Python 字典中的和值小于某个值

我有下面的字典，正试图用它们做一个饼图，但我只想包括前5个（它们已经在这里按值排序），然后将其他的加起来，归入

其他

类别，即替换

出版

，

时尚

，

食物

等，只有一个

另一个

将它们加在一起。坚持如何做到这一点，所以将感谢任何帮助

{'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时确实过于拥挤

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()

---

您可以在使用Pandas之前操作词典：

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}

---

基于@jpp的思想，但使用：

输出

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

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

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': 658334549.8999994, 'Games': 715067930.8599964, 'Film & Video': 379559714.56000066, 'Music': 191227757.8699999}

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

第二个方案（使用numpy）的复杂性为O（n），其中

n

是

d

的键、值对的数量