How to sort file when reading them with dir ?
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Tural
el 17 de Dic. de 2014
Comentada: Chandra Shekhar Lohani
el 29 de Nov. de 2022
I have list of text files , they are ordered like
- doc_1.txt
- doc_2.txt
- ..
- doc_20.txt
when I use
allFiles = dir('*.txt');
it is getting files in this order
- doc_1.txt
- doc_10.txt
- doc_11.txtHow can I get these files in correct order?
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Respuesta aceptada
Stephen23
el 18 de Dic. de 2014
Editada: Stephen23
el 18 de Abr. de 2021
You could download my FEX submission natsortfiles:
and use it to directly sort the output from DIR:
allFiles = natsortfiles(dir('*.txt'));
The function natsortfiles does not perform a naive natural-order sort, but also sorts the filenames and file extensions separately so that the file extension period character does not influence the sort output:
>> A = {'doc_10.txt'; 'doc_20.txt'; 'doc_2.txt'; 'doc_1.txt'; 'doc_11.txt'};
>> sort(A) % gives the wrong order!
ans =
'doc_1.txt'
'doc_10.txt'
'doc_11.txt'
'doc_2.txt'
'doc_20.txt'
>> natsortfiles(A) % the correct number order :)
ans =
'doc_1.txt'
'doc_2.txt'
'doc_10.txt'
'doc_11.txt'
'doc_20.txt'
Más respuestas (4)
per isakson
el 17 de Dic. de 2014
Editada: per isakson
el 17 de Dic. de 2014
See
and there are more in the File Exchange
2 comentarios
per isakson
el 17 de Dic. de 2014
Editada: per isakson
el 18 de Dic. de 2014
"[...] external functions"   why is that?   AFAIK: The alternative is to make your own function (or script). Both these FEX-contributions shows how to do that. They are well documented.
Andrei Bobrov
el 18 de Dic. de 2014
n = dir('doc_*.txt');
n1 = {n.name};
z = regexp(n1,'(?<=doc_)\d*(?=\.txt)','match');
z1 = str2double(cat(1,z{:}));
[~,ii] = sort(z1);
n01 = n1(ii);
z2 = z1(ii);
lga = floor(log10(z2)) + 1;
ml = max(lga);
ptr = ['doc_%0',num2str(ml),'d.txt'];
p = z1(ii);
k = lga < ml;
p1 = p(k);
n02 = n01(k);
if any(k)
for jj = 1:numel(p1)
movefile(n02{jj},sprintf(ptr,p1(jj)));
end
end
0 comentarios
Steve Chavez
el 21 de Mzo. de 2017
Editada: Steve Chavez
el 21 de Mzo. de 2017
Maybe you could use the names of the files
- Choose the order from the name to obtain indexes and then use the indexes to sort the list.
As an example: The names of the variables were chosen assuming that the number in the txts represent days.
Use dir to obtain a list into a struct
list=dir('*.txt')
Convert the struct to cell array
listcell=struct2cell(list)
[a b]=size(listcell)
Create a cell array
dia_string=cell(b,1)
A simple loop to obtain the order (order of days) from the names. If nombre is 'doc_20.txt' nombre(5:6) is equal to 20
for k=1:b
nombre=listcell{1,k};
dia_string{k}=nombre(5:6);
end
Use sort and save the index
[diaorden ind]=sort(dia_string);
Use the index in the cell array
list_2=list(ind) % list_2 is the list ordered as expected
1 comentario
Stephen23
el 21 de Mzo. de 2017
Editada: Stephen23
el 22 de Mzo. de 2017
@Steve Chavez: the original question lists some example filenames:
doc_1.txt
doc_10.txt
doc_11.txt
Notice that the number of digit characters is not constant: in some cases one, some two, and perhaps more. Currently your answer takes a fixed number of characters (two) from a fixed location in the string, which means that it will give incorrect sort orders for any numbers with not-two digits, and that if the number is located elsewhere in the string the output will be rubbish. So it is not a very general solution.
However note also that if the files were named with constant-width numbers, as your answer uses, then there is no point in this answer (or any of the answers) because then a simple sort gives the correct output. Compare your answer:
>> C = {'doc_99.m','doc_10.m','doc_50.m','doc_01.m'};
>> [a,b]=size(C);
>> dia_string=cell(b,1);
>> for k=1:b, nombre=C{1,k}; dia_string{k}=nombre(5:6); end
>> [diaorden,ind]=sort(dia_string);
>> C(ind)
ans =
'doc_01.m' 'doc_10.m' 'doc_50.m' 'doc_99.m'
which gives exactly the same answer as simply calling sort:
>> sort(C)
ans =
'doc_01.m' 'doc_10.m' 'doc_50.m' 'doc_99.m'
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