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how can read filename and rename

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SLIMANE BENYAHIA
SLIMANE BENYAHIA on 4 Sep 2016
file name is : GSM-2_2010060-2010090_0031_EIGEN_G---_005a.gfc
renamed file : 201003.txt
2010 is years
03 is monthe
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Walter Roberson
Walter Roberson on 4 Sep 2016
Is the "2010" part to be extracted from the 2010060 or from the 2010090 ? Is the 03 to be extracted from the middle digits of the "0031" ? Or is the "060" and "090" to be understood as day number within year and the second of the two is to be converted to a month number?
SLIMANE BENYAHIA
SLIMANE BENYAHIA on 5 Sep 2016
year is the same, 060 and 090 are days of 365, i want to convert to month( middle of 60 and 90 is 75, 75 of 365 is day of march(03), i have 1386 files like this

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Answers (2)

R. B.K.
R. B.K. on 4 Sep 2016
Renaming a File in the Current Folder:
In the current folder, rename oldname.m to newname.m:
movefile('oldname.m','newname.m')
Source: movefile function.

Walter Roberson
Walter Roberson on 5 Sep 2016
Use array indexing to exact the day number strings, and use str2double() to convert to numeric. Average and use as a number of days offset into the year. Convert to month number.
For example,
d = datevec(datenum('2011', 'YYYY') + 75 - 1);
month_number = d(2);
The "- 1" is because datenum() applied to a year will already refer to January 1, day 1 in the year.

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