Python Pandas：按另一个合并或过滤 DataFrame。有没有更好的办法？

Question

我有时遇到的一种情况是，我有两个数据框（df1，df2），我想根据df1 和@987654325 之间的多列的交集创建一个新的数据框（df3） @。

例如，我想通过按列Campaign 和Group 过滤df1 来创建df3。

import pandas as pd
df1 = pd.DataFrame({'Campaign':['Campaign 1', 'Campaign 2', 'Campaign 3', 'Campaign 3', 'Campaign 4'], 'Group':['Some group', 'Arbitrary Group', 'Group 1', 'Group 2', 'Done Group'], 'Metric':[245,91,292,373,32]}, columns=['Campaign', 'Group', 'Metric'])
df2 = pd.DataFrame({'Campaign':['Campaign 3', 'Campaign 3'], 'Group':['Group 1', 'Group 2'], 'Metric':[23, 456]}, columns=['Campaign', 'Group', 'Metric'])

df1

     Campaign            Group  Metric
0  Campaign 1       Some group     245
1  Campaign 2  Arbitrary Group      91
2  Campaign 3          Group 1     292
3  Campaign 3          Group 2     373
4  Campaign 4       Done Group      32

df2

     Campaign    Group  Metric
0  Campaign 3  Group 1      23
1  Campaign 3  Group 2     456

我知道我可以通过合并来做到这一点...

df3 = df1.merge(df2, how='inner', on=['Campaign', 'Group'], suffixes=('','_del'))
#df3
     Campaign    Group  Metric  Metric_del
0  Campaign 3  Group 1     292          23
1  Campaign 3  Group 2     373         456

但是我必须弄清楚如何drop 以_del 结尾的列。我猜是这样的：

df3.select(lambda x: not re.search('_del', x), axis=1)
##The result I'm going for but required merge, then select (2-steps)
     Campaign    Group  Metric
0  Campaign 3  Group 1     292
1  Campaign 3  Group 2     373

问题

我主要感兴趣的是返回df1，它只是根据df2 的Campaign|Group 值进行过滤。

1. 有没有一种更好的方式来返回df1而不诉诸merge？

2. 有没有办法merge 但不将df2 的列返回到merge 并且只返回df1 的列？

Answer 1

或者，您可以使用groupby 和filter，如下所示：

# Compute the set of values you're interested in.
# In your example, this will be {('Campaign 3', 'Group 1'), ('Campaign 3', 'Group 2')}
interesting_groups = set(df2[['Campaign', 'Group']].apply(tuple, axis=1))
# Filter df1, keeping only values in that set
result = df1.groupby(['Campaign', 'Group']).filter(
    lambda x: x.name in interesting_groups
)

另一个例子见filter docs。

Answer 2

假设您的 df1 和 df2 具有完全相同的列。您可以先将这些连接键列设置为索引，然后使用df1.reindex(df2.index) 和另一个.dropna() 来生成交集。

df3 = df1.set_index(['Campaign', 'Group'])
df4 = df2.set_index(['Campaign', 'Group'])
# reindex first and dropna will produce the intersection
df3.reindex(df4.index).dropna(how='all').reset_index()

     Campaign    Group  Metric
0  Campaign 3  Group 1     292
1  Campaign 3  Group 2     373

编辑：

当键不唯一时使用.isin。

# create some duplicated keys and values
df3 = df3.append(df3)
df4 = df4.append(df4)

# isin
df3[df3.index.isin(df4.index)].reset_index()

     Campaign    Group  Metric
0  Campaign 3  Group 1     292
1  Campaign 3  Group 2     373
2  Campaign 3  Group 1     292
3  Campaign 3  Group 2     373