字典中小于某个值的总和值

Question

我有以下的解释和我试图让他们从一个饼图，但我想只包括前5名（他们被这里的值已经排序），然后在总结他人一起Other类别即更换Publishing，Fashion，Food等只一个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}

目前我的饼图使用此代码真的人满为患

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()

Answer 1

你可以在使用 Pandas之前操作你的字典：from operator import itemgetter# sort by value descendingitems_sorted = sorted(d.items(), key=itemgetter(1), reverse=True)# calculate sum of othersothers = ('Other', sum(map(itemgetter(1), items_sorted[5:])))# construct dictionaryd = 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}

Answer 2

基于@jpp 的想法，但使用堆：import heapqd = {'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，如果您喜欢 numpy：import numpy as npd = {'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 的键值对的数量d。