字典中的总和值小于某个值答案

【问题标题】：sum values in dictionary less than a certain value字典中的总和值小于某个值
【发布时间】：2018-12-08 12:40:37
【问题描述】：

我有以下字典，并试图从中制作一个饼图，但我只想包括前 5 个（它们已在此处按值排序），然后将其他字典加到 Other 类别中，即替换Publishing、Fashion、Food 等等，只有一个 Other 将它们加在一起。坚持如何做到这一点，所以将不胜感激任何帮助！

{'Games': 715067930.8599964,
 'Design': 705237125.089998,
 'Technology': 648570433.7599969,
 'Film & Video': 379559714.56000066,
 'Music': 191227757.8699999,
 'Publishing': 130763828.65999977,
 'Fashion': 125678824.47999984,
 'Food': 122781563.58000016,
 'Art': 89078801.8599998,
 'Comics': 70600202.99999984,
 'Theater': 42662109.69999992,
 'Photography': 37709926.38000007,
 'Crafts': 13953818.35000002,
 'Dance': 12908120.519999994,
 'Journalism': 12197353.370000007}

目前我的饼图使用这段代码真的人满为患

groupbycategorypledge = df.groupby('main_category')['usd_pledged_real'].sum().sort_values(ascending=False)
plt.figure(figsize=(20, 10))
pie = groupbycategorypledge.plot(kind='pie', startangle=90, radius=0.7, title='Amount Pledged by Main category',autopct='%1.1f%%',labeldistance=1.2)
plt.legend(loc=(1.05,0.75))
plt.ylabel('')

所以我有

dict = groupbycategorypledge.sort_values(ascending=False).to_dict()

【问题讨论】：

标签： python sorting dictionary aggregate

【解决方案1】：

您可以在之前使用 Pandas 操作您的字典：

from operator import itemgetter

# sort by value descending
items_sorted = sorted(d.items(), key=itemgetter(1), reverse=True)

# calculate sum of others
others = ('Other', sum(map(itemgetter(1), items_sorted[5:])))

# construct dictionary
d = dict([*items_sorted[:5], others])

print(d)

{'Games': 715067930.8599964,
 'Design': 705237125.089998,
 'Technology': 648570433.7599969,
 'Film & Video': 379559714.56000066,
 'Music': 191227757.8699999,
 'Other': 658334549.8999995}

【讨论】：

【解决方案2】：

基于@jpp 的想法，但使用heap：

import heapq

d = {'Games': 715067930.8599964,
     'Design': 705237125.089998,
     'Technology': 648570433.7599969,
     'Film & Video': 379559714.56000066,
     'Music': 191227757.8699999,
     'Publishing': 130763828.65999977,
     'Fashion': 125678824.47999984,
     'Food': 122781563.58000016,
     'Art': 89078801.8599998,
     'Comics': 70600202.99999984,
     'Theater': 42662109.69999992,
     'Photography': 37709926.38000007,
     'Crafts': 13953818.35000002,
     'Dance': 12908120.519999994,
     'Journalism': 12197353.370000007}

top_5 = set(heapq.nlargest(5, d, key=d.get))

groups = {}
for category, pledge in d.items():
    new_category = category if category in top_5 else 'Other'
    groups.setdefault(new_category, []).append(pledge)

result = {k: sum(v) for k, v in groups.items()}
print(result)

输出

{'Technology': 648570433.7599969, 'Design': 705237125.089998, 'Other': 658334549.8999994, 'Games': 715067930.8599964, 'Film & Video': 379559714.56000066, 'Music': 191227757.8699999}

或者如果你喜欢 numpy：

import numpy as np

d = {'Games': 715067930.8599964,
     'Design': 705237125.089998,
     'Technology': 648570433.7599969,
     'Film & Video': 379559714.56000066,
     'Music': 191227757.8699999,
     'Publishing': 130763828.65999977,
     'Fashion': 125678824.47999984,
     'Food': 122781563.58000016,
     'Art': 89078801.8599998,
     'Comics': 70600202.99999984,
     'Theater': 42662109.69999992,
     'Photography': 37709926.38000007,
     'Crafts': 13953818.35000002,
     'Dance': 12908120.519999994,
     'Journalism': 12197353.370000007}

categories, pledge_values = map(np.array, zip(*d.items()))
partition = np.argpartition(pledge_values, -5)
top_5 = set(categories[partition][-5:])

groups = {}
for category, pledge in d.items():
    new_category = category if category in top_5 else 'Other'
    groups.setdefault(new_category, []).append(pledge)

result = {k: sum(v) for k, v in groups.items()}
print(result)

输出

{'Technology': 648570433.7599969, 'Design': 705237125.089998, 'Other': 658334549.8999995, 'Music': 191227757.8699999, 'Games': 715067930.8599964, 'Film & Video': 379559714.56000066}

第二个提案（使用 numpy）的复杂度是 O(n)，其中 n 是 d 的键值对的数量。

【讨论】：