How to use regex with optional characters in python?
Question:
Say I have a string
"3434.35353"
and another string
"3593"
How do I make a single regular expression that is able to match both without me having to set the pattern to something else if the other fails? I know d+ would match the 3593, but it would not do anything for the 3434.35353, but (d+.d+) would only match the one with the decimal and return no matches found for the 3593.
I expect m.group(1) to return:
"3434.35353"
or
"3593"
Answers:
You can put a ? after a group of characters to make it optional.
You want a dot followed by any number of digits .d+, grouped together (.d+), optionally (.d+)?. Stick that in your pattern:
import re
print re.match("(d+(.d+)?)", "3434.35353").group(1)
3434.35353
print re.match("(d+(.d+)?)", "3434").group(1)
3434
Use the “one or zero” quantifier, ?. Your regex becomes: (d+(.d+)?).
See Chapter 8 of the TextWrangler manual for more details about the different quantifiers available, and how to use them.
This regex should work:
d+(.d+)?
It matches one ore more digits (d+) optionally followed by a dot and one or more digits ((.d+)?).
Read up on the Python RegEx library. The link answers your question and explains why.
However, to match a digit followed by more digits with an optional decimal, you can use
re.compile("(d+(.d+)?)")
In this example, the ? after the .d+ capture group specifies that this portion is optional.
use (?:<characters>|). replace <characters> with the string to make optional. I tested in python shell and got the following result:
>>> s = re.compile('python(?:3|)')
>>> s
re.compile('python(?:3|)')
>>> re.match(s, 'python')
<re.Match object; span=(0, 6), match='python'>
>>> re.match(s, 'python3')
<re.Match object; span=(0, 7), match='python3'>```
Say I have a string
"3434.35353"
and another string
"3593"
How do I make a single regular expression that is able to match both without me having to set the pattern to something else if the other fails? I know d+ would match the 3593, but it would not do anything for the 3434.35353, but (d+.d+) would only match the one with the decimal and return no matches found for the 3593.
I expect m.group(1) to return:
"3434.35353"
or
"3593"
You can put a ? after a group of characters to make it optional.
You want a dot followed by any number of digits .d+, grouped together (.d+), optionally (.d+)?. Stick that in your pattern:
import re
print re.match("(d+(.d+)?)", "3434.35353").group(1)
3434.35353
print re.match("(d+(.d+)?)", "3434").group(1)
3434
Use the “one or zero” quantifier, ?. Your regex becomes: (d+(.d+)?).
See Chapter 8 of the TextWrangler manual for more details about the different quantifiers available, and how to use them.
This regex should work:
d+(.d+)?
It matches one ore more digits (d+) optionally followed by a dot and one or more digits ((.d+)?).
Read up on the Python RegEx library. The link answers your question and explains why.
However, to match a digit followed by more digits with an optional decimal, you can use
re.compile("(d+(.d+)?)")
In this example, the ? after the .d+ capture group specifies that this portion is optional.
use (?:<characters>|). replace <characters> with the string to make optional. I tested in python shell and got the following result:
>>> s = re.compile('python(?:3|)')
>>> s
re.compile('python(?:3|)')
>>> re.match(s, 'python')
<re.Match object; span=(0, 6), match='python'>
>>> re.match(s, 'python3')
<re.Match object; span=(0, 7), match='python3'>```