Load and name alot of files in loop instead of dynamic variables

I want to load a lot of different .mat files properly instead of using dynamic variables which I have read is pretty bad.. I have more files than shown here but just to use this as an example of what I want to avoid. A loop and a neat way to handle the data (e.g indexing) for further calculation and graphs would be awesome.

Edit: The number corresponds to different windspeeds, so I have a vector with wind = [9 11 17 25 32 56] I have tried to use for naming and loading without any luck.

Thanks in advance!

A9 = load('decayDamp9');
A11 = load('decayDamp11');
A17 = load('decayDamp17');

B9 = load('TpPosDamp9');
B11 = load('TpPosDamp11');
B17 = load('TpPosDamp17');

Damp9 = A9.v;
Damp11 = A11.v;
Damp17 = A17.v;

Tp9 = B9.tp_pos;
Tp11 = B11.tp_pos;
Tp17 = B17.tp_pos;

[Henning Eimstad]

Edit:

I came up with the following code

wind = [9 11 17 25 32 56];

Damp = zeros(6,6000);
Tp_pyMom = zeros(6,52);
Tp_pyTime = zeros(6,52);

for k = 1:6
    A = load(['decayDamp' num2str(wind(k)) '.mat']);
    Damp(k,:) = A.v;
    B = load(['TpPosDamp' num2str(wind(k)) '.mat']);
    TpPy = B.tp_pos;
    if k == 5
        Tp_pyMom(k,:) = TpPy(end-1,2);
        Tp_pyTime(k,:) = TpPy(end-1,1);
    else
        Tp_pyMom(k,:) = TpPy(:,2);
        Tp_pyTime(k,:) = TpPy(:,1);
    end
end

For k=5 the result was +1 longer than the rest. Now everything is stores as in Damp(6x6000), and I can access the different data with Damp(1,:) etc. Is this a better way than the first post?

Answer 1

[Ameer Hamza]

Your new code is definitely better than the first one, and you made the correct choice of using arrays instead of using dynamic variable names. In the following code, I suggest one minor improvement and point out a potential error in your code. Overall, your approach is correct

wind = [9 11 17 25 32 56];
num_files = numel(wind); % always a good idea to use a variable instead of hardcoding values.
                         % It will prevent any trouble if you try to read any other set of files.

Damp = zeros(num_files,6000);
Tp_pyMom = zeros(num_files,52);
Tp_pyTime = zeros(num_files,52);
for k = 1:num_files
    A = load(['decayDamp' num2str(wind(k)) '.mat']);
    Damp(k,:) = A.v;
    B = load(['TpPosDamp' num2str(wind(k)) '.mat']);
    TpPy = B.tp_pos;
    if k == 5
        Tp_pyMom(k,:) = TpPy(1:end-1,2); % I guess you just wanted to exclude the last element.
        Tp_pyTime(k,:) = TpPy(1:end-1,1);
    else
        Tp_pyMom(k,:) = TpPy(:,2);
        Tp_pyTime(k,:) = TpPy(:,1);
    end
end

Note that you can handle matrices of varying sizes using cell arrays, but they can be a little bit slower than numeric arrays.

[Henning Eimstad]

Perfect! Thanks for the feedback, much appreciated! :)

Answer 2

[Florian Floh]

Hi!

For loading multiple '.mat' files, I would suggest using the function dir(). You simple store all the desired '.mat' files in one direction (e.g. where your script file is stored), so that your code can access them all at once.

This is the code I would suggest for loading these '.mat' files:

%% Check for *.mat files in this directory

% the function dir() returns a struct that contains (among others) the name
% of the stored '*.mat' files
a = dir('*.mat');

%% Load all the *.mat files in this directory

% Get the number of rows and columns of the struct a
[nrRowStruct nrColStruct] = size(a);
% Use a for loop to load each file by its name
for i=1:nrRowStruct
   %load each of the files into the workspace 
   A =load(a(i).name); 
   
   % Here you can process the struct A, as you did in your code before
end

Note that my code is not complete, so you have to add whatever you want to do with the struct A, otherwise you would just see the last struct, that was loaded in your workspace ;)

I hope this was helpful, if not, just let me know ;)

[Ameer Hamza]

Although using dir is useful in many cases, It will make the code difficult in this case. The OP wants to load two different types of file names in the same loop

decayDampX.mat

TpPosDampX.mat

With dir(), another if condition will be needed to distinguish between these two files.

[Henning Eimstad]

I do have other .mat files I have loaded manually tho, so this struct might come in handy. However, the other .mat files have completely different names and uses. Anyway, take for example the decayDamp files that are now located in field 10-15 as seen in the image structq (attached). How can I get these as the same Decay (6x6000) matrix I want?

[Ameer Hamza]

Henning, here is an example of using dir() to load the files decayDampX.mat. You can similarly extend to the other file too. It will be useful if you have lot of files with name in same format.

files = dir('decayDamp*.mat'); % load all files with names decayDampX.mat
num_files = numel(files);
Damp = zeros(num_files,6000);
for k = 1:num_files
    A = load(files(k).name);
    Damp(k,:) = A.v;
end

[Henning Eimstad]

Great, thanks alot! Very helpful both of you.