How to use arrayfun with a function having two vector arguments?

28 ビュー (過去 30 日間)
Mr M.
Mr M. 2016 年 10 月 18 日
コメント済み: Andrei Bobrov 2016 年 10 月 18 日
I have a function f(W,x), where W is a matrix and x is a vector. I would like to use arrayfun to calculate the f(W,x) for several x but constant/fix W. What is the syntax?

採用された回答

Andrei Bobrov
Andrei Bobrov 2016 年 10 月 18 日
編集済み: Andrei Bobrov 2016 年 10 月 18 日
fW = @(x) f(W, x);
out = cellfun(fW,{x1,x2,x3},'un',0);
  3 件のコメント
Andrei Bobrov
Andrei Bobrov 2016 年 10 月 18 日
OK. Corrected.

サインインしてコメントする。

その他の回答 (2 件)

Rani V.S
Rani V.S 2016 年 10 月 18 日
A = arrayfun(FUN, B, C, ...) evaluates FUN using elements of arrays B,
C, ... as input arguments. The (I,J,...)th element of A is equal to
FUN(B(I,J,...), C(I,J,...), ...). B, C, ... must all have the same size.

Guillaume
Guillaume 2016 年 10 月 18 日
%W = [.,.,.;.,.,.;.,.,.]; %matrix
%x = [.,.,.,.,.,.]; %vector
result = arrayfun(@(xelem) f(W, xelem), x); %possibly with 'UniformOutput', false, if f output is not scalar
Or if you want to be more explicit
%W = [.,.,.;.,.,.;.,.,.]; %matrix
%x = [.,.,.,.,.,.]; %vector
fW = @(x) f(W, x); %W must exists before this line. Changing W after this line will not affect fW
result = arrayfun(fW, x);
In either case you're creating a new function (with no name in the first case, fW in the second) that binds the matrix W to f and still takes x as an input. You're then using that function in arrayfun to apply it to each element of x.
  2 件のコメント
Guillaume
Guillaume 2016 年 10 月 18 日
Well, first you shouldn't have numbered variables x1, x2, etc. Instead you should have just one variable which is a cell array of vectors. In that case, use cellfun:
x = {x1, x2, x3}; %you really shouldn't create the xi to start with. Put their content in the cell array as you create it.
result = cellfun(@(xv) f(W, xv), x, 'UniformOutput', false);

サインインしてコメントする。

カテゴリ

Find more on Structures in Help Center and File Exchange

Community Treasure Hunt

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

Start Hunting!

Translated by