Python regex for one word

Question

I have this dataframe where I'm trying to delete all one word responses, with/without punctuation and could have spaces in front too. Most of the values are full, long sentences but please find below the kind I am trying to remove.

column
thanks
hello!
really....

My try textonly = re.sub('^.\w+\w+.$' , " " , df.column)

error (even though dtype is string) : expected string or bytes-like object

Another try which seems to go through but doesnt change anything :/

textonly = re.sub('^.\w+\w+.$' , " " , str(df.column))

Please help identify what I'm missing

^[^\n ]*\n(with multiline flag m set) matches lines with no spaces. — Cary Swoveland
– Cary Swoveland, Commented Nov 8, 2021 at 21:24

Wiktor Stribiżew · Accepted Answer · 2021-11-08 21:33:11Z

1

You can use

df['column'] = df['column'].str.replace(r'^\W*\w+\W*$', '', regex=True)

If you mean natural language words by "words", i.e. only consisting of letters, you may use

df['column'] = df['column'].str.replace(r'^[\W\d_]*[^\W\d_]+[\W\d_]*$', '', regex=True)

The regex matches

^ - start of string
\W* - zero or more non-word chars
[\W\d_]* - zero or more non-word chars, digits and _
\w+ - one or more word chars
[^\W\d_]+ - one or more chars other than non-word chars, digits and _
\W* - zero or more non-word chars
$ - end of string.

answered Nov 8, 2021 at 21:33

Wiktor Stribiżew

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Comments

mama · Accepted Answer · 2021-11-08 21:27:28Z

1

You could also not use regex and then check if the string has a space in it

x = [
    'hej med dig',
    'hej',
]

print([x for x in x if ' ' in x.strip()])

answered Nov 8, 2021 at 21:27

mama

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