ValueError:偏移量必须是严格介于-时间增量(小时=24)和时间增量(时间=24)之间的时间增量

Question

我有一个数据帧，它的行34180到34221对于列"MYTIME"是这样的。

df_fl2[‘MYTIME’]。iloc[34180:34221]

34180    2022-03-02 1900-10-22 23:00:00
34181    2022-03-02 1900-10-23 00:00:00
34182    2022-03-02 1900-10-23 01:00:00
34183    2022-03-02 1900-10-23 02:00:00
34184    2022-03-02 1900-10-23 03:00:00
34185    2022-03-02 1900-10-23 04:00:00
34186    2022-03-02 1900-10-23 05:00:00
34187    2022-03-02 1900-10-23 06:00:00
34188    2022-03-02 1900-10-23 07:00:00
34189    2022-03-02 1900-10-23 08:00:00
34190    2022-03-02 1900-10-23 09:00:00
34191    2022-03-02 1900-10-23 10:00:00
34192    2022-03-02 1900-10-23 11:00:00
34193    2022-03-02 1900-10-23 12:00:00
34194    2022-03-02 1900-10-23 13:00:00
34195    2022-03-02 1900-10-23 14:00:00
34196    2022-03-02 1900-10-23 15:00:00
34197    2022-03-02 1900-10-23 16:00:00
34198    2022-03-02 1900-10-23 17:00:00
34199    2022-03-02 1900-10-23 18:00:00
34200    2022-03-02 1900-10-23 19:00:00
34201    2022-03-02 1900-10-23 20:00:00
34202    2022-03-02 1900-10-23 21:00:00
34203    2022-03-02 1900-10-23 22:00:00
34204    2022-03-02 1900-10-23 23:00:00
34205    2022-03-02 1900-10-24 00:00:00
34206    2022-03-02 1900-10-24 01:00:00
34207    2022-03-02 1900-10-24 02:00:00
34208    2022-03-02 1900-10-24 03:00:00
34209    2022-03-02 1900-10-24 04:00:00
34210    2022-03-02 1900-10-24 05:00:00
34211    2022-03-02 1900-10-24 06:00:00
34212    2022-03-02 1900-10-24 07:00:00
34213    2022-03-02 1900-10-24 08:00:00
34214    2022-03-02 1900-10-24 09:00:00
34215    2022-03-02 1900-10-24 10:00:00
34216    2022-03-02 1900-10-24 11:00:00
34217    2022-03-02 1900-10-24 12:00:00
34218    2022-03-02 1900-10-24 13:00:00
34219    2022-03-02 1900-10-24 14:00:00
34220    2022-03-02 1900-10-24 15:00:00
Name: MYTIME, dtype: object

但是，当试图强制转换为日期时间时。

pd.to_datetime(df_fl2['MYTIME'].iloc[34180:34221], errors='coerce')

此错误显示：

ValueError: offset must be a timedelta strictly between -timedelta(hours=24) and timedelta(hours=24).

我怎么能忽略转换中出现的所有错误？我认为错误="错误"总是有用的。

Answer 1

严格解析

提供期望的格式，所有不匹配的内容都将变成NaT:

pd.to_datetime(df['MYTIME'], format='%Y-%m-%d %H:%M:%S', errors='coerce')

转换

假设1900年是一个错误。。。日期在字符串中间，您可以删除它们：

pd.to_datetime(df['MYTIME'].str.replace(r' S+ ', ' ', regex=True),
errors='coerce')

输出：

34180   2022-03-02 23:00:00
34181   2022-03-02 00:00:00
34182   2022-03-02 01:00:00
...
34219   2022-03-02 14:00:00
34220   2022-03-02 15:00:00
Name: MYTIME, dtype: datetime64[ns]

如果你想删除2022年的日期：

pd.to_datetime(df['MYTIME'].str.replace('^S+ ', ' ', regex=True),
errors='coerce')