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:
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):
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!
答案 0 :(得分:2)
使用行的笛卡尔积并检查每一行
代码在线记录
df1 = pd.DataFrame(
{
'ColumnA': ['id1', 'id2'],
'ColumnB': [['a','b','c'], ['a','d','e']],
}
)
df2 = pd.DataFrame(
{
'ColumnA': ['id3'],
'ColumnB': [['a','b','c','x','y', 'z']],
}
)
# Take cartesian product of both dataframes
df1['k'] = 0
df2['k'] = 0
df = pd.merge(df1, df2, on='k').drop('k',1)
# Check the overlap of the lists and find the overlap length
df['overlap'] = df.apply(lambda x: len(set(x['ColumnB_x']).intersection(
set(x['ColumnB_y']))), axis=1)
# Select whoes overlap length > 2
df = df[df['overlap'] > 2]
print (df)
输出:
ColumnA_x ColumnB_x ColumnA_y ColumnB_y overlap
0 id1 [a, b, c] id3 [a, b, c, x, y, z] 3
答案 1 :(得分:0)
如果您使用的是 pandas 1.2.0 或更新版本(2020 年 12 月 26 日发布),笛卡尔积(交叉关节)可以简化如下:
df = df1.merge(df2, how='cross') # simplified cross joint for pandas >= 1.2.0
此外,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)
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)
使用 list(map(...
时:
%%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...
的速度提高了 3 倍!
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)