数据框中的内容:
email user_name sessions ymo
a@a.com JD 1 2015-03-01
a@a.com JD 2 2015-05-01
我需要什么:
email user_name sessions ymo
a@a.com JD 0 2015-01-01
a@a.com JD 0 2015-02-01
a@a.com JD 1 2015-03-01
a@a.com JD 0 2015-04-01
a@a.com JD 2 2015-05-01
a@a.com JD 0 2015-06-01
a@a.com JD 0 2015-07-01
a@a.com JD 0 2015-08-01
a@a.com JD 0 2015-09-01
a@a.com JD 0 2015-10-01
a@a.com JD 0 2015-11-01
a@a.com JD 0 2015-12-01
ymo 列是 pd.Timestamp:
all_ymo
[Timestamp('2015-01-01 00:00:00'),
Timestamp('2015-02-01 00:00:00'),
Timestamp('2015-03-01 00:00:00'),
Timestamp('2015-04-01 00:00:00'),
Timestamp('2015-05-01 00:00:00'),
Timestamp('2015-06-01 00:00:00'),
Timestamp('2015-07-01 00:00:00'),
Timestamp('2015-08-01 00:00:00'),
Timestamp('2015-09-01 00:00:00'),
Timestamp('2015-10-01 00:00:00'),
Timestamp('2015-11-01 00:00:00'),
Timestamp('2015-12-01 00:00:00')]
不幸的是，这个答案:Adding values for missing data combinations in Pandas不好，因为它会为现有的 ymo 值创建副本。
我尝试过类似的方法，但它非常慢:
for em in all_emails:
existent_ymo = fill_ymo[fill_ymo['email'] == em]['ymo']
existent_ymo = set([pd.Timestamp(datetime.date(t.year, t.month, t.day)) for t in existent_ymo])
missing_ymo = list(existent_ymo - all_ymo)
multi_ind = pd.MultiIndex.from_product([[em], missing_ymo], names=col_names)
fill_ymo = sessions.set_index(col_names).reindex(multi_ind, fill_value=0).reset_index()
最佳答案
我尝试使用periods 创建更通用的解决方案:
print (df)
email user_name sessions ymo
0 a@a.com JD 1 2015-03-01
1 a@a.com JD 2 2015-05-01
2 b@b.com AB 1 2015-03-01
3 b@b.com AB 2 2015-05-01
mbeg = pd.period_range('2015-01', periods=12, freq='M')
print (mbeg)
PeriodIndex(['2015-01', '2015-02', '2015-03', '2015-04', '2015-05', '2015-06',
'2015-07', '2015-08', '2015-09', '2015-10', '2015-11', '2015-12'],
dtype='int64', freq='M')
#convert column ymo to period
df.ymo = df.ymo.dt.to_period('m')
#groupby and reindex with filling 0
df = df.groupby(['email','user_name'])
.apply(lambda x: x.set_index('ymo')
.reindex(mbeg, fill_value=0)
.drop(['email','user_name'], axis=1))
.rename_axis(('email','user_name','ymo'))
.reset_index()
print (df)
email user_name ymo sessions
0 a@a.com JD 2015-01 0
1 a@a.com JD 2015-02 0
2 a@a.com JD 2015-03 1
3 a@a.com JD 2015-04 0
4 a@a.com JD 2015-05 2
5 a@a.com JD 2015-06 0
6 a@a.com JD 2015-07 0
7 a@a.com JD 2015-08 0
8 a@a.com JD 2015-09 0
9 a@a.com JD 2015-10 0
10 a@a.com JD 2015-11 0
11 a@a.com JD 2015-12 0
12 b@b.com AB 2015-01 0
13 b@b.com AB 2015-02 0
14 b@b.com AB 2015-03 1
15 b@b.com AB 2015-04 0
16 b@b.com AB 2015-05 2
17 b@b.com AB 2015-06 0
18 b@b.com AB 2015-07 0
19 b@b.com AB 2015-08 0
20 b@b.com AB 2015-09 0
21 b@b.com AB 2015-10 0
22 b@b.com AB 2015-11 0
23 b@b.com AB 2015-12 0
然后如果需要datetimes 使用to_timestamp :
df.ymo = df.ymo.dt.to_timestamp()
print (df)
email user_name ymo sessions
0 a@a.com JD 2015-01-01 0
1 a@a.com JD 2015-02-01 0
2 a@a.com JD 2015-03-01 1
3 a@a.com JD 2015-04-01 0
4 a@a.com JD 2015-05-01 2
5 a@a.com JD 2015-06-01 0
6 a@a.com JD 2015-07-01 0
7 a@a.com JD 2015-08-01 0
8 a@a.com JD 2015-09-01 0
9 a@a.com JD 2015-10-01 0
10 a@a.com JD 2015-11-01 0
11 a@a.com JD 2015-12-01 0
12 b@b.com AB 2015-01-01 0
13 b@b.com AB 2015-02-01 0
14 b@b.com AB 2015-03-01 1
15 b@b.com AB 2015-04-01 0
16 b@b.com AB 2015-05-01 2
17 b@b.com AB 2015-06-01 0
18 b@b.com AB 2015-07-01 0
19 b@b.com AB 2015-08-01 0
20 b@b.com AB 2015-09-01 0
21 b@b.com AB 2015-10-01 0
22 b@b.com AB 2015-11-01 0
23 b@b.com AB 2015-12-01 0
带日期时间的解决方案:
print (df)
email user_name sessions ymo
0 a@a.com JD 1 2015-03-01
1 a@a.com JD 2 2015-05-01
2 b@b.com AB 1 2015-03-01
3 b@b.com AB 2 2015-05-01
mbeg = pd.date_range('2015-01-31', periods=12, freq='M') - pd.offsets.MonthBegin()
df = df.groupby(['email','user_name'])
.apply(lambda x: x.set_index('ymo')
.reindex(mbeg, fill_value=0)
.drop(['email','user_name'], axis=1))
.rename_axis(('email','user_name','ymo'))
.reset_index()
print (df)
email user_name ymo sessions
0 a@a.com JD 2015-01-01 0
1 a@a.com JD 2015-02-01 0
2 a@a.com JD 2015-03-01 1
3 a@a.com JD 2015-04-01 0
4 a@a.com JD 2015-05-01 2
5 a@a.com JD 2015-06-01 0
6 a@a.com JD 2015-07-01 0
7 a@a.com JD 2015-08-01 0
8 a@a.com JD 2015-09-01 0
9 a@a.com JD 2015-10-01 0
10 a@a.com JD 2015-11-01 0
11 a@a.com JD 2015-12-01 0
12 b@b.com AB 2015-01-01 0
13 b@b.com AB 2015-02-01 0
14 b@b.com AB 2015-03-01 1
15 b@b.com AB 2015-04-01 0
16 b@b.com AB 2015-05-01 2
17 b@b.com AB 2015-06-01 0
18 b@b.com AB 2015-07-01 0
19 b@b.com AB 2015-08-01 0
20 b@b.com AB 2015-09-01 0
21 b@b.com AB 2015-10-01 0
22 b@b.com AB 2015-11-01 0
23 b@b.com AB 2015-12-01 0
关于python - 仅在数据框中填充缺失值( Pandas )，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/39466757/