Reordering string arrays if string is >9

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Jason
Jason 2020 年 3 月 18 日
コメント済み: Jason 2020 年 3 月 18 日
Hello, I have a string array that has been sorted but puts 10 next to 1
b =
1×6 string array
"1" "10" "3" "5" "7" "r1"
My strings are upto 12.
I want the last element to remain where it is, and to move any strings > 9 if exist to the end -1
i.e.
"1" "3" "5" "7" "10" "r1"
and if I have
"1" "10" "12" "5" "7" "r1"
then to be
"1" "5" "7" "10" "12" "r1"
This is my attempt:
b
Id10 = find(contains(b,"10"))
if Id10>0
e=b(Id10); b(Id10)=[]; b(end-1)=e;
out{j,1}=b;
b
end
Id11 = find(contains(b,"11"))
if Id11>0
e=b(Id11); b(Id11)=[]; b(end-1)=e;
out{j,1}=b;
b
end
Id12 = find(contains(b,"12"))
if Id12>0
e=b(Id12); b(Id12)=[]; b(end-1)=e;
out{j,1}=b;
end
b
but it seems to be over writing the end-1 element
b =
1×7 string array
"1" "10" "11" "12" "3" "6" "r1"
Id10 =
2
b =
1×6 string array
"1" "11" "12" "3" "10" "r1"
Id11 =
2
b =
1×5 string array
"1" "12" "3" "11" "r1"
Id12 =
2
b =
1×4 string array
"1" "3" "12" "r1"
Thanks for any help
  2 件のコメント
BobH
BobH 2020 年 3 月 18 日
This statement shortens the array (same for Id11/12)
b(Id10)=[];
All you were missing is a statement to restore the size, by copying the last element onto the end.
Then you could overwrite end-1
b(end+1) = b(end); % copy last element onto end
b(Id10)=[]; % remove '10'
b(end-1) = e; % overwrite end-1
Jason
Jason 2020 年 3 月 18 日
Perfect, thankyou very much!

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採用された回答

BobH
BobH 2020 年 3 月 18 日
sort_nat is powerful, and has helped me several times.
An alternative for simple number/not-number values is straightforward
  1. find the locations of the numbers
  2. sort those
  3. append the not-numbers
z = cellfun(@(X) logical(~isempty(str2num(X))),b); % z shows locations of number strings
n = cellfun(@str2num, b(z)); % into numerical
[~, ix] = sort( n ); % ix tells what order to pull from b
r = b(ix); % fill the result using sorted numbers
r{end+1} = b{~z}; % append the not-numbers
  6 件のコメント
Jason
Jason 2020 年 3 月 18 日
Thanks Bob, that will always be the case for me
BobH
BobH 2020 年 3 月 18 日
Just for completeness, for those who might have numbers anywhere, here's a more general solution
  1. find the locations of the numbers
  2. build a new numeric vector, using a placeholder number for the not-numbers. Inf or -Inf are good choices that will sort to the end or the front
  3. use the sort index to pull from b
z = cellfun(@(X) logical(~isempty(str2num(X))),b); % z shows locations of number strings
n(~z) = Inf; % placeholder value; makes not-strings sort to end
n(z) = cellfun(@str2num, b(z));
[~, ix] = sort( n );
r = b(ix);

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その他の回答 (1 件)

dpb
dpb 2020 年 3 月 18 日
  1 件のコメント
Jason
Jason 2020 年 3 月 18 日
Thanks, but Im running into an error
b =
1×6 string array
"1" "3" "6" "8" "9" "r1"
Index in position 1 exceeds array bounds.
Error in sort_nat (line 62)
num_val(i,z(i,:)) = sscanf(sprintf('%s ',digruns{i}{:}),'%f');
Error in PlateSeq2JB/LockPositionsButtonPushed (line 693)
[b,~] = sort_nat(b)

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