2012-03-28 84 views

回答

9

使用列表sort method

In [1]: from datetime import date, timedelta 

In [2]: a=[date.today(), date.today() + timedelta(days=1), date.today() - timedelta(days=1)] 

In [3]: a 
Out[3]: 
[datetime.date(2012, 3, 28), 
datetime.date(2012, 3, 29), 
datetime.date(2012, 3, 27)] 

In [4]: a.sort() 

In [5]: a 
Out[5]: 
[datetime.date(2012, 3, 27), 
datetime.date(2012, 3, 28), 
datetime.date(2012, 3, 29)] 

In [6]: a[0] # min 
Out[6]: datetime.date(2012, 3, 27) 

In [7]: a[-1] # max 
Out[7]: datetime.date(2012, 3, 29) 

更多how to sort

3
list.sort(key=lambda item:item['date'], reverse=True) 
+0

然后你所要做的就是获得第一个和最后一个项目,你是MIN/MAX日期 – 2012-03-28 12:43:48

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