How to delete rows that contain only certain values?

user_13830386 0
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I have a dataframe like this

    column_name 
0   OnePlus phones never fail to meet my expectatiion.  
1   received earlier than expected for local set.   
2   \n  
3   good    
4   must buy!
5   \t
6     
7   awesome product!  
8     \n

I want to remove all rows which contain ONLY \n, \t, , \n in them.

Output should be like this:

    column_name 
0   OnePlus phones never fail to meet my expectatiion.  
1   received earlier than expected for local set.   
2   good    
3   must buy!
4   awesome product!

I tried the following method:

  df = df[df.column_name != '\n'].reset_index(drop=True)
  df = df[df.column_name != ''].reset_index(drop=True)
  df = df[df.column_name != ' '].reset_index(drop=True)
  df = df[df.column_name != '   '].reset_index(drop=True)
  df = df[df.column_name != ' \n '].reset_index(drop=True)

But is there a more elegant way or a pythonic way to do this instead of repeating the code?

pandas delete-row rows dataframe python
3 Answers

You can use Series.str.strip and compare only empty strings:

df1 = df[df.column_name.str.strip() != ''].reset_index(drop=True)

Or convert empty values to boolean:

df1 = df[df.column_name.str.strip().astype(bool)].reset_index(drop=True)

Or filter words, for me was necessary strip (maybe in real data strip should be removed):

df1 = df[df.column_name.str.strip().str.contains('\W', na=False)].reset_index(drop=True)

If need remove missing values and no string values replace these values to NaNs and then use DataFrame.dropna:

df.column_name = df.column_name.replace(r'^\s*$', np.nan, regex=True)
df1 = df.dropna(subset=['column_name']).reset_index(drop=True)
user_29010020

Use df.str.contains() to check if there is any small alpha after forward slash

df[df.Column Name.str.contains('[\\][a-z]+',case=True, na=False, regex=True)]

In your case, Data:

print(pd.DataFrame({'A':['OnePlus phones never fail to meet my expectatiion','received earlier than expected for local set.','\n','good','\t', np.nan,'must buy!','','awesome product!','\n' ]}))

                                               A
0  OnePlus phones never fail to meet my expectatiion
1      received earlier than expected for local set.
2                                                 \n
3                                               good
4                                                 \t
5                                                NaN
6                                          must buy!
7                                                   
8                                   awesome product!
9                                                 \n

Solution

print(df[df.A.str.contains('[\\][a-z]+',case=True, na=False, regex=True)])



                             A
0  OnePlus phones never fail to meet my expectatiion
1      received earlier than expected for local set.
3                                               good
6                                          must buy!
8                                   awesome product!
user_89869750

Another approach, removing rows where the entries match the flagged elements:

df = df[~df['column_name'].isin(['\\n','\\t'])].dropna()

If there are extra spaces in the last row (or others) you can first do:

df['column_name'] = df['column_name'].str.strip()
user_133869790
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