在 Python 列表中合并字典

Question

假设我有这个：

[
    {"name": "bob", "total": 1},
    {"name": "alice", "total": 5},
    {"name": "eve", "total": 2},
    {"name": "bob", "total": 3},
    {"name": "alice", "total": 2},
    {"name": "alice", "total": 2},
]

我想将此列表转换为：

[
    {"name": "bob", "total": 4},
    {"name": "alice", "total": 9},
    {"name": "eve", "total": 2}
]

现在，我遍历整个第二个列表以查找第一个列表的每个循环的键是否存在。

如何以较低的复杂性实现这一目标？

Answer 1

from collections import defaultdict

a = [
    {"name": "bob", "total": 1},
    {"name": "alice", "total": 5},
    {"name": "eve", "total": 2},
    {"name": "bob", "total": 3},
    {"name": "alice", "total": 2},
    {"name": "alice", "total": 2},
]

# calculate the frequency of each key
freq = defaultdict(lambda: 0)
for d in a:
    freq[d['name']] += d['total']

# build the result list
a = list()
for key, val in freq.items():
    a.append({'name': key, 'total': val})
print(a)

Answer 2

如果您只有两条信息（姓名和总数），我建议您稍微更改一下您的架构。而不是字典列表，而是使用单个字典，其中键是名称，值是总数：

>>> values = [
...     {"name": "bob", "total": 1},
...     {"name": "alice", "total": 5},
...     {"name": "eve", "total": 2},
...     {"name": "bob", "total": 3},
...     {"name": "alice", "total": 2},
...     {"name": "alice", "total": 2},
... ]
>>> from collections import defaultdict
>>> totals_by_name = defaultdict(int)
>>> for value in values:
...     totals_by_name[value["name"]] += value["total"]
... 
>>> totals_by_name
defaultdict(<class 'int'>, {'bob': 4, 'alice': 9, 'eve': 2})

即使您有更多要按名称查找的数据也可以使用（将整数值替换为存储总数和其他数据的嵌套字典）。

Answer 3

您可以使用itertools 模块中的groupby：

from itertools import groupby
from operator import itemgetter

# itemgetter(foo) is roughly equivalent to lambda x: x[foo]
get_name = itemgetter('name')
get_total = itemgetter('total')

lst = [
    {"name": "bob", "total": 1},
    {"name": "alice", "total": 5},
    {"name": "eve", "total": 2},
    {"name": "bob", "total": 3},
    {"name": "alice", "total": 2},
    {"name": "alice", "total": 2},
]
grouped = groupby(sorted(lst, key=get_name), get_name)
new_list = [{'name': k, 'total': sum(get_total(x) for x in v)} for k, v in grouped]

groupby 将根据'name' 属性的公共值生成一个新序列，该序列将原始列表中的字典收集到子序列中。对其进行迭代，您可以提取所有 total 值以汇总以用于新的 dict 值列表。

Answer 4

比方说，

your_data = [
    {"name": "bob", "total": 1},
    {"name": "alice", "total": 5},
    {"name": "eve", "total": 2},
    {"name": "bob", "total": 3},
    {"name": "alice", "total": 2},
    {"name": "alice", "total": 2},
]

您可以简单地使用 pandas 来接收所需的输出。

import pandas as pd

df = pd.DataFrame(your_data)    
df = df.groupby(by = 'name', as_index = False).sum('total')
result = df.to_dict(orient = 'records')

输出: [{'name': 'alice', 'total': 9}, {'name': 'bob', 'total': 4}, {'name': “前夕”，“总计”：2}]