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

2 次查看(过去 30 天)
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

采纳的回答

Cris LaPierre
Cris LaPierre 2021-1-18
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;
  8 个评论
Cris LaPierre
Cris LaPierre 2021-1-19
编辑:Cris LaPierre 2021-1-19
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).

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更多回答(1 个)

Jason
Jason 2021-1-19
编辑:Jason 2021-1-19
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
  1 个评论
Cris LaPierre
Cris LaPierre 2021-1-19
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'.

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