基于列表列加入两个 pandas 数据框 [英] Join two pandas dataframes based on lists columns

问题描述

我有 2 个数据框，其中包含列表列.
我想根据列表中的 2+ 共享价值观加入他们.示例:

I have 2 dataframes containing columns of lists.
I would like to join them based on 2+ shared values on the lists. Example:

ColumnA ColumnB        | ColumnA ColumnB        
id1     ['a','b','c']  | id3     ['a','b','c','x','y', 'z']
id2     ['a','d,'e']   | 

在这种情况下，我们可以看到 id1 匹配 id3，因为列表中有 2 个以上的共享值.所以输出将是(列名并不重要，只是举例):

In this case we can see that id1 matches id3 because there are 2+ shared values on the lists. So the output will be (columns name are not important and just for example):

    ColumnA1 ColumnB1     ColumnA2   ColumnB2        
    id1      ['a','b','c']  id3     ['a','b','c','x','y', 'z']
    

我怎样才能达到这个结果?我试图迭代数据帧 #1 中的每一行，但这似乎不是一个好主意.
谢谢！

How can I achieve this result? I've tried to iterate each row in dataframe #1 but it doesn't seem a good idea.
Thank you!

推荐答案

如果您使用的是 pandas 1.2.0 或更高版本(2020 年 12 月 26 日发布)，笛卡尔积(交叉关节)可以简化如下:

If you are using pandas 1.2.0 or newer (released on December 26, 2020), cartesian product (cross joint) can be simplified as follows:

    df = df1.merge(df2, how='cross')         # simplified cross joint for pandas >= 1.2.0

另外，如果您关心系统性能(执行时间)，建议使用 list(map... 而不是较慢的 apply(...axis=1)

Also, if system performance (execution time) is a concern to you, it is advisable to use list(map... instead of the slower apply(... axis=1)

使用apply(...axis=1):

%%timeit
df['overlap'] = df.apply(lambda x: 
                         len(set(x['ColumnB1']).intersection(
                             set(x['ColumnB2']))), axis=1)


800 µs ± 59.1 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

同时使用 list(map(...:

%%timeit
df['overlap'] = list(map(lambda x, y: len(set(x).intersection(set(y))), df['ColumnB1'], df['ColumnB2']))

217 µs ± 25.5 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

请注意，使用 list(map... 快 3 倍！

Notice that using list(map... is 3x times faster!

整套代码供您参考:

    data = {'ColumnA1': ['id1', 'id2'], 'ColumnB1': [['a', 'b', 'c'], ['a', 'd', 'e']]}
    df1 = pd.DataFrame(data)

    data = {'ColumnA2': ['id3', 'id4'], 'ColumnB2': [['a','b','c','x','y', 'z'], ['d','e','f','p','q', 'r']]}
    df2 = pd.DataFrame(data)

    df = df1.merge(df2, how='cross')             # for pandas version >= 1.2.0

    df['overlap'] = list(map(lambda x, y: len(set(x).intersection(set(y))), df['ColumnB1'], df['ColumnB2']))

    df = df[df['overlap'] >= 2]
    print (df)

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