Nested for loop troubles

Hi Everyone,
I am having difficulty troubleshooting a nested for loop. Things work fine on the first pass (when i=49) but break down on the second pass (when i=50). More specifically, instead of appending to consolFut1 on the second pass, it overwrites the data from the first pass. Can anyone spot the error? Thanks for reading.
for i = 49:52
exp=find(tifDate==expDate(i));
for j = [1; 18; 43; 63]
tifCls1 = tifCls(j:exp);
for k = [1112; 1129; 1154; 1174]
consolFut1=consolFut(k:expDateIdx(i),i);
arbTIF=tifCls1-consolFut1;
break;
end
end
end

2 comentarios

James Tursa
James Tursa el 29 de Jun. de 2015
Editada: James Tursa el 29 de Jun. de 2015
What are the dimensions of your variables? The usual procedure is to pre-allocate an array to contain the outputs from each iteration. I assume the break statement is there for debugging, because as written it will always exit the k loop on the first iteration.
Cary
Cary el 30 de Jun. de 2015
I would be really grateful to anyone who can help me solve this issue. I've spent all day and night trying to fix it. On the second pass, consolFut1 overwrites the column instead of adding to the bottom of it. I don't understand why it does this as tifCls1 works perfectly. Thank you for your time. I sincerely appreciate it.

Iniciar sesión para comentar.

Respuestas (1)

Image Analyst
Image Analyst el 29 de Jun. de 2015

0 votos

Nothing on the left hand side of your equations takes an index, so they all get overwritten at each iteration. Add an index if you want to store the different values.
exp is a built in function. Don't use it as the name of a variable!

4 comentarios

Cary
Cary el 29 de Jun. de 2015
I'm sorry, I'm having trouble following. Can you clarify a little more for me please?
Image Analyst
Image Analyst el 29 de Jun. de 2015
consolFut1 is a 2D array. Are you trying to append a whole column to the right hand side of it? If so, then do this
consolFut1 = [consolFut1, consolFut(k:expDateIdx(i),i)]
Of course you have to make sure that the number of rows from consolFut (which is different than consolFut1 because there's no 1 at the end of it) that are specified in k:expDateIdx(i) matches the number of rows in consolFut1.
Or you could preallocate consolFut1 and then just assign a column. The thing that worries me is that k and expDateIdx(i) vary with iteration so how can we be certain that the number of rows match? In other words, that length(k:expDateIdx(i)) matches size(consolFut1, 1).
Cary
Cary el 29 de Jun. de 2015
I'm not trying to append to the right. On the first pass the data is stored in consolFut1 (a column). On the second pass, instead of appending to the bottom of the column, it overwrites the data from the beginning. The column should be getting longer and longer with each pass, while keeping the old data.
Walter Roberson
Walter Roberson el 30 de Jun. de 2015
consolFut1 = [ consolFut1; consolFut(k:expDateIdx(i), i).' ];
Having multiple elements in the first dimension of the consolFut is for selecting part of a row. If you want those elements to be written into a column then you need to transpose the row to become a column.

Iniciar sesión para comentar.

Categorías

Más información sobre Loops and Conditional Statements en Centro de ayuda y File Exchange.

Etiquetas

Preguntada:

el 29 de Jun. de 2015

Comentada:

el 30 de Jun. de 2015

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by