Python - 获得两个字典列表的平均值？

Question

这里我有两个列表：“donlist”是 1 美元到 100 美元之间的随机捐赠金额列表，“charlist”是 1 到 15 之间的随机慈善号码列表。我使用了两个“dict”：“ totals”计算每个慈善机构的捐赠总额，“numdon”计算每个慈善机构的捐赠数量。我现在必须找到每个慈善机构的平均捐款。我尝试将“总数”除以“numdon”，但输出只是“1.0”的列表。我认为这是因为字典中有慈善号码以及其中的捐款总数/数量。请帮我计算每个慈善机构的平均捐款。谢谢！

from __future__ import division
import random
from collections import defaultdict
from pprint import pprint

counter = 0
donlist = []
charlist = []
totals = defaultdict(lambda:0)
numdon = defaultdict(lambda:0)

while counter != 100:
    d = round(random.uniform(1.00,100.00),2)
    c = random.randint(1,15)
    counter +=1
    donlist.append(d)
    donlist = [round(elem,2) for elem in donlist]
    charlist.append(c)
    totals[c] += d
    numdon[c] += 1

    if counter == 100:
        break

print('Charity \t\tDonations')
for (c,d) in zip(charlist,donlist):
    print(c,d,sep='\t\t\t')
print("\nTotal Donations per Charity:") 
pprint(totals)
print("\nNumber of Donations per Charity:")
pprint(numdon)

# The average array doesn't work; I think it's because the "totals" and "numdon" have the charity numbers in them, so it's not just two lists of floats to divide.
avg = [x/y for x,y in zip(totals,numdon)]
pprint(avg)

Answer 1

解决您的问题：

avg = [totals[i] / numdon[i] for i in numdon]

原因：

在字典的python列表理解中，默认迭代将在字典的键上。试试这个：

l = {1: 'a', 2: 'b'}
for i in l:
    print(i) 
# output:
# 1
# 2