从dict创建dict列表的更为pythonic的方法,其中列表元素是`{v:[k1,k2,k3]}形式的dict`

您也可以使用列表理解:

[dict((_, )) for _ in {v: [_k for _k, _v in d.items() if _v ==  v] for v in d.values()}.items()]

来证明它是有效的

>>> d = { 'b': 68, 'c': 68, 'x': 68, 'z': 401, 'aa':401, 'a': 2}
>>> [dict((_, )) for _ in {v: [_k for _k, _v in d.items() if _v ==  v] for v in d.values()}.items()]
[{68: ['b', 'c', 'x']}, {401: ['z', 'aa']}, {2: ['a']}]

我不知道这是什么用法,但如果你可以的话,我建议使用一个dict而不是一个dict列表,因为这种理解比单独使用dict理解要慢。如果可能的话,我建议只使用dict理解,这在我看来会更好,虽然返回一个dict而不是一个dict列表 更像蟒蛇 或者至少更有意义,特别是你可以用同样的方法迭代它 dict.items() .

{v: [_k for _k, _v in d.items() if _v ==  v] for v in d.values()}

这将返回一个dict,如下所示:

{68: ['b', 'c', 'x'], 401: ['z', 'aa'], 2: ['a']}

我建议如下。未经测试。

from collections import defaultdict

a  = { 'b': 66, 'b': 68, 'c': 68, 'x': 68, 'z': 401, 'aa':401, 'a': 2}

t = defaultdict(list)
for k, v in a.items():
    t[v].append(k)

o = [dict((p,)) for p in t.items()]

a = { 'b': 66, 'b': 68, 'c': 68, 'x': 68, 'z': 401, 'aa':401, 'a': 2 }
b = []

for key, val in a.items():
    if (any(val in d for d in b)):
        b[next(i for i, d in enumerate(b) if val in d)][val].append(key)
    else:
        d = {}
        d[val] = [key]
        b.append(d)

print(b)

[{68: ['b', 'c', 'x']}, {401: ['z', 'aa']}, {2: ['a']}]

a = { 'b': 68, 'c': 68, 'x': 68, 'z': 401, 'aa':401, 'a': 2}
b = {}

for k, v in a.items():
    b.setdefault(v, []).append(k)

c = [{k: b[k]} for k in b]
print(c)

输出:

[{68: ['b', 'c', 'x']}, {401: ['z', 'aa']}, {2: ['a']}]

有些相同,但有点优化:

a  = { 'b': 66, 'b': 68, 'c': 68, 'x': 68, 'z': 401, 'aa':401, 'a': 2}
b  = {}
for k,v in a.items():
    val = b.get(v,[])
    val.append(k)
    b[v] = val

lst = []
for k,v in b.items():
    d = dict()
    d[k] = v
    lst.append(d)

print(lst)

[{68: ['b', 'c', 'x']}, {401: ['z', 'aa']}, {2: ['a']}]

使用 defuhaltdict

from collections import defaultdict
dic = defaultdict(list)
k= { 'b': 66, 'b': 68, 'c': 68, 'x': 68, 'z': 401, 'aa':401, 'a': 2}
for ke, v in k.items():
     dic[v]+=[ke]

print(dict(dic))

输出

{68: ['b', 'c', 'x'], 401: ['z', 'aa'], 2: ['a']}

Enumerate()

origdict = { 'b': 66, 'b': 68, 'c': 68, 'x': 68, 'z': 401, 'aa':401, 'a': 2}
my_list = []
my_dict = {}

for index, (key, value) in enumerate(origdict.items()):
    if value in my_dict:
        my_dict[value].append(key)
    else:
        #validate is not first index of list
        if index > 0:
            my_list.append(my_dict)
        my_dict = {}
        my_dict[value] = [key]

    #validate last index of list
    if index == len(origdict)-1:
        my_list.append(my_dict)

print(my_list)

订单号:

[{68: ['b', 'c', 'x']}, {401: ['z', 'aa']}, {2: ['a']}]