我有以下的解釋和我試圖讓他們從一個餅圖，但我想只包括前5名（他們被這里的值已經排序），然后在總結他人一起Other類別即更換Publishing，Fashion，Food等只一個Other將它們加在一起。堅持如何做到這一點，所以將不勝感激任何幫助！{'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}目前我的餅圖使用此代碼真的人滿為患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()
2 回答
慕桂英3389331
TA貢獻2036條經驗 獲得超8個贊
你可以在使用 Pandas之前操作你的字典：
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}
LEATH
TA貢獻1936條經驗 獲得超7個贊
基于@jpp 的想法，但使用堆：
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)
輸出
{'Technology': 648570433.7599969, 'Design': 705237125.089998, 'Other': 658334549.8999994, 'Games': 715067930.8599964, 'Film & Video': 379559714.56000066, 'Music': 191227757.8699999}
或者，如果您喜歡 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)
輸出
{'Technology': 648570433.7599969, 'Design': 705237125.089998, 'Other': 658334549.8999995, 'Music': 191227757.8699999, 'Games': 715067930.8599964, 'Film & Video': 379559714.56000066}
第二個提案（使用 numpy）的復雜度是O(n)，其中n是 的鍵值對的數量d。
添加回答
舉報
0/150
提交
取消