这是这个问题的后续问题'merge' 2 dataframes on elements from list? 扩展是,现在有些名字在df8中可以加倍,所以为了能够区分不同的文本,我创建了一个日期
df8=pd.DataFrame({'Dates':['2017-12-14', '2017-12-14','2017-12-16'],'names':[['Hans','Meier'],['Debby','Harry'],['Hans','Harry']]})
df9=pd.DataFrame({'Date':['2017-12-14','2017-12-14','2017-12-14','2017-12-14','2017-12-14','2017-12-16','2017-12-16'],'caller':['Hans','Meier','Debby','Harry','Peter','Hans','Harry'],'text':[['hi im hans'],['hi im meier'],['hi im debby'],['hi im harry'],['hi im peter'],['my name is hans'],['my name is harry']]})
df9.set_index(df9.Date, inplace = True)
df9.drop('Date', axis = 1, inplace = True)
df9.head(10)
Date names text
2017-12-14 Hans [hi im hans]
2017-12-14 Meier [hi im meier]
2017-12-14 Debby [hi im debby]
2017-12-14 Harry [hi im harry]
2017-12-14 Peter [hi im peter]
2017-12-16 Hans [my name is hans]
2017-12-16 Harry [my name is harry]
结果应该是,如果某人在各自的列表中,那么df9中某人所说的文本将出现在df8中,此外,姓名和日期现在必须匹配
所以输出是
Date names content
2017-12-14 ['Hans','Meier'] ['hi im hans', 'hi im meier']
2017-12-14 ['Debby','Harry',] ['hi im debby', 'hi im harry',]
2017-12-16 ['Hans', 'Harry'] ['my name is hans','my name is harry']
应用上一个问题的结果
df9['text']=df9['text'].str[0]
l=[df9.loc[x,'text'].tolist() for x in df8.names]
df8['cont']=l
产生这个错误
"None of [['Hans', 'Meier']] are in the [index]"
这似乎与日期索引有关,但我不知道如何解决它。
我用loc[]
玩了一会儿,但错误仍然存在
IIUC,你可以这样做:
展平
df8
并分配给m
然后,
merge
和groupby
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