Matlab Array indixing error

Question

Ali 2022-11-18

0
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1855188-matlab-array-indixing-error

评论： Ali 2022-11-18

采纳的回答： KSSV

在 MATLAB Online 中打开

Hi,

On Matlab 2022 i have the following error :

Array indices must be positive integers or logical values.
Error in mdl2List>NomBarreMdl (line 1022)
y1=Nom(end-1:end);

I don't know why it gives me this error because when i use the debugger and i pause on the same line and i use this line on the matlab command it works perfectly. But in the function it doesn't.

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Ali 2022-11-18

在 MATLAB Online 中打开

%% analyse du nom d'une barre pour extraire l'info Tension
function [NomB,Tension,Flag]=NomBarreMdl(Nom)
% NOMBARRE analyse le nom d'une barre. Si une information Tension est présente dans
% le nom de la barre (c'est-à-dire si la dernière ligne du nom est un
% nombre terminé par v, V, kv ou kV, alors Flag est positionné à 1 et la fonction retourne la
% tension de la barre. On ajoute U= dans le nom de la barre pour pouvoir le détecter ultérieurement.
% dans le cas contraire, Flag vaut 0.
%
% Module créé lors de la ModifN14
x=strfind(Nom,sprintf('\n'));
% Recherche de l'unité kV dans le nom de la barre
y1=Nom(end-1:end); % THIS IS WHERE THE ERROR IS
if ~isempty(x)&&(min(y1=='kv')||min(y1=='kV')||min(y1=='KV')||min(y1=='Kv'))
    Tension=1000*str2num(Nom(x(end)+1:end-2));
    if isempty(Tension)
        Flag=0;
        NomB=Nom;
    else
        Flag=1;
        NomB=[Nom(1:x-1) ' U=' num2str(Tension) 'V'];
    end
else
    NomB=Nom;
    Tension=[];
    Flag=0;
end
% Recherche de l'unité V dans le nom de la barre si kV non trouvé.
if Flag == 0
    y=Nom(end);
    if ~isempty(x)&&((y=='v')||(y=='V'))
        Tension=str2num(Nom(x(end)+1:end-1));
        if isempty(Tension)
            Flag=0;
            NomB=Nom;
        else
            Flag=1;
            NomB=[Nom(1:x-1) ' U=' num2str(Tension) 'V'];
        end
    else
        NomB=Nom;
        Tension=[];
        Flag=0;
    end
end

the third line of the function is where the error is. It's supposed to give the result 'kV' with Nom = #TrAF115kV so i don't understand why

Stephen23 2022-11-18

"so i don't understand why"

Your function assumes that NOM has two or more elements, but does not check this in any way. This is a bug.

Ali 2022-11-18

oh ok thx i will try that

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Answer 1

KSSV 2022-11-18

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1855188-matlab-array-indixing-error#answer_1104033

在 MATLAB Online 中打开

In MATLAB the indices should be posittive integers or logicals. If not it will throw the error which you have shown.

% EXample
A = rand(1,10) ;l 
A(1)   % no error 
A(end)  % no error; gives last element 
% logica lindexing 
idx = A>0.5 ; 
A(idx)
% 
A(0.5) % error. Index cannot be fraction 
A(0)   % error index cannot be 0 

So, I suspect in your case, you have a variable on the name end. Check

whos end

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Steven Lord 2022-11-18

在 MATLAB Online 中打开

Is Nom a char vector or a string?

Nom_char = '#TrAF115kV'
Nom_char = '#TrAF115kV'
Nom_string = string(Nom_char)
Nom_string = "#TrAF115kV"
whos Nom_char Nom_string
  Name            Size            Bytes  Class     Attributes

  Nom_char        1x10               20  char                
  Nom_string      1x1               156  string              

Nom_char has 10 characters and so asking for the last two characters works.

Nom_char(end-1:end)
ans = 'kV'

Nom_string is 1 string and so asking for the last two elements using indexing wouldn't work. I'll show that as the last line of code in this comment. You could retrieve the last two characters of either using extractAfter instead of indexing.

c1 = extractAfter(Nom_char, strlength(Nom_char)-2)
c1 = 'kV'
s1 = extractAfter(Nom_string, strlength(Nom_string)-2)
s1 = "kV"
Nom_string(end-1:end) % This doesn't work
Array indices must be positive integers or logical values.

Ali 2022-11-18

ok it was a string that's why it didn't work thx for the answer

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Matlab Array indixing error

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