python - 为日期和时间分配数值

Question

我在 df 中有一个列，其中包含与该行的客户每周时间相关的数值。例如0 = Sunday 12:00am和24 将是 Monday 12:00am和5 将是 Sunday 5:00am。

  Value 
    0
    4
    10
    24

  Value  Expected Output Column
    0           Sunday 12:00am
    4           Sunday 4:00am
    10          Sunday 10:00am
    24          Monday 12:00am
    49          Tuesday 1:00am

如果我希望将所有值分配给一周中正确的相应日期和时间，如何创建新列？值从 0 开始，表示星期天上午 12:00 的第一个值，到 167 结束，即该周的星期六晚上 11:59。

谢谢！

Answer 1

这可以满足您的要求：

Value = pd.Series([0, 4, 10, 24, 49])
AnyGivenSunday = pd.to_datetime('Sunday 2020') # yes this works
(AnyGivenSunday + pd.to_timedelta(Value, 'h')).dt.strftime('%A %I:%M%P')

Answer 2

import pandas as pd
from datetime import datetime
import calendar as cal

df = pd.DataFrame(data={'val': [0, 0, 0, 0, 0], 'date': ["Sunday 12:00am", "Sunday 4:00am", "Sunday 10:00am", "Monday 12:00am", "Tuesday 1:00am"]})

def convertDate(dateString):
    hour = datetime.strptime(dateString, '%A %I:%M%p').hour # Convert to 24 hour format
    day, time = dateString.split(" ") # Get text of day, cannot get datetime.weekday() without full date
    if day.lower() == cal.day_name[6].lower(): # cal.day_name index starts at Monday and not Sunday
        return hour + ((list(cal.day_name).index(day) - 6) * 24)
    else:
        return hour + ((list(cal.day_name).index(day) + 1) * 24)

df['val'] = df['date'].apply(convertDate) # Apply function to all columns