How to convert the name of a table or cell array to string

Dear Experts,

I have a cell array named "cat" and a table named "dog" So cat has a series of data within. Likewise for dog.

I just want to extract the name of the cell array (cat) and name of the table (dog) to be label in title when plotting a figure.

Any ideas? LL

Answer 1

[Star Strider]

See if this does what you want:

cat = {rand(1, 10)};                                % Create Data
dog = table(rand(1, 15));                           % Create Data

varlist = whos;                                     % Workspace Variables
var_class = {varlist.class};                        % Classes
cell_idx = strcmp(var_class, 'cell');               % Logical Indices of Cell Arrays
cell_names = {varlist(cell_idx).name};              % Cell Array Variable Names
tabl_idx = strcmp(var_class, 'table');              % Logical Indices of Table Arrays
tabl_names = {varlist(tabl_idx).name};              % Table Array Variable Names

figure(1)
subplot(2,1,1)
plot(cat{:})
title(cell_names{1})
grid
subplot(2,1,2)
plot(table2array(dog))
title(tabl_names{1})
grid

EDIT — I forgot one was a cell and one was a table.

This works, but you will probably have to adapt it to your application.

[Star Strider]

Cheerful’s ‘Answer’ is duplicated here —

Hi Expert

I think there is a problem. I have many tables when I run my program and it picks up all the table names. So it is quite easily to mess up the names when I simply index it by tabl_names(i)

thanks

[Star Strider]

If there are more than one table arrays in your workspace and you want to choose the one to plot, I would add a listdlg call.

You can’t make it completely automatic if the user has to choose the table to plot. Unless you already know the table array name (in which situation that part of my code is then irrelevant), you have to incorporate some sort of user interaction to choose the correct table, and listdlg is likely the most straightforward to program and use.

I leave the rest to you.

EDIT — (26 Sep 2016 at 03:04 Z)

I had to use eval, but this code will discover all the table arrays in your workspace, and allow you to choose the table array you want to plot. You may need to change the ‘td’ assignment and plot call to work with it:

cat = {rand(1, 10)};                                    % Create Data
dog = table(rand(1, 15));                               % Create Data
cow = table(rand(1,  5));                               % Create Data

varlist = whos;                                         % Workspace Variables
var_class = {varlist.class};                            % Classes
cell_idx = strcmp(var_class, 'cell');                   % Logical Indices of Cell Arrays
cell_names = {varlist(cell_idx).name};                  % Cell Array Variable Names
tabl_idx = strcmp(var_class, 'table');                  % Logical Indices of Table Arrays
tabl_names = {varlist(tabl_idx).name};                  % Table Array Variable Names

[choose_table, v] = listdlg('PromptString','Choose a table to plot:', 'SelectionMode','single', 'ListString',tabl_names);
vnv = {varlist.name};                                   % Variable Names Vector
vnn = strcmp({varlist.name}, tabl_names{choose_table}); % Variable Names Number
plot_table = eval(vnv{vnn});                            % To Others: If There’s A Way To Do This Without ‘eval’, Please Share The Code
td = table2array(plot_table);                           % Evaluate Chosen Table And Return Numeric Values

figure(1)
subplot(2,1,1)
plot(cat{:})
title(cell_names{1})
grid
subplot(2,1,2)
plot(td)
title(tabl_names{choose_table})
grid

I leave it to you to adapt it to your needs.

Answer 2

[Cheerful]

Hi Expert

I think there is a problem. I have many tables when I run my program and it picks up all the table names. So it is quite easily to mess up the names when I simply index it by tabl_names(i)

thanks

[Image Analyst]

Why don't you know what data you're plotting? You must know that in order to plot, so just make up the string. If the user might pick any old arbitrary thing to plot, then get their choice and use it in an "if" or switch statement and inside there (when you'll definitely know what they picked) make up the string.

[Cheerful]

I have data from csv file. Each csv file is already named in other system. So when I convert to csv file to table in matlab, the name of the file remains.

And using the suggestion given, I have to know the index (i) of the tabl_names(i) in order to get the correct names. I would be expanding the program and so, there would be more and more tables in the programs and the required index will be changed. Hence I would retrieve the wrong name.

In summary, the solution suggested required you to know the index or position of the tables' name in advance. You wont know the position or index as a result of modification of program.

I have google hard. Seem like there is no way to get the name of the table ie tableA.name which is "tableA"

[Stephen23]

@Cheerful: do you know why this is so hard? Because this is very bad program design. Even beginners should be able to understand that when a program algorithm and the data are well designed, then the code is simpler and more robust. Imagine that I put my data in nested cell arrays and complicated nested structures: this is going to be much harder to write code for (as well as being slower and buggier) than if I simply put my data into a simple numeric array.

You have chosen to design a program in a very buggy and pointlessly complicated way. Now you are finding out that, because of this bad design, it is hard to write the code for it. This is not something specific to MATLAB, because good design is a requirement for good code in any programming language.

You really need to learn more about good practices for writing code, and how to create robust algorithms.

Using indices is robust, and there are other methods too. Read these:

https://www.mathworks.com/matlabcentral/answers/228557-experts-of-matlab-how-did-you-learn-any-advice-for-beginner-intermediate-users

https://www.mathworks.com/matlabcentral/answers/304528-tutorial-why-variables-should-not-be-named-dynamically-eval

"Each csv file is already named in other system. So when I convert to csv file to table in matlab, the name of the file remains"

There is absolutely no reason why variables that are internal to your program have to have the same names as the files themselves. You could import them into one variable (i.e. better design) and then you entire question (and all of the slow and buggy solutions to it) is no longer needed.

Summary: fix the data import, rather than trying to create buggy code to fix the problems that the import causes.

Answer 3

[Walter Roberson]

function S = name_of_variable(SomeVariable)
   S = inputname(SomeVariable);

To use:

title( name_of_variable( cat ) )

and that would use 'cat' as the title.

[Cheerful]

Doesnt work

>> title(name_of_variable(cat)) Error using inputname Scalar integer value required, but non-numeric value seen

Error in name_of_variable (line 2) S = inputname(SomeVariable);

[Walter Roberson]

 function S = name_of_variable(SomeVariable)
   S = inputname(1);