Function changes its behavior at each iteration of a for loop

Hi everyone, the problem is as follows, i want in a for loop change the behavior of a function at each iteration. Suppose the following for loop F or k = 1:4 B = Min_or_max([k 2 4]); end Let’s assume I have 2 functions called Compute_min and Compute_max computing respectively the maximum and the minimum value of a given vector/array taken as input parameter. I want that at the first iteration of the for loop compute_min is used instead of min_or_max and at the second iteration of the loop the function compute_max is used and so on.. Note that the for loop is placed in a function which takes as input a parameter, which should specify which function of (compute_min and compute_max ) should be used at which step. Example : for a = compute_min and b = compute_max the input parameter of the function in which the for loop is located can take the form [a b a a ] ; which means compute_min is used at the first iteration , compute_max is used at the second iteration , then compute_min at the third and again compute_min at the fourth iteration.

Many Thanks guys

Answer 1

[Stephen23]

If you have two (or more) pre-defined functions that you wish to select between in a loop, then you could consider using function handles to select and pass the functions. Something like this:

fun = {@min,@max};
idx = [1,2,1,1];
mat = [1,2,3,4;5,6,7,8;9,10,11,12];
out = nan(size(idx));
for k = 1:numel(idx)
    out(k) = fun{idx(k)}(mat(:,k));
end

where

>> mat
mat =   1     2     3     4
        5     6     7     8
        9    10    11    12
>> out
out =   1    10     3     4

shows that it has taken the minimum of columns 1, 3, and 4, and the maximum of column 2, exactly as the vector idx selected. How it works:

• fun is a cell array of the function handles that you want to call.
• idx is a vector of the indices of the functions in fun to call on each loop.
• mat is some fake numeric data, with as many columns as idx has elements.
• out is the min/max value of each column of mat, depending on which function was selected for that column by the corresponding index in idx.

[kerin]

Many many many thanks stephen, this is exactly what i was looking for.. Do you have an idea what i can do if the number of input parameter of my two funcion are diferent ? For example

fun = {@rectangle_area,@square_area};
idx = [1,2,1,1]; length = 2 ; width = 3 ; raduis  = 5 ; 
a= rectangle_area(length, width)  % a = 2*(length+width)
b= square_area(raduis)            % b = pi*raduis^2

for k = 1:numel(idx)
  out(k) = fun{idx(k)}(????);
end

what should come at ???? place ?

a lot of thanks again

[Stephen23]

Ah, changing the requirements after the original question has been answered... Oh well :) If the functions require a different number of arguments, then this can be achieved by placing the arguments in a cell array too:

fun = {@rectangle_area,@square_area};
dat = {{2,3},{5}}; % {{length,width},{radius}}
idx = [1,2,1,1];
out = nan(size(idx));
for k = 1:numel(idx)
    x = idx(k);
    out(k) = fun{x}(dat{x});
end

This assumes that the length/width/radius values are the same for all loop iterations. If they are different, then define one element of dat for each iteration:

dat = {{2,3},{5},{7,6},{3,6}};

and call it like so:

      out(k) = fun{x}(dat{k});

Alternatively you could use a switch statement, as per wil's answer, in which case the inputs can be hard-coded.

Note that you should never call a variable length, size, i, j, cd, table, ans, etc, as these are all inbuilt variables or functions. You can use which to identify inbuilt functions and variables.

[kerin]

ohh yessssssssss !!!!!! u're amazing !!! Thanks.. be blessed !!

Answer 2

[wil]

We'll call your input parameter list 'param', you can do something like the following:

for k= 1:4
    switch param(k)
       case 'a'
          B = Compute_min([k 2 4]);
       case 'b'
          B = Compute_max([k 2 4]);
       otherwise
          % handle invalid param
    end
 end

Which checks the index of the parameters and calls the corresponding function depending on the value.

Wil