Search for "does-not-contain" on a DataFrame in pandas

Question

You can use the invert (~) operator (which acts like a not for boolean data):

new_df = df[~df["col"].str.contains(word)]

, where new_df is the copy returned by RHS.

contains also accepts a regular expression...

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If the above throws a ValueError, the reason is likely because you have mixed datatypes, so use na=False:

new_df = df[~df["col"].str.contains(word, na=False)]

Or,

new_df = df[df["col"].str.contains(word) == False]

I was having trouble with the not (~) symbol as well, so here's another way from another StackOverflow thread:

df[df["col"].str.contains('this|that')==False]

You can use Apply and Lambda :

df[df["col"].apply(lambda x: word not in x)]

Or if you want to define more complex rule, you can use AND:

df[df["col"].apply(lambda x: word_1 not in x and word_2 not in x)]

I hope the answers are already posted

I am adding the framework to find multiple words and negate those from dataFrame.

Here 'word1','word2','word3','word4' = list of patterns to search

df = DataFrame

column_a = A column name from from DataFrame df

values_to_remove = ['word1','word2','word3','word4'] 

pattern = '|'.join(values_to_remove)

result = df.loc[~df['column_a'].str.contains(pattern, case=False)]

I had to get rid of the NULL values before using the command recommended by Andy above. An example:

df = pd.DataFrame(index = [0, 1, 2], columns=['first', 'second', 'third'])
df.ix[:, 'first'] = 'myword'
df.ix[0, 'second'] = 'myword'
df.ix[2, 'second'] = 'myword'
df.ix[1, 'third'] = 'myword'
df

    first   second  third
0   myword  myword   NaN
1   myword  NaN      myword 
2   myword  myword   NaN

Now running the command:

~df["second"].str.contains(word)

I get the following error:

TypeError: bad operand type for unary ~: 'float'

I got rid of the NULL values using dropna() or fillna() first and retried the command with no problem.

Additional to nanselm2's answer, you can use 0 instead of False:

df["col"].str.contains(word)==0