如何在python中使用正则表达式组仅获得一行的一部分？

Question

How to get only a part of a line using regex group in python ? I have a database of one entry per line and I want to split it into files according to month and day data at the beginning of a line but I want only to output a line without first 21 characters. Here is a quick sample of the database:

01-01-1989-06:30:00| Stefan Reinartz; 1.1.1989; 06:30; +01; Engelskirchen,Germany; 50n59; 7e24; M;
01-01-1996-08:40:00| Dawid Kwiatkowski; 1.1.1996; 08:40; +01; Gorzów Wielkopolski,Poland; 52n44; 15e15; M;
01-01-2001-01:30:00| Liam Flockhart; 1.1.2001; 01:30; -08; San Diego,California; 32n43; 117w09; M;
01-02-1467-00:20:00| King of Poland Sigismund I the Old; 2.1.1467; 00:20; +00:21:33; Kozienice,Poland; 51n35; 21e33; M;
01-02-1746-09:00:00| Duke of Rambouillet Louis Marie; 2.1.1746; 09:00; -00:03:41; Madrid,Spain; 40n24; 3w41; M;
01-02-1784-01:00:00| Duke of Saxe-Coburg and Gotha Ernst I; 2.1.1784; 01:00; +00:10:58; Coburg,Germany; 50n15; 10e58; M;

Desired output File 01-01.zbs:

Stefan Reinartz; 1.1.1989; 06:30; +01; Engelskirchen,Germany; 50n59; 7e24; M;
Dawid Kwiatkowski; 1.1.1996; 08:40; +01; Gorzów Wielkopolski,Poland; 52n44; 15e15; M;
Liam Flockhart; 1.1.2001; 01:30; -08; San Diego,California; 32n43; 117w09; M;

Output File 01-02.zbs:

King of Poland Sigismund I the Old; 2.1.1467; 00:20; +00:21:33; Kozienice,Poland; 51n35; 21e33; M;
Duke of Rambouillet Louis Marie; 2.1.1746; 09:00; -00:03:41; Madrid,Spain; 40n24; 3w41; M;
Duke of Saxe-Coburg and Gotha Ernst I; 2.1.1784; 01:00; +00:10:58; Coburg,Germany; 50n15; 10e58; M;

I used the beginning to sort them by each day of the year and to split the file accordingly. But I don't want to output the first 21 chars of each line so I am trying to use regex group to do this, like this:

re.search("^[0-9]{2}-[0-9]{2}-[0-9]{4}-[0-9]{2}:[0-9]{2}:[0-9]{2}| (.*)",line[0])
re.search("^.{21}(.*)",line[0])

But, how to use the group (.*) \\1 to only output that part ? Is even regex needed to do this ?

Here is whole code: I am a very beginner to python so the code is probably quite wrong:

import re
with open("database.txt") as f: 
    pstring='' #previous line string beginning
    astring='' #actual line string beginning
    try:
        out = open(re.search("^[0-9]{2}-[0-9]{2}",line[0]) + ".zbs", "w")
        for line in f:
            astring = re.search("^[0-9]{2}-[0-9]{2}-",line[0])
            if not pstring = astring:
                out.write(line)
                pstring = re.search("^[0-9]{2}-[0-9]{2}-",line[0])
                if out: out.close()
                out = open(re.search("^[0-9]{2}-[0-9]{2}",line[0]) + ".zbs", "w")
            else: 
                pstring = re.search("^[0-9]{2}-[0-9]{2}-",line[0])
                out.write(line)
    finally:
        out.close()

Best regards.

Answer 1

Let's consider a single line in your file:

line = "01-01-1989-06:30:00| Stefan Reinartz; 1.1.1989; 06:30; +01; Engelskirchen,Germany; 50n59; 7e24; M;"

If you want to get rid of the first 21 characters of a line, then you can simply use what is referred to as slicing as follows:

>>> print(line[21:])
Stefan Reinartz; 1.1.1989; 06:30; +01; Engelskirchen,Germany; 50n59; 7e24; M;

(Have a look at this site for more details about retrieving substrings via slicing.)

Now, if you need to extract parts of such a line, then you can indeed make use of regular expressions. To get the parts of the date, as you mentioned, you can use, eg, a pattern with named groups as follows:

import re
p = r"[^\;]+; (?P<day>[0-9]+)\.(?P<month>[0-9]+)\.(?P<year>[0-9]+)"
m = re.match(p, line)

The matched groups may then be accessed like this:

>>> m.group("day")
'1'
>>> m.group("month")
'1'
>>> m.group("year")
'1989'

(You can, of course, get the date more easily by extracting it right from the beginning of a line, but this is just an example that demonstrates the use of named group.)