python - 当 str_ 无法提升 datetime64 时，使用 numpy.where

Question

import pandas as pd
from datetime import timedelta
import numpy as np

df = pd.DataFrame({
    'open_local_data':['2022-08-24 15:00:00','2022-08-24 18:00:00'],
    'result':['WINNER','']
})
df['open_local_data'] = pd.to_datetime(df['open_local_data'])
df['clock_now'] = np.where(
    df['result'] != '',
    df['open_local_data'] + timedelta(minutes=150),
    ''
)
print(df[['open_local_data','clock_now']])

由于我必须使用条件进行工作，并且只有在稍后才决定是否处理列中的更改，因此如果收到此错误，我应该怎么做:

    df['clock_now'] = np.where(
  File "<__array_function__ internals>", line 180, in where
TypeError: The DType <class 'numpy.dtype[datetime64]'> could not be promoted by <class 'numpy.dtype[str_]'>. This means that no common DType exists for the given inputs. For example they cannot be stored in a single array unless the dtype is `object`. The full list of DTypes is: (<class 'numpy.dtype[datetime64]'>, <class 'numpy.dtype[str_]'>)

Answer 1

你可以.astype(str)添加，这样NumPy就会很高兴，但最后你会得到字符串。相反，您可以使用df.where:

df["clock_now"] = df["result"].where(df["result"].eq(""),
                                     other=df["open_local_data"].add(pd.Timedelta("150min")))

• 保持“结果”值不变，它们等于空字符串
• 并将 local_data + 150 分钟发送到其他地点

获取

>>> df

      open_local_data  result            clock_now
0 2022-08-24 15:00:00  WINNER  2022-08-24 17:30:00
1 2022-08-24 18:00:00

其中 df.at[0, "clock_now"] 实际上是时间戳，而不是字符串。