How to store values in matrix form for differn iteration

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ASHA PON
ASHA PON 2023 年 1 月 29 日
コメント済み: Stephen23 2023 年 1 月 29 日
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 件のコメント
Dyuman Joshi
Dyuman Joshi 2023 年 1 月 29 日
"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 2023 年 1 月 29 日
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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採用された回答

Dyuman Joshi
Dyuman Joshi 2023 年 1 月 29 日
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
  1 件のコメント
Stephen23
Stephen23 2023 年 1 月 29 日
To make the intent clear and avoid bugs, it is best to use actual transpose:
unless the complex conjugate transpose is specifically required:

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その他の回答 (1 件)

Jan
Jan 2023 年 1 月 29 日
編集済み: Jan 2023 年 1 月 29 日
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).';
  1 件のコメント
ASHA PON
ASHA PON 2023 年 1 月 29 日
Thank you. This is what I needed.

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