从 Pandas Dataframe 列中删除重复的逗号换句话说，我只需要列中的文本用逗号分隔

Question

I have this dataframe with the Text column

Text	Cleaned Col
, , , Apples, , , Hard Work, ,	Apples, Hard Work
, , , , , , , , Apples, , , , ,	Apples
Apples, , Watermelon, , , , , ,	Apples, Watermelon
, , , , , , , , , , , , , , , , ,

I would like to create a column such as Cleaned Col essentially using regex.

I looked at different patterns such as this r'\s*,*([^(a-zA-Z)]*)' but I am not getting the right outcome.

Answer 1

Use Series.str.findall for get words and join by comma:

df['Cleaned Col'] = df['Text'].str.findall('\w+').str.join(', ')
print (df)
                                      Text         Cleaned Col
0      , , , Apples , , , Bananas , , ,        Apples, Bananas
1    , , , , , , , , Apples , , , , ,                   Apples
2        Apples , , Watermelon , , , , , ,  Apples, Watermelon
3  , , , , , , , , , , , , , , , , ,                          

Answer 2

You could try replacing the commas with spaces, then clearing out the left and right spaces and replacing the middle spaces with a comma:

df['Cleaned Col'] = df['Text'].apply(lambda x: x.replace(',', ' ').lstrip().rstrip().replace(' ', ', ')

Answer 3

Since your fields are comma-delimited you can use

# If the fields CANNOT contain whitespace:
df['Cleaned Col'] = df['Text'].str.findall(r'[^\s,]+').str.join(', ')

# If the fields can contain whitespace:
df['Cleaned Col'] = df['Text'].str.findall(r'[^\s,](?:[^,]*[^\s,])?').str.join(', ')

The regex extracts all found matches and .str.join(', ') joins the resulting list items into a single string. The regex ( see its demo ) means:

• [^\s,]+ - one or more chars other than whitespace and comma
• [^\s,] - a single char other than whitespace and comma
• (?:[^,]*[^\s,])? - an optional occurrence of any zero or more chars other than a comma and then a char other than whitespace and comma.

If you have your commas padded with spaces and you really want to use Series.str.replace , you could use

df['Cleaned Col'] = df['Text'].str.replace(r'^[\s,]+|[\s,]+$|(\s)*(,)[\s,]*', r'\2\1', regex=True)

See this regex demo .

Details :

• ^[\s,]+ - one or more whitespaces or commas at the start of string
• [\s,]+$ - one or more whitespaces or commas at the end of string
• (\s)*(,)[\s,]* - zero or more whitespaces (the last one matched is kept in Group 1, \1 ), then a comma (captured into Group 2, \2 ) and then zero or more whitespace or comma chars.

The replacement is Group 2 + Group 1 values.