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help with dynamic variable names set with 'for' incrementer

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I am trying to split up a large numerical dataset into individual days. What I have that works and what I have that failed is;
function [FxStOut, FxStIndex] = FxSt(FxStIn)
%Struct for Forex
UnqDate = unique(FxStIn(:,1));
UnqL = length(UnqDate);
for ii = 1:UnqL
indx = FxStIn(:,1) == UnqDate(ii);
eval(['dataOut.d' num2str(ii) ' = FxStIn(indx, 2:6);']);
%Have tried but died
%dataOut(ii) = FxStIn(indx, 2:6);
%dataOut.d([ii]) = FxStIn(indx, 2:6);
%dataOut.([ii]) = FxStIn(indx, 2:6);
FxStOut = dataOut;
FxStIndex = UnqDate;
Is 'eval' the only way to achieve this, both reading in and out of array. A number/numeral can not be a variable/array name, as I would like;
is this correct.
Can I solve this by using a 3D (:,:,:), however, each depth is a different size (factors different) and I presume that is not allowed.
Scragmore on 27 Nov 2011
Additional information on how I want to expand this.
for ii = 1:length(x)
opMax = zeros(1:length(x), 1:5);
eval(['opMax(' num2str(ii) ') = max(x.d' num2str(ii) ');']);
Each 'x.d1' is a (1:n,1:5) array. I am trying to find the max in each column of 'x.d1' and return the five ans to the corresponding row of new array. This 'eval' and 'num2str' seems really messy and complicated, as I build up complexity I feel this will become less and less readable as code. The above test example does not work, incorrect array size allocation.

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Accepted Answer

Teja Muppirala
Teja Muppirala on 27 Nov 2011
You are right in that using EVAL to dynamically generate variable names is generally a messy approach that's not very readable, slow, and doesn't scale well.
One alternative is using cell arrays.
Cell arrays let you store differently sized data, for example:
M{1} = [1 2 3]
M{2} = [1 2; 3 4]
M{3} = 'hello'
M{1} + 10
M{2} * 5

More Answers (2)

bym on 27 Nov 2011
eval can be evil! avoid if possible.

Walter Roberson
Walter Roberson on 27 Nov 2011
Replace your line
eval(['dataOut.d' num2str(ii) ' = FxStIn(indx, 2:6);']);
dataOut.(['d' num2str(ii)]) = FxStIn(indx, 2:6);
or alternately with
dataOut.(sprintf('d%d',ii)) = FxStIn(indx, 2:6);
  1 Comment
Scragmore on 28 Nov 2011
Thanks. Shame you cant accept two ans. Your response answered my question directly but Teja's ans gave me an alternative view and solution to my problem that I am going to run with.
Thanks All

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