"Index exceeds the number of array elements"

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I've been getting an index exceed the number of array elemnts (2048)
How to I solve this?

Answers (2)

Cris LaPierre
Cris LaPierre on 30 Jul 2020
Use an index that is not greater than the dimension you are indexing. For example, if your array is a 1x2048, you will get this error is you try to index row 2 or column 2049.
  4 Comments
Jarren Berdal
Jarren Berdal on 30 Jul 2020
Edited: Walter Roberson on 30 Jul 2020
Pulling data from a 2048 x 7 array. Isolated some of the data into a 2048 x 1 array. Finding the mean of the data of groups of 5s. Came up with a 409 x 1 array.
error that appears " Index exceeds the number of array elements (2048) "
error in line 24
code:
17 - DATE = D(:,1); % first column, days numbered from 1/1/1900
%cloud of data for summ of the data in one grouping
19- Darray(1,1) = mean(DATE(1:N)); %initial value
20 - PRICE = D(:,5); % 5th column associated with closing price
21- Parray(1,1) = mean(PRICE(1:N));
22 - for i = 1:(R)
23 - p = (i*N);
24 - Darray((i+1),1) = mean(DATE((p+1):(p+N)));
25 - Parray((i+1),1) = mean(PRICE((p+1):(p+N)));
end
Cris LaPierre
Cris LaPierre on 30 Jul 2020
What is the value of
  • R
  • N (5?)
What is the size of DATE

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Walter Roberson
Walter Roberson on 30 Jul 2020
We do not know what R is.
If we suppose R is the 409 and N is 5, then on the last iteration, when i = 409, then p = 409*5 = 2045, and p+1:p+N = 2046 : 2050, but the array is only 2048 long.
Meanwhile, when i = 1, then p=i*N = 5, and p+1:p+N is 5+1 : 5+5 = 6:10 . Notice that you have not accessed index 1:5 .
The solution:
23 - p = (i*N);
should be
23 - p = ((i-1)*N);

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