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>']

what is your expected Output? – styvane Jun 05 '15 at 06:18 — styvane, Jun 05 '15 at 06:18
All of k (first line): ['27*', 'name="namePad">22'] – user2399453 Jun 05 '15 at 06:20 — user2399453, Jun 05 '15 at 06:20

score 1 · Accepted Answer · answered Jun 05 '15 at 06:20

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 05 '15 at 06:20

What is the ? after \* for? – user2399453 Jun 05 '15 at 06:40
1

It will still match the numbers, in case `*` doesn't exist. – Jun 05 '15 at 06:53

score 1 · Answer 2 · edited May 23 '17 at 11:43

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 '17 at 11:43

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answered Jun 05 '15 at 06:21

styvane

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Using regex to grep a string in Python

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