Error using @regress fuction
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Nabil Benhadda
le 24 Fév 2020
Commenté : Walter Roberson
le 25 Fév 2020
Hello, I am trying to create a function as follows with a set of variables that I get from a table named Change1:
X = [Change1.Ones Change1.ReturnsMarket]
func = @regress(Change1.ReturnsCleaned,X);
However, everytime I get the following error message:
Invalid expression. When calling a function or indexing a variable, use parentheses. Otherwise, check for
mismatched delimiters.
Thank you all for your help !
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Star Strider
le 24 Fév 2020
Try this:
func = @(Q)regress(Q,X);
then call it as:
[b,bint] = func(Change1.ReturnsCleaned);
(It seems a bit inefficient to me to wrap regress in an anonymous function. Personal opinion only.)
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Walter Roberson
le 24 Fév 2020
The @ operator can be used in four related syntaxes:
@NAME : create handle to function indicated by the name
@NAME1.NAME2 : create handle to function whose name inside its Package is as given
@() EXPRESSION : create handle to anonymous function with no parameters that evaluates to the given expression when invoked
@(NAME LIST) EXPRESSION : create handle to anonymous function with dummy parameters named inside the ()
You however have @EXPRESSION which is not valid syntax.
@ always has to do with creating handles, except possibly for some class definition purposes. If you want to invoke regress function instead of creating a handle then leave out the @
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Nabil Benhadda
le 24 Fév 2020
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Walter Roberson
le 25 Fév 2020
Q = Change1.ReturnsCleaned;
You are using that as your grouping variable, but it is not obvious to us that it contains only positive integers.
Change1.Groups = findgroups(Change1.Numero);
X = [Change1.Groups Change1.Ones Change1.ReturnsMarket];
findgroups does return the kind of positive integers needed for a grouping variable, but you are folding that information in as part of X and using X as the variable to be split rather than as the grouping variable. That seems odd.
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