sprintf of a variable and string name

it's needed to display a variable string name. for example:
for i=1:w I have these:
i=1 >>>> name='t56757w_e_l_l'
i=2 >>>> name='h_a_h_a_h_a067'
i=3 >>>> name='a56754no_the_r'
.
.
.
't56757w_e_l_l' and 'h_a_h_a_h_a067' and 'a56754no_the_r' ,... are the names of matrix.
Problem: its needed to plot a figure in these names that i syaed above but when i use:
figure('Name',sprintf('%s',name))
sprintf('%s',name) gives numbers of arrays of matrix and not the name of that.

2 commentaires

Fangjun Jiang
Fangjun Jiang le 17 Sep 2011
You might have many variables. How do you know 't56757w_e_l_l' is corresponding to i=1, 'h_a_h_a_h_a067' is corresponding to i=2, etc.?
mohammad
mohammad le 17 Sep 2011
because I give these names manually in each loop.
in other words in each loop, it asks to import a .xls file and read this file and that must put in a matrix that name of this matrix must be same with .xls file name.
so these name in each loop are the names of .xls files

Connectez-vous pour commenter.

 Réponse acceptée

Fangjun Jiang
Fangjun Jiang le 17 Sep 2011
In that case, the name is actually given manually. To avoid entering the same name twice, maybe you can do this.
Name='t56757w_e_l_l';
Data=xlsread([Name,'.xls']);
assignin('base',Name,Data);
sprintf('%s',Name)
You know the name of the variable. You need to use this variable. You also want to use the variable name in its string format.
The name sounds very irregular. You have to specify it manually. Why can't you also specify the string since you have already specified the variable name? It doesn't make a difference if you specify it once, or twice. Like.
plot(t56757w_e_l_l); title('t56757w_e_l_l');
Or, if you are willing to make a function, you can do this
function PlotAndTitle(a) plot(a); title(inputname(1));
You can call it this way:
PlotAndTitle(t56757w_e_l_l)

6 commentaires

mohammad
mohammad le 17 Sep 2011
I did exactly this in another function m-file. now 't56757w_e_l_l' is exist in workspace. but in this function( plotting(Name)) that i want plot 't56757w_e_l_l', 'Name' converts to data!!! and this is really strange. there is no problem in this command:plot(Name)
because 'Name' is data (i mean a matrix)!
but it must be a string. problem there is where i want to determine name of figure that i want name of figure to be 't56757w_e_l_l'
Fangjun Jiang
Fangjun Jiang le 17 Sep 2011
see update
mohammad
mohammad le 17 Sep 2011
very nice
i need to make function because there are more than 300 .xls files the names of these must automatically be the names of matrix too, in a for-loop.
with your helping its solved:
name=[inputname(1),'and' inputname(2)]
set(gcf,'Name',name)
Image Analyst
Image Analyst le 17 Sep 2011
I presume by now you've noticed that underlines in the string convert the following characters to subscripts in the title() function. Doesn't happen for figure() though - the underlines show up in that case.
Fangjun Jiang
Fangjun Jiang le 17 Sep 2011
Good catch, Image Analyst! I didn't pay attention to the underscore effect in the title() string.
Image Analyst
Image Analyst le 1 Nov 2011
If you want to have an actual underline in the string, you can replace the underline with a backslash underline with this code:
yourString = strrep(yourString , '_', '\_');
before you call title() or text().
This will make sure you get the underline to show up when you use title() and text().

Connectez-vous pour commenter.

Plus de réponses (1)

Malcolm Lidierth
Malcolm Lidierth le 17 Sep 2011

0 votes

No. sprintf(%s,name) gives:
Error: Expression or statement is incorrect--possibly unbalanced (, {, or [.
(In fact missing single quotes). Try pasting the actual code.
figure('Name',sprintf('%s',name)) works, and will make sense if name is a char array.

1 commentaire

mohammad
mohammad le 17 Sep 2011
Thanks dear, I used '%s', here i wrote wrong.

Connectez-vous pour commenter.

Catégories

En savoir plus sur Characters and Strings dans Centre d'aide et File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by