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Problems in accessing 2nd column of matrix

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Mich
Mich 2015-4-11
关闭： MATLAB Answer Bot 2021-8-20
I know this is a very simple problem, but I cant seem to find what is the issue as it seems logical.
I tried instances when xx is a column vector or horizontal vector, but it doesnt make a difference.
Count is the final number of data points i want to obtain. E.g.
count = 10
for k = 1:(length(xx)-1)
temp = *data* (xx(k):xx(k+1));
mat(:,k) = interrp(temp, count);
end
interrp is a function that was written to use the i
interp1 function to use the method 'spline'.
My code always stop after finishing filling up the 1st column. This is the error that I receive: Subscripted assignment dimension mismatch.
  2 个评论
Mich
Mich 2015-4-11
编辑：Mich 2015-4-11
just to make things clearer, this is my interrp function
function result=interrp(data,Max_number)
n=length(data);
xx=1:n/Max_number:n;
y=data;
result=interp1(y,xx,'spline');
Mich
Mich 2015-4-13
I realised that the errors lies in the in-built interp1 function. It does not seem to be able to interpolate and obtain the length desired (e.g. length of xx). Any one knows how to solve this error? the error seems to occur randomly as it works when i run the code on different datasets

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回答（2 个）

Geoff Hayes
Geoff Hayes 2015-4-11
编辑：Geoff Hayes 2015-4-11
Mich - what are the dimensions of the array that the interrp function is returning? The error message Subscripted assignment dimension mismatch typically means that the code is trying to insert more data than it can fit into the destination array. For example,
x = zeros(3,42);
x(:,2) = randi(255,4,1);
returns the same error because I am trying to assign a 4x1 matrix into a 3x1 destination. Put a breakpoint at the line
mat(:,k) = interrp(temp, count);
then run your code and verify that what interrp is returning is too large for the mat(:,k) destination.
You may also want to pre-size mat prior to entering the for loop so that you know exactly what you are trying (and what you are able) to assign to this matrix.
  2 个评论
Mich
Mich 2015-4-11
Hi Geoff,
interrp function return a horizontal array of dimensions (1, count).
I've tried to presize mat and clearing any earlier allocation within mat by doing this:
clearvars mat mat = []; mat = zeros(count,length(xx)-1);
I'm not sure how to put a break at the code with the "interrp'. However, I include size(mat) before and after the line with interrp function, and realised that the main problem lies because mat(:,2) cant be accessed when k = 2.
pfb
pfb 2015-4-11
编辑：pfb 2015-4-11
Hi,
Your code is very obscure. What exactly do you want do attain? In particular, this line
temp = xx(xx(k):xx(k+1));
looks weird, especially if xx does not contain positive integers. Even in that case, I kind of see why matlab complains about the dimension mismatch. What is xx?
  1 个评论
Mich
Mich 2015-4-11
Hi pfb,
this is a simplified version of the code. xx is actually a row array with positive integer. It includes the indices which i'll like to split each column in mat.
E.g. if xx = [3 6 9 15];
for k = 1:(length(xx)-1) temp = data(xx(k):xx(k+1)); mat(:,k) = interrp(temp, count); end
what i want to get is : data(3) data(6) data(9) : : : : : : data(6) data(9) data(15)
but the interrp function would have made everything the same lenght, so it can be put into a matrix. Now the problem i have is getting the for loop to access mat(:,2)

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