sec_column']];
end
data_new
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