How to store values in matrix form for differn iteration

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I am having 7 decimal input data. This data varies for 500 iteration. Now, I need to store the 7 input data obtained in each iteration in a matrix form of 500*7. Thank you in advance.
For example:
A= [6 3 4 5 2 7 1]
Expected output:
[1 3 7 5 6 4 2 % iteration 1
2 4 7 6 5 3 1 % iteration 2
.
.
.
.
7 4 5 1 6 3 2 %iteration 500]
  4 Comments
Dyuman Joshi
Dyuman Joshi on 29 Jan 2023
"Actually my requirement is, "
Then mention it clearly in the question. What your requirement is quite different from what your question asks.
"Now, I need to split this array into 500*7 matrix"
How do you want to split the array?
ASHA PON
ASHA PON on 29 Jan 2023
I am already having array (3500 element) with me. Just, I want to convert it to a matrix of 500*7. That is first 7 element from my array has to be in first row of new matrix. The eigth to fourteenth element from my array has to come in second row of matrix. Then, fifteenth to twenty first elemnt from my array has to come in third row of matrix. Like wise, i will have 500 rows and 7 columns in my final matrix

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Accepted Answer

Dyuman Joshi
Dyuman Joshi on 29 Jan 2023
Use reshape()
%in case of row array
x=1:3500;
y=reshape(x,7,500)'
y = 500×7
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70
%in case of column array
a=(1:3500)';
b=reshape(a,7,500)'
b = 500×7
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70

More Answers (1)

Jan
Jan on 29 Jan 2023
Edited: Jan on 29 Jan 2023
A = [6 3 4 5 2 7 1];
Collected = zeros(500, 7);
for k = 1:500
A = rem(A + randi([0, 100], 1, 7), 10); % A random test function
Collected(k, :) = A;
end
" I am having an array stored with 3500 elemnts. Now, I need to split this array into 500*7 matrix.":
X = rand(1, 3500);
Y = reshape(X, 500, 7);
% Or:
Y = reshape(X, 7, 500).';

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