Print 10 characters before a string

Question

I'm currently trying to print a string that usually comes before another string, for example "12/07/201911 : 00Type of Occurrence" . Here I'm trying to print the time "11 : 00" which always come before the text "Type of Occurrence" .

I have tried this which print everything before the identifier. I would like to only print 7 characters before the identifier.

import re
A="12/07/2019 11 : 00Type of Occurrence"
print(A.split('Type of Occurrence', 1)[0].replace('.', '').upper())

It prints :

12/07/2019 11 : 00

Answer 1

Try this:

A="12/07/2019 11 : 00Type of Occurrence"
print(" ".join(A.split("Type of Occurrence")[0].split()[1:]))

OUTPUT :

11 : 00

Process :

1. A.split("Type of Occurrence")[0] gives "12/07/2019 11 : 00"

2. "12/07/2019 11 : 00".split()[1:] gives ['11', ':', '00']

3. " ".join(['11', ':', '00']) gives 11 : 00

Answer 2

Try this :

>>> A="12/07/2019 11 : 00Type of Occurrence"
>>> i = A.find('Type of Occurrence')
>>> i
18
>>> A[i-7:i]
'11 : 00'

Answer 3

If the whitespaces between the colon can vary, you might use a capturing group for the time part and match Type of Occurrence:

((?:0[0-9]|1[0-9]|2[0-3])\s*:\s*[0-5][0-9])Type of Occurrence

Regex demo

Answer 4

Basic Idea:

A="12/07/2019 11 : 00Type of Occurrence"

A = A.split('Type of Occurrence', 1)[0].replace('.', '').upper()
A[-7:]
>>> '11 : 00'

Is this what you want?

Answer 5

I would like to only print 7 characters before the identifier.

You may use this regex:

>>> import re
>>> A="12/07/2019 11 : 00Type of Occurrence"
>>> print re.findall(r'(.{7})Type of Occurrence', A)[0]
11 : 00

re.findall prints captured group if they are available in regex.