如何按降序对这个列表进行排序?
timestamps = [
"2010-04-20 10:07:30",
"2010-04-20 10:07:38",
"2010-04-20 10:07:52",
"2010-04-20 10:08:22",
"2010-04-20 10:08:22",
"2010-04-20 10:09:46",
"2010-04-20 10:10:37",
"2010-04-20 10:10:58",
"2010-04-20 10:11:50",
"2010-04-20 10:12:13",
"2010-04-20 10:12:13",
"2010-04-20 10:25:38"
]
当前回答
由于您的列表已经是升序的，我们可以简单地反转列表。
>>> timestamps.reverse()
>>> timestamps
['2010-04-20 10:25:38',
'2010-04-20 10:12:13',
'2010-04-20 10:12:13',
'2010-04-20 10:11:50',
'2010-04-20 10:10:58',
'2010-04-20 10:10:37',
'2010-04-20 10:09:46',
'2010-04-20 10:08:22',
'2010-04-20 10:08:22',
'2010-04-20 10:07:52',
'2010-04-20 10:07:38',
'2010-04-20 10:07:30']
其他回答
这是另一种方法
timestamps.sort()
timestamps.reverse()
print(timestamps)
你简单的类型:
timestamps.sort()
timestamps=timestamps[::-1]
这将为您提供一个排序版本的数组。
sorted(timestamps, reverse=True)
如果你想就地排序:
timestamps.sort(reverse=True)
检查文档排序如何
由于您的列表已经是升序的，我们可以简单地反转列表。
>>> timestamps.reverse()
>>> timestamps
['2010-04-20 10:25:38',
'2010-04-20 10:12:13',
'2010-04-20 10:12:13',
'2010-04-20 10:11:50',
'2010-04-20 10:10:58',
'2010-04-20 10:10:37',
'2010-04-20 10:09:46',
'2010-04-20 10:08:22',
'2010-04-20 10:08:22',
'2010-04-20 10:07:52',
'2010-04-20 10:07:38',
'2010-04-20 10:07:30']
在一行中，使用lambda:
timestamps.sort(key=lambda x: time.strptime(x, '%Y-%m-%d %H:%M:%S')[0:6], reverse=True)
传递一个函数给list.sort:
def foo(x):
return time.strptime(x, '%Y-%m-%d %H:%M:%S')[0:6]
timestamps.sort(key=foo, reverse=True)
