[英]Python - getting the average of two dict lists?
在这里,我有两个列表:“ donlist”是一个介于$ 1和$ 100之间的随机捐赠金额列表,“ charlist”是一个介于1和15之间的随机慈善编号列表。我使用了两个“ dict”:“ totals”计算每个慈善机构的捐款总额,“ numdon”计算每个慈善机构的捐款数量。 我现在必须找到每个慈善机构的平均捐款额。 我尝试将“总计”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编号以及捐款的总数/数量。 请帮助我计算每个慈善机构的平均捐款额。 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)
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。 尝试这个:
l = {1: 'a', 2: 'b'}
for i in l:
print(i)
# output:
# 1
# 2
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