Nsryjdtyk

How can I add a new line every time there is a pattern of a regex-list found in a string ?

I am using python 3.6.

I got the following input:

12.13.14 Here is supposed to start a new line.

12.13.15 Here is supposed to start a new line.

Here is some text. It is written in one lines. 12.13. Here is some more text. 2.12.14. Here is even more text.

I wish to have the following output:

12.13.14

Here is supposed to start a new line.

12.13.15

Here is supposed to start a new line.

Here is some text. It is written in one lines.

12.13.

Here is some more text.

2.12.14.

Here is even more text.

My first try returns as the output the same as the input:

in_file2 = 'work1-T1.txt'

out_file2 = 'work2-T1.txt'





start_rx = re.compile('|'.join(

    ['dd.dd.', 'd.dd.dd','dd.dd.dd']))





with open(in_file2,'r', encoding='utf-8') as fin2, open(out_file2, 'w', encoding='utf-8') as fout2:

    text_list = fin2.read().split()

    fin2.seek(0)



    for string in fin2:

        if re.match(start_rx, string):

            string = str.replace(start_rx, 'nn' + start_rx + 'n')



        fout2.write(string)

My second try returns an error 'TypeError: unsupported operand type(s) for +: '_sre.SRE_Pattern' and 'str''

in_file2 = 'work1-T1.txt'

out_file2 = 'work2-T1.txt'





start_rx = re.compile('|'.join(

            ['dd.dd.', 'd.dd.dd','dd.dd.dd']))



with open(in_file2,"r") as fin2, open(out_file2, 'w') as fout3:

    for line in fin2:

        start = False

        if re.match(start_rx, line):

            start = True

        if start == False:

            print ('do something')

        if start == True:

            line = 'n' + line ## leerzeichen vor Pos Nr

            line = line.replace(start_rx, start_rx + 'n')

        fout3.write(line)

First of all, to search and replace with a regex, you need to use re.sub, not str.replace.

Second, if you use a re.sub, you can't use the regex pattern inside a replacement pattern, you need to group the parts of the regex you want to keep and use backreferences in the replacement (or, if you just want to refer to the whole match, use g<0> backreference, no capturing groups are required).

Third, when you build an unanchored alternation pattern, make sure longer alternatives come first, i.e. start_rx = re.compile('|'.join(['dd.dd.dd', 'd.dd.dd', 'dd.dd.'])). However, you may use a more precise pattern here manually.

Here is how your code can be fixed:

with open(in_file2,'r', encoding='utf-8') as fin2, open(out_file2, 'w', encoding='utf-8') as fout2:

    text = fin2.read()

    fout2.write(re.sub(r's*(d+(?:.d+)+.?)s*', r'nn1n', text))

See the Python demo

The pattern is

s*(d+(?:.d+)+.?)s*

See the regex demo

Details

• 
  s* - 0+ whitespaces


• 
  (d+(?:.d+)+.?) - Group 1 (1 in the replacement pattern):
  
  
  
  
  • 
    d+ - 1+ digits
  
  
  • 
    (?:.d+)+ - 1 or more repetitions of . and 1+ digits
  
  
  • 
    .? - an optional .
    
  
  
  
  


• 
  s* - 0+ whitespaces

Try this

out_file2=re.sub(r'(d+) ', r'1n', in_file2)

out_file2=re.sub(r'(w+).', r'1.n', in_file2)