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                ## 問題 你需要執行簡單的時間轉換,比如天到秒,小時到分鐘等的轉換。 ## 解決方案 為了執行不同時間單位的轉換和計算,請使用datetime模塊。比如,為了表示一個時間段,可以創建一個timedelta實例,就像下面這樣: >>> from datetime import timedelta >>> a = timedelta(days=2, hours=6) >>> b = timedelta(hours=4.5) >>> c = a + b >>> c.days 2 >>> c.seconds 37800 >>> c.seconds / 3600 10.5 >>> c.total_seconds() / 3600 58.5 >>> 如果你想表示指定的日期和時間,先創建一個datetime實例然后使用標準的數學運算來操作它們。比如: >>> from datetime import datetime >>> a = datetime(2012, 9, 23) >>> print(a + timedelta(days=10)) 2012-10-03 00:00:00 >>> >>> b = datetime(2012, 12, 21) >>> d = b - a >>> d.days 89 >>> now = datetime.today() >>> print(now) 2012-12-21 14:54:43.094063 >>> print(now + timedelta(minutes=10)) 2012-12-21 15:04:43.094063 >>> 在計算的時候,需要注意的是datetime會自動處理閏年。比如: >>> a = datetime(2012, 3, 1) >>> b = datetime(2012, 2, 28) >>> a - b datetime.timedelta(2) >>> (a - b).days 2 >>> c = datetime(2013, 3, 1) >>> d = datetime(2013, 2, 28) >>> (c - d).days 1 >>> ## 討論 對大多數基本的日期和時間處理問題,datetime模塊以及足夠了。如果你需要執行更加復雜的日期操作,比如處理時區,模糊時間范圍,節假日計算等等,可以考慮使用 [dateutil模塊](http://pypi.python.org/pypi/python-dateutil) 許多類似的時間計算可以使用 `dateutil.relativedelta()` 函數代替。但是,有一點需要注意的就是,它會在處理月份(還有它們的天數差距)的時候填充間隙。看例子最清楚: >>> a = datetime(2012, 9, 23) >>> a + timedelta(months=1) Traceback (most recent call last): File "<stdin>", line 1, in <module> TypeError: 'months' is an invalid keyword argument for this function >>> >>> from dateutil.relativedelta import relativedelta >>> a + relativedelta(months=+1) datetime.datetime(2012, 10, 23, 0, 0) >>> a + relativedelta(months=+4) datetime.datetime(2013, 1, 23, 0, 0) >>> >>> # Time between two dates >>> b = datetime(2012, 12, 21) >>> d = b - a >>> d datetime.timedelta(89) >>> d = relativedelta(b, a) >>> d relativedelta(months=+2, days=+28) >>> d.months 2 >>> d.days 28 >>>
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