Using regex to grep a string in Python

Question

I am trying to grep two kinds of patterns in a list using re in python:

'<xyz>number followed by optional *</xyz>'
'name="namepad">number</xyz>

Using regex in python, I am not able to get the data with asterisk. Here is a sample session, what can I do so that the filter also returns the first element?

>>> k = ['<xyz>27*</xyz>', 'name="namePad">22</xyz>']
>>> f = filter(lambda x:re.search('^name="namePad"|^<xyz>[0-9]{1,3}\*"  <\/xyz>',x), k)
>>> f
['name="namePad">22</xyz>']

All of k (first line): ['<xyz>27*</xyz>', 'name="namePad">22</xyz>'] — user2399453
– user2399453, Commented Jun 5, 2015 at 6:20

user142650 · Accepted Answer · 2015-06-05 06:20:38Z

1

Your regex has mismatched " quotes. Try this:

filter(lambda x:re.search(r'^name="namePad"|^<xyz>[\d]{1,3}\*?</xyz>',x), k)

It will give you the following:

['27*', 'name="namePad">22']

answered Jun 5, 2015 at 6:20

user142650

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2 Comments

user2399453 Over a year ago

What is the ? after * for?

user142650 Over a year ago

It will still match the numbers, in case * doesn't exist.

Community · Accepted Answer · 2017-05-23 11:43:42Z

1

You can use re.match since to check for a match only at the beginning of the string. Also you don't need filter use list comprehensions instead.

>>> [i for i in k if re.match(r'(<xyz>|name="namePad">)\d+\*?', i)]
['<xyz>27*</xyz>', 'name="namePad">22</xyz>']

The ? after * mean that * is optional you can read more about quantifiers Here

edited May 23, 2017 at 11:43

CommunityBot

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answered Jun 5, 2015 at 6:21

Sede

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