regex for non alphabets or nothing

Question

I want to identify "how" from python strings.

string1
how to find
it is a show
okay how to

Used:

df[df['string1'].str.contains("how",case=False)]

Output coming:

string1
how to find
it is a show
okay how to

As 'show' contains 'how'

output needed:

string1
how to find
okay how to

after that i used

df[df['string1'].str.contains(r"\Whow",case=False)]

but output i got is:

string1
okay how to

which is again worng.
regex '\W' is not including nothing.
how to achieve this?

@kaleidawave then there should be a space before how. Which i dont want — lil-wolf, Apr 17 '20 at 08:25
it should help you: https://stackoverflow.com/a/60712160/1138192 — A l w a y s S u n n y, Apr 17 '20 at 08:27

score 2 · Accepted Answer · answered Apr 17 '20 at 08:24

2

You need boundaries (\b) for the pattern, otherwise it will also match substrings contained in words:

df[df['string1'].str.contains(r"\bhow\b",case=False)]

      string1
1    how to find
3    okay how to
dtype: object

answered Apr 17 '20 at 08:24

yatu

score 2 · Answer 2 · answered Apr 17 '20 at 08:25

2

Make it as word boundary using \b using regex https://regex101.com/r/2Dlnxj/1

df['string1'].str.contains("\bhow\b",case=False)

answered Apr 17 '20 at 08:25

2 Answers2