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How to get column values using indexes

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Stephen john
Stephen john 2022-5-13
编辑： Voss 2022-5-13
Hello everyone, I hope you are doing well.
I have the following data, i want to seperate the four column based on the fifth column
for example if the fifth column has value 1 it the first four column values store in other matrix.
Similarly for fifth column value = 2 the first four column related to values=2 store in other matrix.

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回答（2 个）

KSSV
KSSV 2022-5-13
编辑：KSSV 2022-5-13
LEt A be your matrix.
C5 = A(:,5); % 5th column
M1 = A(C5==1,1:4) ;
M2 = A(C5==2,1:4) ;
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Stephen john
Stephen john 2022-5-13
@KSSV Using a Single line you manully type
M1 = A(C5==1,1:4) ;
M2 = A(C5==2,1:4) ;
M3=A(C5==3,1:4) ;
Stephen23
Stephen23 2022-5-13
" i want it to be automatic"
Numbered variable names are a sign that you are doing something wrong.
https://www.mathworks.com/matlabcentral/answers/304528-tutorial-why-variables-should-not-be-named-dynamically-eval
A container array (e.g. a cell array) would probably be the best approach.

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Voss
Voss 2022-5-13
Ran in:
S = load('matlab.mat');
A = S.Dataset23;
result = splitapply(@(x){x(:,1:4)},A,findgroups(A(:,5)))
result = 5×1 cell array
{ 96×4 double} {303×4 double} { 83×4 double} { 87×4 double} { 17×4 double}
  2 个评论
Stephen john
Stephen john 2022-5-13
@_ Now you created a cell thats great, Then how can i save this as a matrix?
Voss
Voss 2022-5-13
编辑：Voss 2022-5-13
Ran in:
The cell array contains the matrices you need. There is no reason to store them as separate matrix variables. Just access them from the cell array as needed.
S = load('matlab.mat');
A = S.Dataset23;
result = splitapply(@(x){x(:,1:4)},A,findgroups(A(:,5)))
result = 5×1 cell array
{ 96×4 double} {303×4 double} { 83×4 double} { 87×4 double} { 17×4 double}
result{1} % get the 1st matrix
ans = 96×4
1.0e+09 * 0.0000 2.6001 -0.0000 0.0000 0.0000 2.6001 -0.0000 0.0000 0.0000 2.6001 -0.0000 0.0000 0.0000 2.5997 -0.0000 0.0000 0.0000 2.5998 -0.0000 0.0000 0.0000 2.6002 -0.0000 0.0000 0.0000 2.6003 -0.0000 0.0000 0.0000 2.6004 -0.0000 0.0000 0.0000 2.6005 -0.0000 0.0000 0.0000 2.5998 -0.0000 0.0000
result{5} % get the 5th matrix
ans = 17×4
1.0e+09 * 0.0000 2.6037 -0.0000 0.0000 0.0000 2.6042 -0.0000 0.0000 0.0000 2.6044 -0.0000 0.0000 0.0000 2.6038 -0.0000 0.0000 0.0000 2.6040 -0.0000 0.0000 0.0000 2.6043 -0.0000 0.0000 0.0000 2.6039 -0.0000 0.0000 0.0000 2.6036 -0.0000 0.0000 0.0000 2.6038 -0.0000 0.0000 0.0000 2.6045 -0.0000 0.0000
3*result{5} % get the 5th matrix and multiply it by 3 (or do whatever you need to do with it)
ans = 17×4
1.0e+09 * 0.0000 7.8111 -0.0000 0.0000 0.0000 7.8126 -0.0000 0.0000 0.0000 7.8131 -0.0000 0.0000 0.0000 7.8114 -0.0000 0.0000 0.0000 7.8121 -0.0000 0.0000 0.0000 7.8129 -0.0000 0.0000 0.0000 7.8117 -0.0000 0.0000 0.0000 7.8109 -0.0000 0.0000 0.0000 7.8114 -0.0000 0.0000 0.0000 7.8134 -0.0000 0.0000

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