Python 中时间日期对象之间的秒数差异

Question

我有以下代码可以根据实际时间和日期以及 Excel 电子表格中的 sunset/sunrise 时间和日期更改 Windows 10 上的背景图片.

from datetime import datetime
import pandas
import ctypes

file_path = "myfile.xlsx"
data = pandas.read_excel(file_path, header=0) #Header on line 0

#Today as day number in reference to 1st of Jan
day = datetime.now().timetuple().tm_yday

#Today's parameters
#sr and ss are column names in the Excel spreadsheet
#Minus 1 to account for 0 based indexing
**sunrise = data["sr"][day-1]
sunset = data["ss"][day-1]** 

#Time right now
**now = datetime.now().time()**

#Setting up the day_night variable depending on the now variable
if now > sunrise and now < sunset:
    day_night = 'day'
else:
    day_night = 'night'

#The path to the wallpapers being used
path = 'C:\wallpapers\'+ day_night +'.jpg'
SPI_SETDESKWALLPAPER = 20

#Function to change the wallpaper
def changeBG(path):
    ctypes.windll.user32.SystemParametersInfoW(SPI_SETDESKWALLPAPER, 0, path, 3) #SystemParametersInfoW for x64 architecture

changeBG(path)

有没有办法计算 now 变量和 sunrise 或 sunset 变量之间的时间差（以秒为单位）？到目前为止，我尝试了一个像 print (now-sunset) 这样的简单减法，它会 return 给我一个像这样的错误：

---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)
<ipython-input-4-c713ddbffa2f> in <module>
     18 now = datetime.now().time()
     19 
---> 20 print (now-sunset)

TypeError: unsupported operand type(s) for -: 'datetime.time' and 'datetime.time'

Answer 1

最直接的方法是手动将时间转换为秒：

from datetime import datetime, timedelta, time


def seconds_in_time(time_value: time):
    return (time_value.hour * 60 + time_value.minute) * 60 + time_value.second


# Examples of time objects
t1 = datetime.now().time()
t2 = (datetime.now() + timedelta(minutes=2)).time()

print(seconds_in_time(t2) - seconds_in_time(t1))  # 120

不确定是否有更好的方法，就您使用 time 个对象而言。

Python 中时间日期对象之间的秒数差异

Difference in seconds between timedate objects in Python

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datetime

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