functional programming - Map function in MATLAB?

Question

I'm a little surprised that MATLAB doesn't have a Map function, so I hacked one together myself since it's something I can't live without. Is there a better version out there? Is there a somewhat-standard functional programming library for MATLAB out there that I'm missing?

function results = map(f,list)
% why doesn't MATLAB have a Map function?
results = zeros(1,length(list));
for k = 1:length(list)
    results(1,k) = f(list(k));
end

end

usage would be e.g.

map( @(x)x^2,1:10)

Answer 1

The short answer: the built-in function arrayfun does exactly what your map function does for numeric arrays:

>> y = arrayfun(@(x) x^2, 1:10)
y =

     1     4     9    16    25    36    49    64    81   100

There are two other built-in functions that behave similarly: cellfun (which operates on elements of cell arrays) and structfun (which operates on each field of a structure).

However, these functions are often not necessary if you take advantage of vectorization, specifically using element-wise arithmetic operators. For the example you gave, a vectorized solution would be:

>> x = 1:10;
>> y = x.^2
y =

     1     4     9    16    25    36    49    64    81   100

Some operations will automatically operate across elements (like adding a scalar value to a vector) while others operators have a special syntax for element-wise operation (denoted by a . before the operator). Many built-in functions in MATLAB are designed to operate on vector and matrix arguments using element-wise operations (often applied to a given dimension, such as sum and mean for example), and thus don't require map functions.

To summarize, here are some different ways to square each element in an array:

x = 1:10;       % Sample array
f = @(x) x.^2;  % Anonymous function that squares each element of its input

% Option #1:
y = x.^2;  % Use the element-wise power operator

% Option #2:
y = f(x);  % Pass a vector to f

% Option #3:
y = arrayfun(f, x);  % Pass each element to f separately

Of course, for such a simple operation, option #1 is the most sensible (and efficient) choice.