Generating sequences from data

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Mahendran Subramanian
Mahendran Subramanian am 25 Sep. 2019
Hi All,
I have a data set (x1, x2, x3, x4, x5, x6, x7, .... xn) from which I want to generate sequences like
x1, x2, x3, x4, x5
x2, x3, x4, x5, x6
x3, x4, x5, x6, x7
x4, x5, x6, x7, x8
..................... xn
Thank you

Akzeptierte Antwort

Andrei Bobrov
Andrei Bobrov am 26 Sep. 2019
xy = [5 14
6 16
1 16
6 11
4 16
1 16
2 13];
n = 5;
[m,k] = size(xy);
out = xy(reshape(hankel(1:n,n:m),1,n,[]) + m*(0:k-1)')
  5 Kommentare
Andrei Bobrov
Andrei Bobrov am 26 Sep. 2019
XY = readtable('XY.xlsx');
xy = XY{:,:};
n = 5;
[m,k] = size(xy);
out = xy(reshape(hankel(1:n,n:m),1,n,[]) + m*(0:k-1)');
Mahendran Subramanian
Mahendran Subramanian am 26 Sep. 2019
works thank you

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Weitere Antworten (2)

John D'Errico
John D'Errico am 25 Sep. 2019
So, given a vector x, of length n, you want to create the array with rows that are the sub-sequences of length 5? The result will be a (n-4) x 5 array.
A trivial solution would just concatenate columns to create the array. However, that would not be easily fixed if you then wnted to create sub-sequences of length 6 or 4.
So far better is to create an index array, then use that to index into the vector.
n = length(x);
m = 5;
ind = (1:n-m+1)' + (0:m-1);
A = x(ind);
This works for any length vector, and any size of sub-sequences.
It does use a feature of MATLAB that was introduced in R2016b, to create the index array ind. Earlier releases might use this instead:
ind = bsxfun(@plus,(1:n-m+1)',0:m-1);
  4 Kommentare
Mahendran Subramanian
Mahendran Subramanian am 26 Sep. 2019
I have already done that. I need the data as separate sequences for running through various analyses. And I am looking into it.
Stephen23
Stephen23 am 26 Sep. 2019
"I need the data as separate sequences for running through various analyses."
Then you can easily use an ND array or a cell array, exactly as John D'Errico wrote.

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Dwarka Sahu
Dwarka Sahu am 25 Sep. 2019
for i=1:5
sprintf('%g, %g, %g, %g, %g',i, i+1, i+2, i+3, i+4)
end
  3 Kommentare
Dwarka Sahu
Dwarka Sahu am 26 Sep. 2019
For the first sequence of (nx2)
for i=1:5
sprintf('x%g, y%g', i, i)
end
For the next set of sequence of (2x5)
for i=1:5
sprintf('x%g, x%g, x%g, x%g, x%g', i, i+1,i+2,i+3,i+4)
sprintf('y%g, y%g, y%g, y%g, y%g', i, i+1,i+2,i+3,i+4)
end
Mahendran Subramanian
Mahendran Subramanian am 26 Sep. 2019
RelativeX = 2500 x 1 double
RelativeY = 2500 x 1 double
or
XY = 2500 x 2 double
Thank you for the answer, however, Iam getting the following
>> for i=1:5
sprintf('RelativeX%g, RelativeY%g', i, i)
end
for i=1:5
sprintf('RelativeX%g, RelativeX%g, RelativeX%g, RelativeX%g, RelativeX%g', i, i+1,i+2,i+3,i+4)
sprintf('RelativeY%g, RelativeY%g, RelativeY%g, RelativeY%g, RelativeY%g', i, i+1,i+2,i+3,i+4)
end
ans =
'RelativeX1, RelativeY1'
ans =
'RelativeX2, RelativeY2'
ans =
'RelativeX3, RelativeY3'
ans =
'RelativeX4, RelativeY4'
ans =
'RelativeX5, RelativeY5'
ans =
'RelativeX1, RelativeX2, RelativeX3, RelativeX4, RelativeX5'
ans =
'RelativeY1, RelativeY2, RelativeY3, RelativeY4, RelativeY5'
ans =
'RelativeX2, RelativeX3, RelativeX4, RelativeX5, RelativeX6'
ans =
'RelativeY2, RelativeY3, RelativeY4, RelativeY5, RelativeY6'
ans =
'RelativeX3, RelativeX4, RelativeX5, RelativeX6, RelativeX7'
ans =
'RelativeY3, RelativeY4, RelativeY5, RelativeY6, RelativeY7'
ans =
'RelativeX4, RelativeX5, RelativeX6, RelativeX7, RelativeX8'
ans =
'RelativeY4, RelativeY5, RelativeY6, RelativeY7, RelativeY8'
ans =
'RelativeX5, RelativeX6, RelativeX7, RelativeX8, RelativeX9'
ans =
'RelativeY5, RelativeY6, RelativeY7, RelativeY8, RelativeY9'

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