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How to extract information from the filename?

Asked by SGMukherjee on 9 May 2018
Latest activity Commented on by Pawel Jastrzebski on 9 May 2018
I have almost 6000 files with names like Arenosillo.2005.344.13.49.G13.txt where 2005 is the year, 344 is the number of day of that year and 13:49 is the time. I want to extract all these information from the filename. Please help me.

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2 Answers

Answer by Pawel Jastrzebski on 9 May 2018
Edited by Pawel Jastrzebski on 9 May 2018

This will get you the structure of all of the text files in your current folder:
x = dir('*.txt') % structure
In my case it's:
x =
6×1 struct array with fields:
name
folder
date
bytes
isdir
datenum
And this will extract all of the text file names from the structure to the cell:
y = {x.name}
Once you've got this far, it should be easy to extract the name of the file i.e with the FOR loop and break it down to the information that you need.

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I already got this. list=dir('*.txt'); for n=1:length(list); filename=list(n).name; end But I want to extract the year, day number and time from the filename.
Try strsplit :
>> s = 'Arenosillo.2005.344.13.49.G13.txt'
s =
'Arenosillo.2005.344.13.49.G13.txt'
>> c = strsplit(s,'.')
c =
1×7 cell array
Columns 1 through 6
{'Arenosillo'} {'2005'} {'344'} {'13'} {'49'} {'G13'}
Column 7
{'txt'}

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Answer by jonas
on 9 May 2018

If you know something about the structure, then this is quite simple. Let's say the structure is [year.day.hour.min], with some arbitrary string before.
Use regexp to find the separate digits:
string='Arenosillo.2005.344.13.49.G13.txt';
[ind1,ind2]=regexp(string,'\d+')
This gives you the index at start and end of all digits, respectively. Then extract that information:
y=string(a(1):b(1))
d=string(a(2):b(2))
h=string(a(3):b(3))
m=string(a(4):b(4))
It's a bit more complicated if you have other numbers in your string, but can still be solved with regexp

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