Help with matrix indexing.
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C is a matrix with i rows and 5 columns. i=1:16
I am interested in finding the value of C(i-1,:).
Using the following matlab code, I try to direct matrix c row i-1 any column to c row 16 any column when i equals 1.
if i==1
c(i-1,:)=c(16,:);
end
Error: Subscript indices must either be real positive integers or logicals.
Problem: It gives me above error message as it seems always recognize the c(i-1,:)=c(0,:) first and therefore ignore the statement after that which is equalling to the c(16,:).
Any idea how to solve it?
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the cyclist
am 26 Jan. 2011
You seem to be treating this in a sorta symbolic way trying to tell MATLAB that the c(i-1) element is really element c(16,0) when i==1. However, what you have done in practice is an assignment statement,
>> c(0,:) = c(16,:)
which is not valid because you can't assign to element c(0,:).
I think the easiest way to accomplish what you want is to use the CIRCSHIFT command:
>> index = (1:16)'
>> down_one_index = circshift(index,1)
The guts of CIRCSHIFT, in your case, is the following use of MOD:
>> mod((0:15)-1, 16)+1
This creates the periodic indexing that you want. Hope that helps.
the cyclist
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Weitere Antworten (3)
Siddharth Shankar
am 26 Jan. 2011
if "i" is 1, then you can not index into c with the row index "i-1". MATLAB uses 1-based indexing. You need to "special case" your code to avoid indexing into row 0, or a row value that is negative, as the error message suggests.
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Lucy
am 26 Jan. 2011
2 Kommentare
the cyclist
am 26 Jan. 2011
Glad you found a solution. Be aware that applying logic functions to handle special cases is in general going to be slower than vectorized solutions like the one I proposed.
Paulo Silva
am 26 Jan. 2011
Lucy there's no need for any if statements or other fancy functions
nrows=16 %choose the number of rows (horizontal lines)
indrow=2:nrows+1 %create the index of the rows, the trick is here (+1)
C(indrow-1,:) %no problem now, indrow-1 is never zero
2 Kommentare
the cyclist
am 26 Jan. 2011
Lucy's trying tell MATLAB that the row "below" row 1 should be row 16. (Periodic boundary condition.) I don't see how your indexing does this. "indrow-1" seems to just be 1:16. Do I misunderstand?
Paulo Silva
am 26 Jan. 2011
Now I understand, thanks for the good explanation :) forget about my code, it just rotates the rows 180º
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