# How to make dynamic variable names (A1, A2, ..., An) with "for" loop? Using "eval" "num2str"?

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Hello community,

my knowledge of Matlab is limited, I admit it. So, I am sorry if I am going to make mistakes.

I have to create a series of variable using a "for" loop.

I make this:

for i=1:3

eval(['A' num2str(i) '= i'])

end

and it works well, it makes 3 variables A1, A2, A3.

But I need to use this variables to make other variables B1, B2, B3 where Bi=Ai*i. So I should have B1=A1*1=1, B2=A2*2=2*2=4, B3=A3*3=3*3=9

I tried something like this:

for i=1:3

eval(['A' num2str(i) '= i'])

eval(['B' num2str(i) '= 'A' num2str(i) '*i])

end

but it gives me error and it doesn't work. Of course I simplified things, the real code is a bt more complicated but I need to know the basics of how this could work.

Thank you

##### 11 Comments

Steven Lord
on 19 Jan 2022

I think of dynamic file names as somewhat different from dynamic variable names, as the file names are usually themselves data (passed into a function that reads the file contents then returns a variable whose name is not necessarily tied to the file name, or performs some other operation on the file.)

t = fullfile(tempdir, 'writeASeriesOfFiles');

if ~isfolder(t)

mkdir(t);

end

dir(t) % Nothing in this directory yet

for k = 1:5

fn = fullfile(t, "file" + k + ".txt");

% Now use fn as data in several function calls

fclose(fopen(fn, 'wt')); % Open and immediately close the file

doesItExist = isfile(fn);

if doesItExist

fprintf("File %s does exist!\n", fn)

else

fprintf("File %s does NOT exist!\n", fn)

end

end

dir(t)

### Accepted Answer

Stephen23
on 12 Jan 2016

Edited: Stephen23
on 19 Jun 2019

##### 2 Comments

Giovanni Barbarossa
on 27 Jun 2018

### More Answers (5)

Amit
on 22 Jan 2014

Edited: Amit
on 22 Jan 2014

Italo, I am glad you're trying out things on matlab.

Matlab is very good with matrixes. What you're doing in these can be done much easily. For example:

A = 1:3;

B = A.*A; %

You can find many tutorials online on how to use MAtlab. For example: http://www.mathworks.com/academia/student_center/tutorials/launchpad.html

Jerry Olup
on 28 Apr 2018

Edited: Jerry Olup
on 28 Apr 2018

I did find dynamically named variables useful for a specific need. I have several files in a directory. Each file has a few columns that are read into a structure. It's convenient to read and simultaneously name the variables per a filename. For instance I read to a temp var then use the ill-advised exec capability to assign this to a variable specific to the file I read in. Looping, I read all files in the dir efficiently. 'files' is a cell array of file names discovered in that directory....

tmp = table(dataArray{1:end-1}, 'VariableNames', {'max_mag','max_noise','max_cnt'});

eval([files{i} '=tmp'])

I do see where this is not good for general variable naming, but I believe there's a niche usage in this case. Opinions? Best, Jerry

##### 3 Comments

Jerry Olup
on 12 Aug 2018

Edited: Stephen23
on 12 Aug 2018

Hi Stephen, I'll give the idea of provide parallel constructs where the file-specific variable references are in one cell while data is collected in structure. I've shifted in the last month or so to use python for preprocessing - so this problem has largely gone away and been replaces with xlsread, csvread etc.

>>files data are not restricted to valid field/variable names

My file names are conventionally, ascii with underscores and timestamps yyyymmdd_hhmmss type, so I do see where this is problem. They are pretty specific so I thought I was giving a concrete counter example in contrast to @Italo and @Muna, who started this thread naming generic matrices.

>>you have not given any reason why none of those methods would work.

You're absolutely correct. I latched onto that solution as it was elegant for what I needed.

Thanks for the interaction, best, Jerry

Gabor
on 20 Sep 2021

T=table;

Date=datetime(2014,12,31);

eval(['Dynamic_var_name_' datestr(Date,'mm_dd_yyyy') '=T;']);

This is how you name dynamically a variable or a table or a cell or anything regardless if it is recommended or not.

