how can i obtain this?

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RADWAN A F ZEYADI
RADWAN A F ZEYADI 2021 年 11 月 8 日
編集済み: Cris LaPierre 2021 年 11 月 8 日
HI
ihave data with dimension 153*1 and i want to perform for loop in order to obtain 153*71 where 71 is position along horiziontal axis but i tried to perform it and i got 153*71 but duplicated rows and columns and this is doesnt make sense so how is it linspace function useful for this?
  3 件のコメント
RADWAN A F ZEYADI
RADWAN A F ZEYADI 2021 年 11 月 8 日
i have obtained from % d_pred = reshape(d_pred(:,1,:),[],1,1);%convert data from 3d to column vector
RADWAN A F ZEYADI
RADWAN A F ZEYADI 2021 年 11 月 8 日
it was 51 71 3

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回答 (2 件)

Cris LaPierre
Cris LaPierre 2021 年 11 月 8 日
編集済み: Cris LaPierre 2021 年 11 月 8 日
You have not shared your code so it's a little challenging to say what is happening, but if I were to start with a 51x71x3 matrix, and wanted to change it to 153x71, where each column is all the data in the indicated column across all 'sheets', I'd use permute then reshape.
d_pred = rand(51,71,3);
D_pred = permute(d_pred,[1 3 2]);
D_Pred = reshape(D_pred,153,[]);
size(D_Pred)
ans = 1×2
153 71

DGM
DGM 2021 年 11 月 8 日
編集済み: DGM 2021 年 11 月 8 日
I'm assuming that you want all 71 columns to be extracted in the same fashion. Consider the example:
% 5x5x3 array
A = cat(3,reshape(1:25,5,5),reshape(101:125,5,5),reshape(201:225,5,5))
A =
A(:,:,1) = 1 6 11 16 21 2 7 12 17 22 3 8 13 18 23 4 9 14 19 24 5 10 15 20 25 A(:,:,2) = 101 106 111 116 121 102 107 112 117 122 103 108 113 118 123 104 109 114 119 124 105 110 115 120 125 A(:,:,3) = 201 206 211 216 221 202 207 212 217 222 203 208 213 218 223 204 209 214 219 224 205 210 215 220 225
B = reshape(permute(A,[1 3 2]),[],size(A,2),1) % 15x5 array
B = 15×5
1 6 11 16 21 2 7 12 17 22 3 8 13 18 23 4 9 14 19 24 5 10 15 20 25 101 106 111 116 121 102 107 112 117 122 103 108 113 118 123 104 109 114 119 124 105 110 115 120 125

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