topic Re: How to use regular expression to extract Name of a File and it's extension ? in SAS Programming

How to use regular expression to extract Name of a File and it's extension ?

Midi — Tue, 19 Nov 2019 18:27:52 GMT

Hi everyone , i have this file name:

20191102_Registre_Distributions_isan.csv;

and i want to extract the first part which would be ' Registre_Distributions ' and the second part which would be ' isan ' , and i'm wondering how it could be achieved using the prxchange function.

any help would be much appreciated , thank you .

Re: How to use regular expression to extract Name of a File and it's extension ?

Tom — Tue, 19 Nov 2019 18:41:47 GMT

To write a regular expression you need rules. What are the rules that describe how the filenames are constructed? And what parts you want to extract?

The easiest would be if the string always consists of four parts separated by underscore. Then there would be no need for regular expressions at all. Simple SCAN() function will work.

first_part = catx(('_',scan(filename,2,'_'),scan(filename(3,'_'));
second_part = scan(filename,4,'_.');

Re: How to use regular expression to extract Name of a File and it's extension ?

PGStats — Tue, 19 Nov 2019 22:24:30 GMT

Assuming that the file name consists of a numeric field followed by an underscore, and then an alphanumeric field, an underscore, and an alphabetic word:

data _null_;
txt = "20191102_Registre_Distributions_isan.csv";
firstþart = prxchange("s/(\d+_)(\w+)_([a-z]+)\.csv/\2/io", 1, txt);
secondPart = prxchange("s/(\d+_)(\w+)_([a-z]+)\.csv/\3/io", 1, txt);
put (_all_) (=/);
run;

 txt=20191102_Registre_Distributions_isan.csv
 firstþart=Registre_Distributions
 secondPart=isan

Re: How to use regular expression to extract Name of a File and it's extension ?

Reeza — Tue, 19 Nov 2019 23:16:26 GMT

SCAN() is infinitely easier.

file_name = scan (var_name, 1, ".");
ext = scan(var_name, 2, ".");