Question

我有一个嵌套的字典，其中包含有关书籍的数据：

UID
条件
价格

这是定义：

books = {
    'uid1':
        {'price': '100',
        'condition': 'good'},
    'uid2':
        {'price': '80',
        'condition': 'fair'},
    'uid3':
        {'price': '150',
        'condition': 'excellent'},
    'uid4':
        {'price': '70',
        'condition': 'fair'},
    'uid5':
        {'price': '180',
        'condition': 'excellent'},
    'uid6':
        {'price': '60',
        'condition': 'fair'}
    }

我需要获取按条件分组的平均价格。因此，预期结果是：

{'fair': 70, 'good': 100, 'excellent': 165}

最Python化的方法是什么？

Answer 1

使用collections.defaultdict

演示：

from collections import defaultdict

res = defaultdict(list)
for k,v in books.items():
    res[v['condition']].append(int(v['price'])) 

print({k: sum(v)/len(v) for k, v in res.items() })

输出：

{'good': 100, 'fair': 70, 'excellent': 165}

Answer 2

我想用熊猫图书馆回答这个问题。

import pandas as pd
books = {
    'uid1':
        {'price': '100',
        'condition': 'good'},
    'uid2':
        {'price': '80',
        'condition': 'fair'},
    'uid3':
        {'price': '150',
        'condition': 'excellent'},
    'uid4':
        {'price': '70',
        'condition': 'fair'},
    'uid5':
        {'price': '180',
        'condition': 'excellent'},
    'uid6':
        {'price': '60',
        'condition': 'fair'}
   }
data = pd.DataFrame.from_dict(books, orient='index')
data['price'] = data[['price']].apply(pd.to_numeric)
data.groupby(['condition'])['price'].mean()

输出：

condition
excellent    165
fair          70
good         100

Answer 3

这是一种方法：

from statistics import mean
result = {condition: mean(float(book['price']) for book in books.values() if book['condition'] == condition) for condition in ('fair','good','excellent')}

#result = {'fair': 70.0, 'good': 100.0, 'excellent': 165.0}

Answer 4

除了不使用defaultdict之外，我看不到为什么需要Try Except-

for k, v in books.items():
    try:
        avg[v['condition']].append(int(v['price']))
    except KeyError:
        avg[v['condition']] = [int(v['price'])]
avg = {k: sum(v)/len(v) for k, v in avg.items()}

一种“查询”字典的Python方法

4 个答案: