Completing elements of a matrix

2021 3 1
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更新时间：2021 3 5
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I have a matrix a
from a I have generated a new matrix b which have 2 new elements for 1 element of a
a = [2;3]
[m,n]=size(a)
for i=1:m
b=[1.5*(a(i)) 2*(a(i))]
end
I want to generate a new matrix 'c' such that it includes all the sets
expected results as
c = [2 3;2 4;3 4.5;3 6]

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KSSV
KSSV 2021-3-1
What is C? How you got it?
Karanvir singh Sohal
编辑：Karanvir singh Sohal 2021-3-1
I want to generate c matrix. For a(1,1) i have 2 values in b i.e. b=[1.5*2 2*2] Which gives b=[3 4] And similarly for a(2,1) I'll have b=[4.5 6]
I want new matrix c
C=[a(1) 1.5*a(1); a(1) 2*a(1); a(2) 1.5*a(2); a(2) 2*a(2)]

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Well I got what i was trying to do :)
Thanks @Jan for your time and help
L=4
X_Range=[L*1000/16 L*1000/10];
Shut= [152.4; 228.6; 304.8; 381.0; 457.2; 533.4; 609.6; 685.8; 762.0; 838.2; 914.4; 990.6; 1066.8];
X = Shut(Shut>=X_Range(1,1) & Shut<= X_Range(1,2));
[m,n]=size(X);
index = 0;
for i=1:m
Y_Range =[1.5*X(i) 2*X(i)];
Y = Shut(Shut>=Y_Range(1,1) & Shut<= Y_Range(1,2));
[mY,nY]=size(Y);
for j=1:mY
index=index+1;
XD(index)=X(i);
YD(index)=Y(j);
end
end
D= [reshape(XD,[],1) [reshape(YD,[],1)]]

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Jan
Jan 2021-3-3
If this works, here is a simplified version:
L = 4;
X_Range = [L*1000/16 L*1000/10];
Shut = [152.4; 228.6; 304.8; 381.0; 457.2; 533.4; 609.6; ...
685.8; 762.0; 838.2; 914.4; 990.6; 1066.8];
X = Shut(Shut>=X_Range(1,1) & Shut<= X_Range(1,2));
XD = [];
YD = [];
for i = 1:numel(X)
Y = Shut(Shut >= 1.5*X(i) & Shut <= 2*X(i));
mY = numel(Y);
XD(end + 1:end + mY) = X(i);
YD(end + 1:end + mY) = Y;
end
D = [XD(:), YD(:)];
Thanks this is a really simple version.
can you please explain me this part:
XD(end + 1:end + mY) = X(i);
YD(end + 1:end + mY) = Y;
Jan
Jan 2021-3-5
These lines do the same as your loop:
for j=1:mY
index=index+1;
XD(index)=X(i);
YD(index)=Y(j);
end
but more efficiently.

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更多回答（1 个）

Jan
Jan 2021-3-1

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a = [2; 3];
b = [1.5; 2];
c = kron(a, [ones(size(a,1), 1), b])

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This won't work if i have more number of rows in a matrix as compared to b matrix
Jan
Jan 2021-3-1
编辑：Jan 2021-3-2
My code does solve the problem, which you have asked for in your question.
It would be useful, if you explain, what the actual problem is. Posting a specific problem, when you want to solve a general one wastes the time of the persons, who want to help you to solve your problem.
"Wont work" does not allow to understand, what you want to get as result instead.
Maybe you want:
c = kron(a, [ones(size(a)), b])
% or
c = [a, a(:) * b(:).']
I do understand that I need to post general problem rather than be specific. Sorry for that
Here I'm attaching my code
L=4
X_Range=[L*1000/16 L*1000/10];
Shut= [152.4; 228.6; 304.8; 381.0; 457.2; 533.4; 609.6; 685.8; 762.0; 838.2; 914.4; 990.6; 1066.8]
X = Shut(Shut>=X_Range(1,1) & Shut<= X_Range(1,2))
[m,n]=size(X)
Shut(:,1);
for i=1:m
Y_Range =[1.5*X(i) 2*X(i)];
Y = Shut(Shut>=Y_Range(1,1) & Shut<= Y_Range(1,2));
c = kron(X, [ones(size(X,1), 1), Y])
endfor
Desired output:
c = [304.80 533.40; 304.80 609.60; 381.00 609.60; 381.00 685.80; 381.00 762.00]
When I used
c = kron(X, [ones(size(X,1), 1), Y])
elements of 2nd column are not as per required(Numerical value changed).
While using
c = kron(XDim, [ones(size(XDim), YDim)])
following erorr is recived
error: conversion of 533.4 to int64_t value failed
if we use
c = [XDim, XDim(:) * YDim(:).']
error: horizontal dimensions mismatch (2x1 vs 3x1)
and also values of Y in c matrix are changed.
I used for loop to obtain the desired results
L=4
X_Range=[L*1000/16 L*1000/10];
Shut= [152.4; 228.6; 304.8; 381.0; 457.2; 533.4; 609.6; 685.8; 762.0; 838.2; 914.4; 990.6; 1066.8];
X = Shut(Shut>=X_Range(1,1) & Shut<= X_Range(1,2));
[m,n]=size(X);
Shut(:,1);
for i=1:m
Y_Range =[1.5*X(i) 2*X(i)];
Y = Shut(Shut>=Y_Range(1,1) & Shut<= Y_Range(1,2));
[mY,nY]=size(Y);
for j=1:mY
c(j,1)=X(i);
endfor
c(:,2)=Y
endfor
This gives me the correct matrix but, I want to combine each c matrix.
when loop runs for first time it gives me c matrix of 2x2 and on second run i have 3x2
I want to combine c matrixs to get 5x2 in above example
Jan
Jan 2021-3-2
I do not get an idea of what you want to achieve.

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