在 Python 中合并（同时保留值）多个字典的最快方法？

Question

我搜索了合并字典，但我查看的结果都假设 替换 的值。也就是说，如果你有一个像{'config_prop': 2} 这样的dict，另一个像{'config_prop': 7} 这样的dict，则合并的最终结果是{'config_prop': 7}。我要的是{'config_prop': 9}。

我的幼稚方法如下，虽然有效，但速度很慢。

split_output = [{'some_prop': 1}, {'some_prop': 2, 'other_prop': 19}]
combined_output = {}
    for d in split_output:
        if combined_output == {}:
            combined_output = d.copy()
        else:
            for key, value in d.items():
                if key in combined_output:
                    combined_output[key] = combined_output[key] + value  # add to existing val
                else:
                    combined_output[key] = value

我很想听听有关更好的方法的建议。谢谢！

更新：我试过了，但它比我的原始代码慢得多：

final_count = Counter() 
for d in split_output:
    final_count += Counter(d)   
final_output = dict(final_count)

Answer 1

您可以使用以下内容：

from collections import Counter

A = Counter({'config_prop': 2})
B = Counter({'config_prop': 7})
A + B

Answer 2

可以使用字典理解

combo = [{'some_prop': 1}, {'some_prop': 2, 'other_prop': 19}]

d = {k: sum([i[k] for i in combo]) if k in combo[0] else i[k] for i in combo for k in i}

{'some_prop': 3, 'other_prop': 19}

Answer 3

我很快介绍了几种不同的方法来解决您的问题：

combined_output = [{'some_prop': 1}, {'some_prop': 2, 'other_prop': 19}]

# set key or append value to new dictionary object:

def dict_sum(dicts):
    output = {}
    for d in dicts:
        for key, value in d.iteritems():
            if key in output:
                output[key] += value
            else:
                output[key] = value
     return output
    
# Implement a reducer function using functools:

from functools import reduce

def reducer(accumulator, element):
    for key, value in element.items():
        accumulator[key] = accumulator.get(key, 0) + value
    return accumulator
    
# Use a Counter:

from collections import Counter

def sum_dicts_values_by_key(dicts):
    return dict(sum([Counter(x) for x in combined_output], Counter()))

# Using dictionary comprehension:

def sum_dict_comprehension(dicts):
    return {k: sum([i[k] for i in dicts]) 
    if k in dicts[0] else i[k] for i in dicts for k in i}

使用timeit 运行快速基准测试以进行比较：

res_1 = dict_sum(combined_output)
1000000 loops, best of 3: 457 ns per loop

res_2 = reduce(reducer, combined_output, {})
1000000 loops, best of 3: 1.35 µs per loop

res_3 = sum_dicts_values_by_key(combined_output)
100000 loops, best of 3: 12.8 µs per loop

res_4 = sum_dict_comprehension(combined_output)
1000000 loops, best of 3: 1.53 µs per loop