如何有效地找到两个具有 pandas 的大型数据帧之间的逆交集?

How to efficiently find the inverse intersection between two large dataframes with pandas?

我试图找到两个 large 数据帧之间的逆交集。 我得到它与此后剪断的代码一起工作。不幸的是,这种方法在大型数据帧上“太慢了”,如下所述。你能想出一种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,数据帧的大小具有以下尺寸,在我的本地计算机上大约需要 30 分钟来计算:

variable shape
df1 (3300,77)
df2 (642000,77)
l_columns list 12
df_out (611000,77)

(这意味着 df_1 中出现的每一行大约是 df_2 中的 10 次)

你能想出一个更快的方法来计算这个结果吗?

您可以尝试替换为以下几行:

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]

波浪号运算符:

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

应该会大大减少时间。