根据公共键值汇总字典列表

Question

我有一个字典的列表，长得像这样：

dictlist = [{'day': 0, 'start': '8:00am', 'end': '5:00pm'},
            {'day': 1, 'start': '10:00am', 'end': '7:00pm'},
            {'day': 2, 'start': '8:00am', 'end': '5:00pm'},
            {'day': 3, 'start': '10:00am', 'end': '7:00pm'},
            {'day': 4, 'start': '8:00am', 'end': '5:00pm'},
            {'day': 5, 'start': '11:00am', 'end': '1:00pm'}]

我想把那些有相同的 'start' 和 'end' 时间的天数总结在一起。

举个例子，

summarylist = [([0,2, 4], '8:00am', '5:00pm'),
               ([1, 3], '10:00am', '7:00pm')
               ([5], '11:00am', '1:00pm')]

我试过用其他人在StackOverflow上提供的关于集合和交集的解决方案来实现这个目标，但都没有成功。我还尝试过重新利用这个问题的解决方案，但也没能解决。希望有人能给我指个明路。

Answer 1

你可以这样使用 itertools.groupby。

源代码：

from itertools import groupby
for k, grp in groupby(sorted(dictlist, key=lambda x:(x['end'], x['start'])), key=lambda x:(x['start'], x['end'])):
    print [i['day'] for i in grp], k

输出结果：

[5] ('11:00am', '1:00pm')
[0, 2, 4] ('8:00am', '5:00pm')
[1, 3] ('10:00am', '7:00pm')

不过我觉得在这个特定的情况下，使用 defaultdict（@Akavall 的回答）是更合适的做法。

Answer 2

使用 itertools.groupby：

In [1]: %paste
dictlist = [{'day': 0, 'start': '8:00am',  'end': '5:00pm'},
            {'day': 1, 'start': '10:00am', 'end': '7:00pm'},
            {'day': 2, 'start': '8:00am',  'end': '5:00pm'},
            {'day': 3, 'start': '10:00am', 'end': '7:00pm'},
            {'day': 4, 'start': '8:00am',  'end': '5:00pm'},
            {'day': 5, 'start': '11:00am', 'end': '1:00pm'}]

## -- End pasted text --

In [2]: from itertools import groupby

In [3]: tuplist = [(d['day'], (d['start'], d['end'])) for d in dictlist]

In [4]: key = lambda x: x[1]

In [5]: summarylist = [(sorted(e[0] for e in g),) + k
   ...:        for k, g in groupby(sorted(tuplist, key=key), key=key)]

In [6]: summarylist
Out[6]:
[([1, 3], '10:00am', '7:00pm'),
 ([5], '11:00am', '1:00pm'),
 ([0, 2, 4], '8:00am', '5:00pm')]

Answer 3

如果你不需要提供的格式完全一样的话，可以使用 defaultdict。

dictlist = [{'day': 0, 'start': '8:00am', 'end': '5:00pm'},
            {'day': 1, 'start': '10:00am', 'end': '7:00pm'},
            {'day': 2, 'start': '8:00am', 'end': '5:00pm'},
            {'day': 3, 'start': '10:00am', 'end': '7:00pm'},
            {'day': 4, 'start': '8:00am', 'end': '5:00pm'},
            {'day': 5, 'start': '11:00am', 'end': '1:00pm'}]

from collections import defaultdict

dd = defaultdict(list)

for d in dictlist:
    dd[(d['start'],d['end'])].append(d['day'])

结果：

>>> dd
defaultdict(<type 'list'>, {('11:00am', '1:00pm'): [5], ('10:00am', '7:00pm'): [1, 3], ('8:00am', '5:00pm'): [0, 2, 4]})

如果格式对你很重要的话，你可以这样做：

>>> my_list = [(v, k[0], k[1]) for k,v in dd.iteritems()]
>>> my_list
[([5], '11:00am', '1:00pm'), ([1, 3], '10:00am', '7:00pm'), ([0, 2, 4], '8:00am', '5:00pm')]
>>> # If you need the output sorted:  
>>> sorted_my_list = sorted(my_list, key = lambda k : len(k[0]), reverse=True)
>>> sorted_my_list
[([0, 2, 4], '8:00am', '5:00pm'), ([1, 3], '10:00am', '7:00pm'), ([5], '11:00am', '1:00pm')]