Question

我在数据框df下，其中stamp B有时为空。必须用Stamp A的日期和Time列中的相应时间填充此类空值

              stamp A             stamp B      Time
0 2012-10-08 18:15:05 2012-10-08 18:15:05  19:00:01
1 2012-10-09 12:15:05                 NaT  18:45:09
2 2012-10-11 18:13:00                 NaT  12:20:20
3 2012-10-11 08:15:15 2012-10-11 18:15:05  22:10:05
4 2012-10-12 18:15:20 2012-10-12 17:10:20  19:34:12

这是我的解决方法-

>>>from datetime import dateime as dtm    
>>>result = df[df['stamp B'].isnull()].apply(lambda x: dtm.combine(x['stamp A'].date(), dtm.strptime(x["Time"], "%H:%M:%S").time()), axis=1)

它返回result如下：

1   2012-10-09 18:45:09
2   2012-10-11 12:20:20
dtype: datetime64[ns]

但不确定，如何用原始数据帧result中的NaT值将此df['stamp B']替换为

Answer 1

我将从stamp A中提取日期，添加Time，然后在fillna上进行stamp B：

s = df['stamp A'].dt.normalized() + pd.to_timedelta(df['Time'])

df['stamp B'] = df['stamp B'].fillna(s)

Answer 2

使用Series.dt.floor删除时间，并用to_timedelta添加时间增量，然后用Series.combine_first替换缺少的值：

dates = df['stamp A'].dt.floor('d').add(pd.to_timedelta(df['Time']))
df['stamp B'] = df['stamp B'].combine_first(dates)

print (df)
              stamp A             stamp B      Time
0 2012-10-08 18:15:05 2012-10-08 18:15:05  19:00:01
1 2012-10-09 12:15:05 2012-10-09 18:45:09  18:45:09
2 2012-10-11 18:13:00 2012-10-11 12:20:20  12:20:20
3 2012-10-11 08:15:15 2012-10-11 18:15:05  22:10:05
4 2012-10-12 18:15:20 2012-10-12 17:10:20  19:34:12

熊猫用df.apply结果替换原始列空值

2 个答案: