I am having problems in anonymous function (y=@(x)(...))

1 view (last 30 days)
Let me re-word my querry.
Let's say there is a square matrix of v=[1 2 3 4 5 6 78 9 10; 11 12 13 14 15 16 17 18 19 20;21 22 23 24 25 26 27 28 29 30;31 32 33 34 35 36 37 38 39 40;41 42 43 44 45........ 99 100] and a is a variable that prompt the user to input. It is easy to obtain a square matrix of size a by a of the original matrix v (left hand corner) with the use of w=v(1:a,1:a), of course a is less than length(v).
However, I would like to know how to do the same for anonymous functions. v in this case will be a column matrix a quantities of "@(x)(B.*[x x^2 x^3 ......]);", where [x x^2 x^3 ......] is a row matrix with a elements.
Here is an example:
Let's say v={@(x)([x x^2 x^3 x^4]; @(x)([x^2 x^3 x^4 x^5]); @(x)([x^3 x^4 x^5 x^6]); @(x)([x^4 x^5 x^6 x^7])};
if a = 3, the new v would be v={@(x)([x x^2 x^3]; @(x)([x^2 x^3 x^4]); @(x)([x^3 x^4 x^5])};
  4 Comments
Kevin Phung
Kevin Phung on 29 Apr 2019
ok, just wanted to make sure. Also, what is x supposed to be? I think maybe it'll be easier to construct the matrix if x was an input to your function fun, instead of creating so many anonymous functions.
Angus Wong
Angus Wong on 29 Apr 2019
Edited: Angus Wong on 29 Apr 2019
x is a variable.
The script is something like this:
clear; close all; clear;
dat=load(data);
leadingcoe=coef(dat);
yo=fun(leadingcoe);
k=input('Input a number for x');
for a=1:k
disp(yo{a}(k));
end
This is a simplified version.

Sign in to comment.

Answers (1)

Kevin Phung
Kevin Phung on 29 Apr 2019
Edited: Kevin Phung on 29 Apr 2019
let me know if this is what youre trying to do with your anon functions:
function y = fun(a,x)
y = sum(triu(repmat(flip(x.^(1:a)),numel(a),1)),2)
%uncomment this if you want to see how the matrix looks like before summing the rows
%y = triu(repmat(flip(x.^(1:a)),numel(a),1))
end
  3 Comments
Angus Wong
Angus Wong on 29 Apr 2019
This function is intended to output an anonymous function.

Sign in to comment.

Categories

Find more on Loops and Conditional Statements in Help Center and File Exchange

Products


Release

R2019a

Community Treasure Hunt

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

Start Hunting!