使用公共键-值对对字典列表中的值求和

Sum values in a list of lists of dictionaries using common key-value pairs(使用公共键-值对对字典列表中的值求和)

问题描述

如何对词典列表中的重复元素求和？

样例列表：

data = [
        [
            {'user': 1, 'rating': 0},
            {'user': 2, 'rating': 10},
            {'user': 1, 'rating': 20},
            {'user': 3, 'rating': 10}
        ],
        [
            {'user': 4, 'rating': 4},
            {'user': 2, 'rating': 80},
            {'user': 1, 'rating': 20},
            {'user': 1, 'rating': 10}
        ],
    ]

预期输出：

op = [
        [
            {'user': 1, 'rating': 20},
            {'user': 2, 'rating': 10},
            {'user': 3, 'rating': 10}
        ],
        [
            {'user': 4, 'rating': 4},
            {'user': 2, 'rating': 80},
            {'user': 1, 'rating': 30},
        ],
    ]

推荐答案

您可以尝试：

from itertools import groupby

result = []
for lst in data:
    sublist = sorted(lst, key=lambda d: d['user'])
    grouped = groupby(sublist, key=lambda d: d['user'])
    result.append([
        {'user': name, 'rating': sum([d['rating'] for d in group])}
        for name, group in grouped])

# Sort the `result` `rating` wise:
result = [sorted(sub, key=lambda d: d['rating']) for sub in result]

# %%timeit
# 7.54 µs ± 220 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

---

更新(更高效的解决方案)：

result = []
for lst in data:
    visited = {}
    for d in lst:
        if d['user'] in  visited:
            visited[d['user']]['rating'] += d['rating'] 
        else:
            visited[d['user']] = d

    result.append(sorted(visited.values(), key=lambda d: d['rating']))

# %% timeit
# 2.5 µs ± 54 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

结果：

# print(result)
[
    [
        {'user': 2, 'rating': 10},
        {'user': 3, 'rating': 10},
        {'user': 1, 'rating': 20}
    ],
    [
        {'user': 4, 'rating': 4},
        {'user': 1, 'rating': 30},
        {'user': 2, 'rating': 80}
    ]
]

这篇关于使用公共键-值对对字典列表中的值求和的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持编程学习网！