Building a set of simple moving averages using a loop

Hi everyvody,

I have to build a set of simple moving averages with different periods; may be that my problem is pretty tricky, but I really do not know how to solve that.

I thought to use the loop for as it follows:

for i = 5:1:30
    SMAvector(:,i) = tsmovavg(vector,'s',i,2);
end

but matlab gives an error out as:

 Undefined function 'le' for input arguments of type 'cell'.

Can someone help me to find a solution or explain me what does the error mean? Thanks in advance for the help.

[Geoff Hayes]

I think that the error refers to the less than or equal operator not being defined for inputs of type cell. What is the data type on vector? In the Command Window type

 class(vector)

Does it show cell as the answer?

Looking at tsmovavg, the first input can only be a financial time-series object, or a vector (or matrix) of observations (so not a cell array).

[Quantopic]

thanks for the answer Geoff. Typing

class(vector_name)

the ans was double, so I think it is not that problem. Also because if I run that code without using a loop for, as follows:

SMAvector = tsmovavg(vector,'s',10,2);

I get a simple moving average on 10 time periods. I think that the problem is in the loop syntax.

[Geoff Hayes]

Ok - what have you defined SMAvector to be? Is it a cell array or an empty matrix or..? What happens if you do

 SMAvector = zeros(length(vector),30-5+1);
 for k = 5:1:30
    SMAvector(:,k-5+1) = tsmovavg(vector,'s',k,2);
 end

Note that I replaced your i with k since both i and j are used as representations for the imaginary number, and so it is good practice to avoid using either as indices in loops.

[Quantopic]

Ok. It seems to work. Thanks a lot Geoff.

Answer 1

[Geoff Hayes]

Quantopic verified that the vector variable was of type double. As it was unclear how the SMAvector variables was declared, it was suggested that the following code be tried

 SMAvector = zeros(length(vector),30-5+1);
 for k = 5:1:30
    SMAvector(:,k-5+1) = tsmovavg(vector,'s',k,2);
 end

That seemed to work.