Understanding :error To RESHAPE the number of elements must not change while manipulating matrix

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Hi, i need some explaination, i run this code and the input matrix 40*11 and resulted 41*11 matrix. I try to put it into a new matrix by using reshape and has error.
function [] = cacl_mean(G)
k = 4
j = 1
[numRows,numCols] = size(G) %numRows =40,numCols =11
while j <= k
for n = i:40 %iteration dr data row 1 hingga 40
if (G(i,11)) == j %check data col ke 11 ada value j=1@2@3@4
kk = reshape(L,41,11); % error To RESHAPE the number of elements must not change.

Answers (1)

Guillaume on 26 Oct 2019
Edited: Guillaume on 26 Oct 2019
It's very unclear what your code is meant to do, it makes no sense.
Anyway, in your loop your creating (sometimes!) L as a vector consisting of a row of your G matrix. If the condition is true several times in your loop, this L will contain only the last row for which it is true. Also note that if the condition is never true, then L won't exist and you'll have an error telling you it's undefined later on.
So, if L exists, it's a vector corresponding to one row of G. According to your comments G has 11 columns, so L is 1x11. Obviously, it can't be reshaped into a 41x11 matrix.
You should learn to debug your programs. By stepping through your code one line at a time, you'd have seen the problem immediately.
  • You query the size of the input matrix, a good thing to do, but then completely ignore that and hardcode it in the loop. So in the end your code only works for a 40x11 matrix.
  • Your function doesn't return anything
  • it's unclear why you're using a while loop, when for j = 1:k would be clearer.
  • It's unclear why the i=i+1 is in both branches of the if. May as well put it afterward and write the line only once.
  • More importantly it's very unclear what you meant to do with your for n=i:40, which here is exactly equivalent to for n = 1:40 since the value of i is always 1 when the loop starts. Additionally you never use n. Your loops are exactly equivalent to:
for j = 1:4
for i = 1:40 %would be better as for i = 1:numRows
if G(i, 11) == j
L = G(i, :);
Again, this does the same as what you wrote, only it's a lot clearer. Of course, it's unlikely to be what you meant to do.


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