How to remove a pandas dataframe from another dataframe

How to remove a pandas dataframe from another dataframe, just like the set subtraction:

a=[1,2,3,4,5]
b=[1,5]
a-b=[2,3,4]

And now we have two pandas dataframe, how to remove df2 from df1:

In [5]: df1=pd.DataFrame([[1,2],[3,4],[5,6]],columns=['a','b'])
In [6]: df1
Out[6]:
   a  b
0  1  2
1  3  4
2  5  6


In [9]: df2=pd.DataFrame([[1,2],[5,6]],columns=['a','b'])
In [10]: df2
Out[10]:
   a  b
0  1  2
1  5  6

Then we expect df1-df2 result will be:

In [14]: df
Out[14]:
   a  b
0  3  4

How to do it?

Thank you.

Solution

Use pd.concat followed by drop_duplicates(keep=False)

pd.concat([df1, df2, df2]).drop_duplicates(keep=False)

It looks like

   a  b
1  3  4

Explanation

pd.concat adds the two DataFrames together by appending one right after the other. if there is any overlap, it will be captured by the drop_duplicates method. However, drop_duplicates by default leaves the first observation and removes every other observation. In this case, we want every duplicate removed. Hence, the keep=False parameter which does exactly that.

A special note to the repeated df2. With only one df2 any row in df2 not in df1 won't be considered a duplicate and will remain. This solution with only one df2 only works when df2 is a subset of df1. However, if we concat df2 twice, it is guaranteed to be a duplicate and will subsequently be removed.

You can use .duplicated, which has the benefit of being fairly expressive:

%%timeit
combined = df1.append(df2)
combined[~combined.index.duplicated(keep=False)]

1000 loops, best of 3: 875 µs per loop

For comparison:

%timeit df1.loc[pd.merge(df1, df2, on=['a','b'], how='left', indicator=True)['_merge'] == 'left_only']

100 loops, best of 3: 4.57 ms per loop


%timeit pd.concat([df1, df2, df2]).drop_duplicates(keep=False)

1000 loops, best of 3: 987 µs per loop


%timeit df2[df2.apply(lambda x: x.value not in df2.values, axis=1)]

1000 loops, best of 3: 546 µs per loop

In sum, using the np.array comparison is fastest. Don't need the .tolist() there.

A set logic approach. Turn the rows of df1 and df2 into sets. Then use set subtraction to define new DataFrame

idx1 = set(df1.set_index(['a', 'b']).index)
idx2 = set(df2.set_index(['a', 'b']).index)

pd.DataFrame(list(idx1 - idx2), columns=df1.columns)

   a  b
0  3  4

To get dataframe with all records which are in DF1 but not in DF2

DF=DF1[~DF1.isin(DF2)].dropna(how = 'all')

My shot with merge df1 and df2 from the question.

Using 'indicator' parameter

In [74]: df1.loc[pd.merge(df1, df2, on=['a','b'], how='left', indicator=True)['_merge'] == 'left_only']
Out[74]: 
   a  b
1  3  4

A masking approach

df1[df1.apply(lambda x: x.values.tolist() not in df2.values.tolist(), axis=1)]

   a  b
1  3  4

I think the first tolist() needs to be removed, but keep the second one:

df1[df1.apply(lambda x: x.values() not in df2.values.tolist(), axis=1)]

An easiest option is to use indexes.

1. Append df1 and df2 and reset their indexes.

   df = df1.concat(df2)
df.reset_index(inplace=True)

2. e.g:
   This will give df2 indexes

   indexes_df2 = df.index[ (df["a"].isin(df2["a"]) ) & (df["b"].isin(df2["b"]) ) result_index = df.index[~index_df2] result_data = df.iloc[ result_index,:]

Hope it will help to new readers, although the question posted a little time ago :)

Solution if df1 contains duplicates + keeps the index.

A modified version of piRSquared's answer to keep the duplicates in df1 that do not appear in df2, while maintaining the index.

df1[df1.apply(lambda x: (x == pd.concat([df1.drop_duplicates(), df2, df2]).drop_duplicates(keep=False)).all(1).any(), axis=1)]

If your dataframes are big, you may want to store the result of

pd.concat([df1.drop_duplicates(), df2, df2]).drop_duplicates(keep=False)

in a variable before the df1.apply call.

This solution works when your df_to_drop is a subset of main data frame data.

data_clean = data.drop(df_to_drop.index)