如何有效地找到两个具有panda的大型数据帧之间的反向交集

Question

我正试图找到两个大型数据帧之间的逆交集。我用它来处理以后剪下的代码。不幸的是，这种方法是"；太慢"；如下面进一步描述的。你能想出一个更快的方法来计算这个结果吗？

import pandas as pd
df_1 = pd.DataFrame({'a': [8, 2, 2],
'b': [0, 1, 3],
'c': [0, 2, 2],
'd': [0, 2, 2],
'e': [0, 2, 2]})
df_2 = pd.DataFrame({'a': [8, 2, 2, 2, 8, 2],
'b': [0, 1, 1, 6, 0, 1],
'c': [0, 3, 2, 2, 0, 2],
'd': [0, 4, 2, 2, 0, 4],
'e': [0, 1, 2, 2, 0, 2]})
l_columns = ['a','b','e']
def df_drop_df(df_1, df_2, l_columns):
"""
Eliminates all equal rows present in dataframe 1 (df_1) from dataframe 2 (df_2) depending on a subset of columns (l_columns)
:param df_1: dataframe that defines which rows to be removed
:param df_2: dataframe that is reduced
:param l_columns: list of column names, present in df_1 and df_2, that is used for the comparison
:return df_out: final dataframe
"""
df_1r = df_1[l_columns]
df_2r = df_2[l_columns].reset_index()
df_m = pd.merge(df_1r, df_2r, on=l_columns, how='inner')
row_indexes_m = df_m['index'].to_list()
row_indexes_df_2 = df_2.index.to_list()
row_indexes_out = [x for x in row_indexes_df_2 if x not in row_indexes_m]
df_out = df_2.loc[row_indexes_out]
return df_out

给出以下正确结果：

#row_indexes_out = [1,3]
df_output = df_drop_df(df_1, df_2, l_columns)
df_output
({'a': [2, 2],
'b': [1, 6],
'c': [3, 2],
'd': [4, 2],
'e': [1, 2]})

然而，对于实际应用程序，数据帧的大小有以下维度，在我的本地机器上计算大约需要30分钟：

变量shapedf1(3300,77(df2(642000,77(列表12df_out(611000,77(

Answer 1

您可以尝试替换以下行：

row_indexes_df_2 = df_2.index.to_list()
row_indexes_out = [x for x in row_indexes_df_2 if x not in row_indexes_m]
df_out = df_2.loc[row_indexes_out]

由波浪号操作员：

df_out = df_2.loc[~df_2.index.isin(row_indexes_m)]

这将大大缩短时间。