问题描述
我遇到了一种情况,我试图使用标准日期来估算日期列中的缺失值。我正在使用以下代码,但缺失值仍然保持原样,不会被我使用的日期替换。
df:
termination_date
2020-06-28 00:00:00
2020-07-13 00:00:00
2020-08-11 00:00:00
2020-08-11 00:00:00
现在要替换缺失值,我想使用日期“2020-07-31 00:00:00”,我使用以下代码:
df['termination_date'] = df['termination_date'].fillna(value=pd.to_datetime('2020-07-31 00:00:00'))
输出应该是这样的:
termination_date
2020-06-28 00:00:00
2020-07-31 00:00:00
2020-07-13 00:00:00
2020-08-11 00:00:00
2020-07-31 00:00:00
2020-08-11 00:00:00
解决方法
将非日期时间的值转换为 NaT
,因此可能替换为 fillna
:
df['termination_date'] = (pd.to_datetime(df['termination_date'],errors='coerce')
.fillna(pd.to_datetime('2020-07-31')))
#because same times 00:00:00 are not shown
print (df)
termination_date
0 2020-06-28
1 2020-07-31
2 2020-07-13
3 2020-08-11
4 2020-07-31
5 2020-08-11
print(df['termination_date'].tolist())
[Timestamp('2020-06-28 00:00:00'),Timestamp('2020-07-31 00:00:00'),Timestamp('2020-07-13 00:00:00'),Timestamp('2020-08-11 00:00:00'),Timestamp('2020-08-11 00:00:00')]
print (df.termination_date.dtypes)
datetime64[ns]
,
来自您的DataFrame
:
>>> df = pd.DataFrame({'termination_date': ["2020-06-28 00:00:00",... "",... "2020-07-13 00:00:00",... "2020-08-11 00:00:00",... "2020-08-11 00:00:00"]},... index = [0,1,2,3,4,5])
>>> df
termination_date
0 2020-06-28 00:00:00
1
2 2020-07-13 00:00:00
3 2020-08-11 00:00:00
4
5 2020-08-11 00:00:00
我们可以使用 loc
将缺失值替换为 pd.to_datetime('2020-07-31 00:00:00')
以获得预期结果:
>>> df.loc[df['termination_date'] == '','termination_date'] = pd.to_datetime('2020-07-31 00:00:00')
>>> df
termination_date
0 2020-06-28 00:00:00
1 2020-07-31 00:00:00
2 2020-07-13 00:00:00
3 2020-08-11 00:00:00
4 2020-07-31 00:00:00
5 2020-08-11 00:00:00
最后,我们可以将列转换为 Datetime
格式以确保我们没有 string
值:
df['termination_date'] = pd.to_datetime(df['termination_date'])