递归合并字典，以便将具有共享键的元素组合成一个列表

Question

我有两个想要合并的字典：

a = {"name": "john",
     "phone":"123123123",
     "owns": {"cars": "Car 1", "motorbikes": "Motorbike 1"}}

b = {"name": "john",
     "phone":"123",
     "owns": {"cars": "Car 2"}}

如果

a

和

b

在同一嵌套级别上有一个公共键，则结果应该是一个列表，其中包含两个值，该列表被指定为共享键的值。

结果应该是这样的：

{"name": "john",
 "phone":["123123123","123"],
 "owns": {"cars": ["Car 1", "Car 2"], "motorbikes": "Motorbike 1"}}

使用

a.update(b)

不起作用，因为它会用

a

的共享值覆盖

b

的共享值，结果如下所示：

{'name': 'john', 'phone': '123', 'owns': {'cars': 'Car 2'}}

目标是合并字典而不覆盖并保留与特定键相关的所有信息（在任一字典中）。

Answer 1

通过递归，您可以构建一个字典理解来实现这一点。

此解决方案还考虑到您可能希望稍后合并两个以上的字典，从而在这种情况下展平值列表。

def update_merge(d1, d2):
    if isinstance(d1, dict) and isinstance(d2, dict):
        # Unwrap d1 and d2 in new dictionary to keep non-shared keys with **d1, **d2
        # Next unwrap a dict that treats shared keys
        # If two keys have an equal value, we take that value as new value
        # If the values are not equal, we recursively merge them
        return {
            **d1, **d2,
            **{k: d1[k] if d1[k] == d2[k] else update_merge(d1[k], d2[k])
            for k in {*d1} & {*d2}}
        }
    else:
        # This case happens when values are merged
        # It bundle values in a list, making sure
        # to flatten them if they are already lists
        return [
            *(d1 if isinstance(d1, list) else [d1]),
            *(d2 if isinstance(d2, list) else [d2])
        ]

示例：

a = {"name": "john", "phone":"123123123",
     "owns": {"cars": "Car 1", "motorbikes": "Motorbike 1"}}
b = {"name": "john", "phone":"123", "owns": {"cars": "Car 2"}}

update_merge(a, b)
# {'name': 'john',
#  'phone': ['123123123', '123'],
#  'owns': {'cars': ['Car 1', 'Car 2'], 'motorbikes': 'Motorbike 1'}}

合并两个以上对象的示例：

a = {"name": "john"}
b = {"name": "jack"}
c = {"name": "joe"}

d = update_merge(a, b)
d = update_merge(d, c)

d # {'name': ['john', 'jack', 'joe']}

Answer 2

您可以使用

itertools.groupby

和递归：

import itertools, sys
a = {"name": "john", "phone":"123123123", "owns": {"cars": "Car 1", "motorbikes": "Motorbike 1"}}
b = {"name": "john", "phone":"123", "owns": {"cars": "Car 2"}}
def condense(r):
  return r[0] if len(set(r)) == 1 else r

def update_dict(c, d):
  _v = {j:[c for _, c in h] for j, h in itertools.groupby(sorted(list(c.items())+list(d.items()), key=lambda x:x[0]), key=lambda x:x[0])}
  return {j:update_dict(*e) if all(isinstance(i, dict) for i in e) else condense(e) for j, e in _v.items()}

print(update_dict(a, b))

输出：

{'name': 'john', 'owns': {'cars': ['Car 1', 'Car 2'], 'motorbikes': 'Motorbike 1'}, 'phone': ['123123123', '123']}

Answer 3

使用集合和事物，还可以合并任意数量的字典：

from functools import reduce
import operator

# Usage: merge(a, b, ...)
def merge(*args):
    # Make a copy of the input dicts, can be removed if you don't care about modifying
    # the original dicts.
    args = list(map(dict.copy, args))

    # Dict to store the result.
    out = {}

    for k in reduce(operator.and_, map(dict.keys, args)):  # Python 3 only, see footnotes.
        # Use `.pop()` so that after the all elements of shared keys have been combined,
        # `args` becomes a list of disjoint dicts that we can merge easily.
        vs = [d.pop(k) for d in args]

        if isinstance(vs[0], dict):
            # Recursively merge nested dicts
            common = merge(*vs)
        else:
            # Use a set to collect unique values
            common = set(vs)
            # If only one unique value, store that as is, otherwise use a list
            common = next(iter(common)) if len(common) == 1 else list(common)

        out[k] = common

    # Merge into `out` the rest of the now disjoint dicts
    for arg in args:
        out.update(arg)

    return out

假设要合并的每个字典具有相同的“结构”，因此

"owns"

不能是

a

中的列表和

b

中的字典。字典的每个元素也需要是可哈希的，因为此方法使用集合来聚合唯一值。

---

以下内容仅适用于 Python 3，因为在 Python 2 中，

dict.keys()

返回一个普通的旧列表。

reduce(operator.and_, map(dict.keys, args))

另一种方法是添加额外的

map()

将列表转换为集合：

reduce(operator.and_, map(set, map(dict.keys, args)))

Answer 4

这是支持任意数量参数的通用解决方案：

def _merge_dicts(dict_args):
    if not isinstance(dict_args[0], dict):
        return list(set(dict_args)) if len(set(dict_args)) > 1 else dict_args[0]
    keys = set().union(*dict_args)
    result = {key: 
              _merge_dicts(([d.get(key, None) for d in dict_args if d.get(key, None) is not None])) 
              for key in keys}
    return result
def merge_dicts(*dict_args):
    return _merge_dicts(dict_args)

a = {"name": "john",
     "phone":"123123123",
     "owns": {"cars": "Car 1", "motorbikes": "Motorbike 1"}}

b = {"name": "john",
     "phone":"123",
     "owns": {"cars": "Car 2"}}

merge_dicts(a, b)

产量

{'name': 'john',
 'owns': {'motorbikes': 'Motorbike 1', 'cars': ['Car 2', 'Car 1']},
 'phone': ['123123123', '123']}