How to sort file when reading them with dir ?
60 vues (au cours des 30 derniers jours)
Afficher commentaires plus anciens
Tural
le 17 Déc 2014
Commenté : Chandra Shekhar Lohani
le 29 Nov 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?
0 commentaires
Réponse acceptée
Stephen23
le 18 Déc 2014
Modifié(e) : Stephen23
le 18 Avr 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'
Plus de réponses (4)
per isakson
le 17 Déc 2014
Modifié(e) : per isakson
le 17 Déc 2014
See
and there are more in the File Exchange
2 commentaires
per isakson
le 17 Déc 2014
Modifié(e) : per isakson
le 18 Déc 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
le 18 Déc 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 commentaires
Steve Chavez
le 21 Mar 2017
Modifié(e) : Steve Chavez
le 21 Mar 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 commentaire
Stephen23
le 21 Mar 2017
Modifié(e) : Stephen23
le 22 Mar 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'
Voir également
Catégories
En savoir plus sur Shifting and Sorting Matrices dans Help Center et File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!