Read files in a specific order
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Isma_gp
am 23 Mai 2018
Bearbeitet: Stephen23
am 22 Jun. 2021
Hi, I have a series of files named with a date, for every third hour i.e. "NUM_1985_6_4_0_0.dat", "NUM_1985_6_4_3_0.dat", "NUM_1985_6_4_6_0.dat" ,....., "NUM_1985_6_4_21_0.dat",......, "NUM_1985_6_10_21_0.dat", "NUM_1985_6_11_0_0.dat". I'm using the dir command before using the fopen command. Is there any option when using the dir command to be able to list the files on a specific order. I would like to read the files by dates, however, the dir command gives me a list of files where the first one is "NUM_1985_6_10_0_0.dat" following a different logic, where the number 1 goes first in the list.
Thanks
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Stephen23
am 23 Mai 2018
Bearbeitet: Stephen23
am 18 Apr. 2021
Download my FEX submission natsorfiles, which was designed to fix the exact problem that you are having. Use it to simply sort the output from DIR:
D = '.'; % folder path
S = dir(fullfile(D,'*.dat'));
S = natsortfiles(S); % alphanumeric sort by filename
for k = 1:numel(S)
F = fullfile(D,S(k).name)
... import/process file F
end
Here is natsortfiles used on your example filenames:
>> C = {'NUM_1985_6_4_0_0.dat';'NUM_1985_6_4_3_0.dat';'NUM_1985_6_4_6_0.dat';'NUM_1985_6_4_21_0.dat';'NUM_1985_6_10_21_0.dat';'NUM_1985_6_11_0_0.dat'};
>> C = C(randperm(numel(C))) % "random" order
C =
NUM_1985_6_11_0_0.dat
NUM_1985_6_4_0_0.dat
NUM_1985_6_4_21_0.dat
NUM_1985_6_4_3_0.dat
NUM_1985_6_4_6_0.dat
NUM_1985_6_10_21_0.dat
>> D = natsortfiles(C) % alphanumeric sort
D =
NUM_1985_6_4_0_0.dat
NUM_1985_6_4_3_0.dat
NUM_1985_6_4_6_0.dat
NUM_1985_6_4_21_0.dat
NUM_1985_6_10_21_0.dat
NUM_1985_6_11_0_0.dat
4 Kommentare
Luke Riddell
am 22 Jun. 2021
Hi Stephen,
I've tried this but I get an error returning.
When using:
S = dir(fullfile(FinalFolderloc,'*.ASC'))
S = natsortfiles(S)
It returns the error in line with the natsortfiles function:
" Error using bsxfun
Non-singleton dimensions of the two input arrays must match each other.
mat(logical(bsxfun(@eq,vec,nmn).')) = fnm; % TRANSPOSE bug loses type (R2013b)"
Do you have any advice?
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Rik
am 23 Mai 2018
You can sort the output or dir any way you like. See below for an example.
filelist=dir('NUM_*.dat');
dates=[filelist.datenum];
[~,order]=sort(dates);
filelist=filelist(order);
3 Kommentare
Stephen23
am 23 Mai 2018
Bearbeitet: Stephen23
am 23 Mai 2018
This is a very fragile solution, because it assumes that the files themselves were written in the same order as the data was recorded: this will break easily, e.g. any time a file is edited, or if the files are copied from one folder to another.
"Can we sort the name field from dir?"
Yes. See my answer to know how.
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