How can I rearrange this matrix into new one?
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HyoJae Lee
am 25 Jan. 2023
Kommentiert: HyoJae Lee
am 25 Jan. 2023
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.
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Akzeptierte Antwort
Dyuman Joshi
am 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 Kommentare
Askic V
am 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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