Question

我有一个类似于上周问过的问题： "Sorting dict items by key, beyond alphanumeric sorting"

问题是正确的，但现在我的dict的键只是整数：

>>>lis_nodiz = list(nodiz.items()) #list of dict items as tuples
[(2, 316),
(3, 66),
(4, 37),
(5, 15),
(6, 10),
(7, 4),
(8, 3),
(9, 1),
(10, 2),
(11, 1),
(12, 1),
(45, 1),
(109, 1),
(16, 1),
(126, 1)]

正如您猜测的那样，我正在搜索这个有序的元组列表：

[(2, 316),
(3, 66),
(4, 37),
(5, 15),
(6, 10),
(7, 4),
(8, 3),
(9, 1),
(10, 2),
(11, 1),
(12, 1),
(16, 1),
(45, 1),
(109, 1),
(126, 1)]

我试图重用“key_func”Padraic Cunningham建议我，不幸的是这个函数似乎只能用字符串类型的dict键：

def key_func(x):
"""'a0p12' -> (0, 12)"""
return tuple(int("".join(v)) for k,v in groupby(x[0], key=str.isdigit) if k)

>>>lis_nodiz_od = sorted(nodiz.items(), key=key_func)
TypeError: expected string or buffer

如何修改功能，或者我如何以其他方式来实现我的目的？如果有人解释“key_func”是如何工作的，并且为了使它与整数一起工作而做的正确修改，我将非常感激。通过这种方式，我希望更多地了解Python编程，而不仅仅是解决我今天的问题！非常感谢你们！

Answer 1

将@app.task(bind=True) def myfunc(): # notifications.send_message_to_admin('sdaa','dsadasdsa') with open('text.txt','a') as f: f.write('sa')或> celery -A dolava beat -l info celery beat v3.1.23 (Cipater) is starting. __ - ... __ - _ Configuration -> . broker -> amqp://guest:**@localhost:5672// . loader -> celery.loaders.app.AppLoader . scheduler -> djcelery.schedulers.DatabaseScheduler . logfile -> [stderr]@%INFO . maxinterval -> now (0s) [2016-10-26 17:46:50,135: INFO/MainProcess] beat: Starting... [2016-10-26 17:46:50,138: INFO/MainProcess] Writing entries... [2016-10-26 17:46:51,433: INFO/MainProcess] DatabaseScheduler: Schedule changed. [2016-10-26 17:46:51,433: INFO/MainProcess] Writing entries... [2016-10-26 17:46:51,812: INFO/MainProcess] Scheduler: Sending due task schedule-name (dolava_app.tasks.myfunc) [2016-10-26 17:46:51,864: INFO/MainProcess] Writing entries... [2016-10-26 17:46:57,138: INFO/MainProcess] Scheduler: Sending due task schedule-name (dolava_app.tasks.myfunc) [2016-10-26 17:47:02,230: INFO/MainProcess] Scheduler: Sending due task schedule-name (dolava_app.tasks.myfunc)称为：

sorted()

其中list.sort()是元组列表，如上所述。

注意：此处无需指定# sorted(): Creates new list with sorted order >>> sorted(lis_nodiz) [(2, 316), (3, 66), (4, 37), (5, 15), (6, 10), (7, 4), (8, 3), (9, 1), (10, 2), (11, 1), (12, 1), (16, 1), (45, 1), (109, 1), (126, 1)] # list.sort(): sort the existing list >>> lis_nodiz.sort() >>> lis_nodiz [(2, 316), (3, 66), (4, 37), (5, 15), (6, 10), (7, 4), (8, 3), (9, 1), (10, 2), (11, 1), (12, 1), (16, 1), (45, 1), (109, 1), (126, 1)]。默认情况下，在第0个索引上进行词典排序;在相同值的情况下，对第一个索引进行排序。

Answer 2

无需在第一个键上使用lambda排序：这是元组的默认值。

>>> sorted(x)
[(2, 316), (3, 66), (4, 37), (5, 15), (6, 10), (7, 4), (8, 3), (9, 1), (10, 2), (11, 1), (12, 1), (16, 1), (45, 1), (109, 1), (126, 1)]

它将自动对第一个成员进行排序，如果它们相等则将按第二个成员排序等。

通过整数键ad hoc函数对dict项进行排序

2 个答案: