我正在寻找一种方法,通过包含最多值的dict对dict进行排序。
这是原文:
test = {u'Beta': {
u'SW Engineering': {
u'Resolved': {u'2017-10-06 08:04:15': 1, u'2017-10-15 00:19:35': 3, u'2017-10-11 00:19:29': 2, u'2017-10-09 00:19:00': 1, u'2017-10-14 00:21:09': 3, u'2017-10-12 00:19:52': 3, u'2017-10-08 00:18:58': 1, u'2017-10-10 00:19:31': 2, u'2017-10-13 00:19:28': 3, u'2017-10-07 09:54:33': 1},
u'In Progress': {u'2017-10-06 08:04:15': 9, u'2017-10-15 00:19:35': 12, u'2017-10-11 00:19:29': 11, u'2017-10-09 00:19:00': 9, u'2017-10-14 00:21:09': 12, u'2017-10-12 00:19:52': 11, u'2017-10-08 00:18:58': 9, u'2017-10-10 00:19:31': 10, u'2017-10-13 00:19:28': 11, u'2017-10-07 09:54:33': 9},
u'In Testing': {u'2017-10-06 08:04:15': 7, u'2017-10-15 00:19:35': 7, u'2017-10-11 00:19:29': 7, u'2017-10-09 00:19:00': 7, u'2017-10-14 00:21:09': 7, u'2017-10-12 00:19:52': 7, u'2017-10-08 00:18:58': 7, u'2017-10-10 00:19:31': 7, u'2017-10-13 00:19:28': 7, u'2017-10-07 09:54:33': 7},
u'Reopened': {u'2017-10-06 08:04:15': 1, u'2017-10-15 00:19:35': 1, u'2017-10-11 00:19:29': 1, u'2017-10-09 00:19:00': 1, u'2017-10-14 00:21:09': 1, u'2017-10-12 00:19:52': 1, u'2017-10-08 00:18:58': 1, u'2017-10-10 00:19:31': 1, u'2017-10-13 00:19:28': 1, u'2017-10-07 09:54:33': 1},
u'Closed': {u'2017-10-06 08:04:15': 17, u'2017-10-15 00:19:35': 18, u'2017-10-11 00:19:29': 18, u'2017-10-09 00:19:00': 17, u'2017-10-14 00:21:09': 18, u'2017-10-12 00:19:52': 18, u'2017-10-08 00:18:58': 17, u'2017-10-10 00:19:31': 18, u'2017-10-13 00:19:28': 18, u'2017-10-07 09:54:33': 17},
u'Open': {u'2017-10-06 08:04:15': 5, u'2017-10-15 00:19:35': 8, u'2017-10-11 00:19:29': 8, u'2017-10-09 00:19:00': 5, u'2017-10-14 00:21:09': 8, u'2017-10-12 00:19:52': 7, u'2017-10-08 00:18:58': 5, u'2017-10-10 00:19:31': 8, u'2017-10-13 00:19:28': 8, u'2017-10-07 09:54:33': 5}},
u'DSP Engineering': {
u'In Progress': {u'2017-10-15 00:19:35': 1, u'2017-10-14 00:21:09': 1}}}}
这就是我想要的样子:
test = {u'Beta': {
u'DSP Engineering': {
u'In Progress': {u'2017-10-15 00:19:35': 1, u'2017-10-14 00:21:09': 1}},
u'SW Engineering': {
u'Resolved': {u'2017-10-06 08:04:15': 1, u'2017-10-15 00:19:35': 3, u'2017-10-11 00:19:29': 2, u'2017-10-09 00:19:00': 1, u'2017-10-14 00:21:09': 3, u'2017-10-12 00:19:52': 3, u'2017-10-08 00:18:58': 1, u'2017-10-10 00:19:31': 2, u'2017-10-13 00:19:28': 3, u'2017-10-07 09:54:33': 1},
u'In Progress': {u'2017-10-06 08:04:15': 9, u'2017-10-15 00:19:35': 12, u'2017-10-11 00:19:29': 11, u'2017-10-09 00:19:00': 9, u'2017-10-14 00:21:09': 12, u'2017-10-12 00:19:52': 11, u'2017-10-08 00:18:58': 9, u'2017-10-10 00:19:31': 10, u'2017-10-13 00:19:28': 11, u'2017-10-07 09:54:33': 9},
u'In Testing': {u'2017-10-06 08:04:15': 7, u'2017-10-15 00:19:35': 7, u'2017-10-11 00:19:29': 7, u'2017-10-09 00:19:00': 7, u'2017-10-14 00:21:09': 7, u'2017-10-12 00:19:52': 7, u'2017-10-08 00:18:58': 7, u'2017-10-10 00:19:31': 7, u'2017-10-13 00:19:28': 7, u'2017-10-07 09:54:33': 7},
u'Reopened': {u'2017-10-06 08:04:15': 1, u'2017-10-15 00:19:35': 1, u'2017-10-11 00:19:29': 1, u'2017-10-09 00:19:00': 1, u'2017-10-14 00:21:09': 1, u'2017-10-12 00:19:52': 1, u'2017-10-08 00:18:58': 1, u'2017-10-10 00:19:31': 1, u'2017-10-13 00:19:28': 1, u'2017-10-07 09:54:33': 1},
u'Closed': {u'2017-10-06 08:04:15': 17, u'2017-10-15 00:19:35': 18, u'2017-10-11 00:19:29': 18, u'2017-10-09 00:19:00': 17, u'2017-10-14 00:21:09': 18, u'2017-10-12 00:19:52': 18, u'2017-10-08 00:18:58': 17, u'2017-10-10 00:19:31': 18, u'2017-10-13 00:19:28': 18, u'2017-10-07 09:54:33': 17},
u'Open': {u'2017-10-06 08:04:15': 5, u'2017-10-15 00:19:35': 8, u'2017-10-11 00:19:29': 8, u'2017-10-09 00:19:00': 5, u'2017-10-14 00:21:09': 8, u'2017-10-12 00:19:52': 7, u'2017-10-08 00:18:58': 5, u'2017-10-10 00:19:31': 8, u'2017-10-13 00:19:28': 8, u'2017-10-07 09:54:33': 5}}}}
这不是最好的解决办法,但你可以试试。 为了测试的目的,我在dict中又增加了一个项目,它的最大元素名为“swengineering33”
如何使用
OrderedDict
执行此操作的示例。你知道吗相关问题 更多 >
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