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"Index exceeds the number of array elements"

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Jarren Berdal
Jarren Berdal il 30 Lug 2020
Commentato: Cris LaPierre il 30 Lug 2020
I've been getting an index exceed the number of array elemnts (2048)
How to I solve this?

Risposte (2)

Cris LaPierre
Cris LaPierre il 30 Lug 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 Commenti
Jarren Berdal
Jarren Berdal il 30 Lug 2020
Modificato: Walter Roberson il 30 Lug 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 il 30 Lug 2020
What is the value of
  • R
  • N (5?)
What is the size of DATE

Accedi per commentare.


Walter Roberson
Walter Roberson il 30 Lug 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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