Explanation of cellfun()

Explaination of cellfun() call with @(data):

Hi everyone,

I have a rather stupid question I think, but I do not understand a specific call of the cellfun().

My question came up when I was working with the MATLAB example: Similarity-Based Remaining Useful Life Estimation. There is a function applied to each cell of a cell array, for example in line 34:

trainDataNormalized = cellfun(@(data) regimeNormalization(data, centers, centerstats), ...
    trainData, 'UniformOutput', false);

[...] line 176: (for info)

function data = regimeNormalization(data, centers, centerstats)

For me the content of the cellfun() and its function is clear, exept of the expression @(data). Cellfun() applies the function regimeNormalization individual to each cell of the cell array trainData.

Looking in the doku of cellfun() they call a funktion like this and leave out the additional function like above. Which I think I undestand, see below.

A = cellfun(@mean,C)
p = cellfun(@plot,X,Y);

But in the documentation they also do this, which is exactly like my problem, but the explaination is not sufficient for me. (MATLAB advanced beginner) What is the expression @(x) mean?

B = cellfun(@(x) x(1:3),str,'UniformOutput',false)

Does cellfun() accessing one cell of the cell array temporarily store the data inside the cell in the variable x?

[Stephen23]

https://www.mathworks.com/help/matlab/matlab_prog/anonymous-functions.html

https://www.mathworks.com/help/matlab/matlab_prog/creating-a-function-handle.html

Answer 1

[Rik]

cellfun will apply your function to the contents of you cell array. There are 3 ways to specify the function:

1. as a char array (only a handfull of functions, but they are faster when called like this)\
2. with a function handle, e.g. @mean
3. with an anonymous function

That last option is what you see here.

%instead of this
function output=MyFun(in1,in2)
output=in1.*in2;
end

%you do
MyFun=@(in1,in2) in1.*in2;

[Marvin Eckert]

Tanks for your quick answer. :)

Your answer were helpflull, because I did not thought about anonymous functions. I think I undestand now what the cellfun()

trainDataNormalized = cellfun(@(data) regimeNormalization(data, centers, centerstats), ...
    trainData, 'UniformOutput', false);

above is doing.

To be shure I explain it in my words and you give a short YES or NO? That would be very helpful to me.

The function temporarily stores the information within a single cell of the trainData cell array in the variable data, executes the regimeNormalization() function using the variable data as input, and writes the result of regimeNormalization() function back to this cell.

[John Navarro]

Hello. I think the answer is NO. cellfun is required because of the format or the variable, or the data type. So far that I understand is that regimenNormalization() is being evaluated for each cell of the variable (data).

Chech this for more details

https://www.mathworks.com/help/matlab/matlab_prog/anonymous-functions.html

https://www.mathworks.com/help/matlab/ref/cellfun.html?s_tid=doc_ta

[Rik]

Actually the answer is yes.

Cellfun in this case is only hiding the loop.

%the first step is to create an anonymous function with the input variable
%called data. The values of the variables centers and centerstats are
%captured with their current value.
anon_function=@(data) regimeNormalization(data, centers, centerstats);
%the second step is to execute the anonymous function for each cell and
%store the result in a cell array.
trainDataNormalized = cellfun(anon_function, ...
    trainData, 'UniformOutput', false);

The cellfun call is equivalent to this:

trainDataNormalized=cell(size(trainData));
for n=1:numel(trainData)
    trainDataNormalized{n}=anon_function(trainData{n});
end

The loop will probably be marginally faster.

It is the anonymous function that is evaluated for every cell. Inside that anonymous function there is a call to regimenNormalization, which is called with the data in each cell, but also with two other variables that were captured when the anonymous function was created.

[Marvin Eckert]

Hi Rik,

very cool. Thanks for your comment much appreciated.