How to Efficiently Identify Unique Rows in Pandas DataFrames When Comparing Two DataFrames?

Obtaining Unique Rows in Pandas DataFrames

Given two Pandas dataframes, it is often necessary to identify rows that exist in only one of them. This can be achieved effectively by utilizing a merge operation.

For instance, consider the following dataframes:

df1 = pd.DataFrame(data={'col1': [1, 2, 3, 4, 5, 3], 'col2': [10, 11, 12, 13, 14, 10]})
df2 = pd.DataFrame(data={'col1': [1, 2, 3], 'col2': [10, 11, 12]})

To obtain rows from df1 that are not present in df2, we can perform a left-join between df1 and df2. To ensure that each row in df1 is matched with exactly one row in df2, we need to first remove duplicate rows from df2. We can do this by using the drop_duplicates() function.

df_all = df1.merge(df2.drop_duplicates(), on=['col1', 'col2'], how='left', indicator=True)

The resulting df_all dataframe will have an additional column named _merge that indicates whether each row came from both df1 and df2 ('both'), from df1 only ('left_only'), or from df2 only ('right_only').

   col1  col2     _merge
0     1    10       both
1     2    11       both
2     3    12       both
3     4    13  left_only
4     5    14  left_only
5     3    10  left_only

To extract the rows from df1 that are not present in df2, we can simply select the rows where _merge is equal to 'left_only':

rows_not_in_df2 = df_all[df_all['_merge'] == 'left_only']

   col1  col2
0     4    13
1     5    14
2     3    10

Avoiding Incorrect Approaches

It is important to avoid incorrect solutions that fail to consider the rows as a whole. Some solutions only check if each individual value in a row exists in the other dataframe, which can lead to incorrect results.

For example, if we had added another row to df1 with data [3, 10], which is also present in df2, the incorrect approaches would still identify it as not being present in df2 because it has different values in the two columns. However, our approach will correctly identify it as not being present because it is already in df2 with the same values for both columns.

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