Reversing parts of (even) items in a listbox (cellarray)

Hello, I have items in a listbox (so cell array) - the integers before the "r" can be any number from 1-12.

My aim is to take any line containing an even number after r (r2, r4 etc) and reverse the other numbers and then put back into the list box

So line 2 & 4 would change to

1 2 3 r2  -> 3 2 1 r2
1 2 3 r4  -> 3 2 1 r4

This was my attempt.

               items=app.ListBoxPos.Items
                  nl=numel(items);
                  rstr=["r2","r4","r6","r8"];
                  s1=[];
                  out={};
                  for i=1:nl                      
                      line=char(items(i));
                      if contains(line, rstr)                       
                        tokens = strsplit(line, 'r' )                                                                   
                        t=fliplr(tokens{1,1})
                        new=[t,' r',tokens{1,2}]
                        app.ListBoxPos.Items{i}=new
                        %Not sure what to do now!
                      end
                  end    

Not sure if this is the most elegant way, and it falls over trying to overwrite the line in the listbox.

Thanks for any help

Jason

Answer 1

[Cris LaPierre]

You can use mod to determine if the last number is even or not. I would then use indexing to reverse the order of the rows that end in an even number.

items=app.ListBoxPos.Items
nl=numel(items);
for i=1:nl              
  if mod(items{i}(end),2)==0
      out{i}=items{i}([end-3:-1:1 end-2:end]);
  else
      out(i)=items(i);
  end
end 

app.ListBoxPos.Items = out;

[Jason]

Hi, thanks for this. Im getting all rows satisfying the condition

if mod(items{i}(end),2)==0

[Cris LaPierre]

How do you populate your listbox? This is how I tested it.

fig = uifigure;
ListBoxPos = uilistbox(fig,'Items',{'1 2 r1','1 2 3 r2','1 2 r3','1 2 3 r4'}); 

Save your variable items to a mat file and attach it to your post using the paperclip icon.

[Jason]

I found if I do

L=str2num(items{i}(end) Then do If mod(L,2)==0

It works. (Sorry for not formatting in code as Im using a phone)

[Cris LaPierre]

This update will handle double digit numbers at the end (e.g. 'r10' and 'r12').

r = strfind(items{i},' r');
out{i}=items{i}([r-1:-1:1 r:end]);

[Jason]

Hi, do you mean use that instead of

out{i}=items{i}([end-3:-1:1 end-2:end]);

[Jason]

This seems to work (have to convert to integers first

                           item=items{i};
                           n=item(end)
                           token = strtok(item,'r');
                           tn=str2num(token);
                           t=fliplr(tn);
                           out{i}=[num2str(t),' r',n]

[Cris LaPierre]

What version of MATLAB are you using?

It's good to see you coming up with your own approach. Nice job. Just be aware that this solution suffers from the original problem - it doesn't work when the final number has 2 digits (10 or 12).

[Jason]

Im using 2020b

Answer 2

[Jason]

Here is my solution based on Cris' answer:

            val = app.SNAKESCANCheckBox.Value;            
            switch val
                case 1
                  items=app.ListBoxPos.Items;
                  nl=numel(items);
                  out={};
                  for i=1:nl                    
                      L=str2num(items{i}(end));                      
                      if mod(L,2)==0                           
                         out{i}=items{i}([end-3:-1:1 end-2:end]);
                      else                          
                          out{i}=items{i};
                      end
                  end 

                  app.ListBoxPos.Items=out;
                case 0    
            end
   

I thought this worked, but when I have a number 10,11 or 12 (my numbers can be 1-12), then it doesn't quite work - Its reversing the order of the double digits

[Cris LaPierre]

When the number at the end is 2 characters, it would require a different approach. Perhaps finding the 'r' and using that to set the index rather the 'end'.