Keeping the indexed matrix the same shape as its index

Why does f(index) get transposed exclusively when f contains three variables?

f=[1 2 3 4];		
n=numel(f)-1;		
p=(1:n)';		
q=(1:n-1);		
index=(floor((p+n.*(q-1)-1)./(n-1))+1)		

index = 3×2

1 2 1 3 2 3

●

f(index)		

ans = 3×2

1 2 1 3 2 3

	

f=[1 2 3 ];		
n=numel(f)-1;		
p=(1:n)';		
q=(1:n-1);		
index=(floor((p+n.*(q-1)-1)./(n-1))+1)		

index = 2×1

1 2

●

f(index)		

ans = 1×2

1 2

●

Answer 1

[Matt J]

Why does f(index) get transposed exclusively when f contains three variables?

It is not exclusive to f having three elements. It is related to the fact that f and index are both vectors, in which case the orientation of f decides the shape,

f=[1,2,3,4,5,6];			

f([1,2,5,6])

ans = 1×4

1 2 5 6

●

f([1,2,5,6].')

ans = 1×4

1 2 5 6

●

f([1,2;5 6])		

ans = 2×2

1 2 5 6

●

[Matt J]

In other words, if you want the result to consistently have the shape of index, you must do,

reshape(f(index),size(index))