Assignment has more non-singleton rhs dimensions than non-singleton subscripts

3 vues (au cours des 30 derniers jours)
nt
nt le 13 Déc 2012
Réponse apportée : Tree le 19 Jan 2014
I am trying to figure out to format the numbers in the uitable to have thousands seperator. I am adding this undocumentd java code lines to achive it but gives me the below error. I can tell if i m using the undocumented code wrong or not? I appreciate any help on this.
c = tp(:,1);
c(1,:)=[];
g = spend_and_elas(:,1);
inf=java.text.DecimalFormat;
ho=char(inf.format(hist_out));
hr=char(inf.format(hist_rev));
k=grps(:,1);
k(1,:)=[];
k(:,2)=num2cell(ho);
% k(:,2)= mat2cell(ho);
k(:,3)=num2cell(hr);
set(handles.tpoints,'data',[ c num2cell(g)]);
set(handles.model,'data',k);
Assignment has more non-singleton rhs dimensions than non-singleton subscripts
Error in Top_Down_Optimizer_vn>pushbutton_Load_Data_vn_Callback (line 122)
k(:,2)=num2cell(ho);
  2 commentaires
Matt Fig
Matt Fig le 13 Déc 2012
Modifié(e) : Matt Fig le 13 Déc 2012
What is ho?
whos ho
And what is grps?
whos grps
nt
nt le 13 Déc 2012
i m tring to format " hist_out " and create it ho which will be formatted version of hist_out
grps is a matrix

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Réponse acceptée

Matt J
Matt J le 14 Déc 2012
If num2cell(ho) has more than 2 dimensions (you can check ndims(num2cell(ho))), then
k(:,2)=num2cell(ho);
makes no sense. The left hand side is being treated as 2-dimensional
  2 commentaires
nt
nt le 17 Déc 2012
thanks, also have this error when i use java to format thousand seperator for hist_rev. It works if when hist_rev is 1,1 but not when it s one dim vector. I can not come up with a solution to this. I appriciate if you suggest one. Regards,Nazmi
>> ho= char(inf.format(hist_rev)) Java exception occurred: java.lang.IllegalArgumentException: Cannot format given Object as a Number
at java.text.DecimalFormat.format(Unknown Source)
at java.text.Format.format(Unknown Source)
Matt J
Matt J le 17 Déc 2012
Modifié(e) : Matt J le 17 Déc 2012
Post it as a new question where Java SMEs can see it.

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Plus de réponses (1)

Tree
Tree le 19 Jan 2014
Has this been identified as a bug? If so can it be fixed?
I've run across the same error while trying to convert cells to chars via a for loop.

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