César Silva Proaño
on 12 Aug 2018

Hi, look I was doing something similar in order to create a function with the Krammer Method, for solving any size of matrix-equation problem. Look I created a function in order to do this, therefore, I needed to do it automatically, creating variables dinamically. Look at this fucntion.

function [ X ] = funcionMetodoKrammer( A )

%UNTITLED Summary of this function goes here

% Detailed explanation goes here

% ex: A = [-3 2 1 10;5 -8 3 15;-8 -9 10 20];

sizeA = size(A);

sizeFilasA = sizeA(1,1);

sizeColumnasA = sizeA(1,2);

matrizASinResultado = A(1:sizeFilasA,1:sizeFilasA);

matrizResultados = A(:,sizeColumnasA);

if (sizeFilasA == sizeColumnasA - 1)

for (i = 1:sizeColumnasA)

genvarname('matrizColumna', num2str(i));

eval(['matrizColumna' num2str(i) '=A(:,i)']);

end

for (i = 1:sizeFilasA)

genvarname('matrizTemporalColumna', num2str(i));

eval(['matrizTemporalColumna' num2str(i) '=matrizASinResultado(:,i)']);

end

matrizIncognita1 = [matrizResultados,matrizASinResultado(:,2:sizeFilasA)];

for (i = 2:sizeFilasA)

genvarname('matrizIncognita', num2str(i));

eval(['matrizIncognita' num2str(i) '=[matrizASinResultado(:,1:i-

1),matrizResultados,matrizASinResultado(:,i+1:sizeFilasA)]']);

end

else

msgbox('La matriz ingresada para el método Krammer es incorrecta');

end

for (i = 1:sizeFilasA)

genvarname('X', num2str(i));

genvarname('matrizIncognita', num2str(i));

eval(['X' num2str(i) '= det(matrizIncognita' num2str(i) ')/det(matrizASinResultado)']);

end

X = zeros(1,sizeFilasA);

for (i = 1:sizeFilasA)

genvarname('X', num2str(i));

eval(['X' '(1,' num2str(i) ')=X' num2str(i)]);

end

end

%{

Asuminedo que existen variables B1, B2 y B3 se pueden acceder a las

variables de la siguiente manera

B1 = 10; B2 = 11; B3 = 12;

for (i = 1:3)

genvarname('Prueba', num2str(i));

genvarname('B', num2str(i));

eval(['Prueba' num2str(i) '=' 'B' num2str(i) '+10']);

end

%}

I strongly disagree with the idea that you shouldn't use this because, if it exists I mean why not? It is not prohibited, and additionally it is very helpful in some cases.

Regards,

Sebastián

##### 5 Comments

jonas
on 11 Sep 2018

Edited: Stephen23
on 20 Sep 2021

"you just attack this method and you do not present an formal solution"

This is simply untrue. On the contrary, from what I've seen Stephen always provides a concrete solution when possible. Here's an example from 5 minutes ago ( link ) which is nothing out of the ordinary.

"But please make us all a favor and present a code, not just words or links..."

This is the mindset I see in a lot of younger students today - avoid instructions like the plague and feel entitled to a solution. Remember that no one is getting paid to read about your dynamic variables, so you should appreciate any replies including links and words (whatever that means).

João
on 4 May 2022

Edited: João
on 4 May 2022

##### 14 Comments

João
on 6 May 2022

@Stephen23 Yes, equations must be abe to be defined systematically depending on the number of objects. And yes, for each time increment I know which equations to use and which x, y, z, ... to define.

For two objects I would have something like:

x, y, z, j = vars # variables

eq1 = (m*g - k*(y - x)) * dt - m * (j - c) # where m, k, g and c are known values

eq2 = (k*(y - x) + m*g - k*x * dt - m * (z - d) # where m, k, g and d are known values

eq3 = m * (c + d) - m * (z + j)

eq4 = m*g*y - 1/2*k*(y-x)**2 + m*g*x - 1/2*k*(x-l)**2 + m*g*l - 1/2*k*(l)**2

- 1/2*m*(j**2 - c**2) - 1/2*m*(z**2 - d**2)

For three objects I would have something like:

x, y, z, j, l, f = vars # variables

eq1 = (m*g - k*(y - x)) * dt - m * (j - c) # where m, k, g and c are known values

eq2 = (k*(y - x) + m*g - k*(x - l) * dt - m * (z - d) # where m, k, g and d are known values

eq3 = (k*(x - l) + m*g - k*l) * dt - m * (f - e) # where m, k, g and e are known values

eq4 = m * (c + d + e) - m * (z + j + f)

eq5 = m*g*y - 1/2*k*(y-x)**2 + m*g*x - 1/2*k*(x-l)**2 + m*g*l - 1/2*k*(l)**2

- 1/2*m*(j**2 - c**2) - 1/2*m*(z**2 - d**2) - 1/2*m*(f**2 - e**2)

eq6 = j - c # assuming that the velocities of object 1 and object 2 are the same during impact, i.e. when object 1 hits objects 2 they have the same velocity

So this made me realise I have to keep making assumptions as the number of objects grows.

So I need to solve the equations explicitly by Runge-Kutta or Central Difference methods...

I need to restructure my code. Thank you for the constructive input!

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