我有这样的字典
d = {1:'Bob', 2:'Joe', 3:'Bob', 4:'Bill', 5:'Bill'}
我想保留每个名称作为字典值出现的次数。所以,输出应该是这样的:
d = {1:['Bob', 1], 2:['Joe',1], 3:['Bob', 2], 4:['Bill',1] , 5:['Bill',2]}
答案 0 :(得分:6)
要强制执行订单(dict
本身没有订单),假设您按键上的已排序顺序。然后你可以做 - 假设值是可以清除的,如你的例子......:
import collections
def enriched_by_count(somedict):
countsofar = collections.defaultdict(int)
result = {}
for k in sorted(somedict):
v = somedict[k]
countsofar[v] += 1
result[k] = [v, countsofar[v]]
return result
答案 1 :(得分:6)
计算所需值的一种方法如下所示:
from collections import Counter
d = {1:'Bob',2:'Joe',3:'Bob', 4:'Bill', 5:'Bill'}
c = Counter()
new_d = {}
for k in sorted(d.keys()):
name = d[k]
c[name] += 1;
new_d[k] = [name, c[name]]
print(new_d)
# {1: ['Bob', 1], 2: ['Joe', 1], 3: ['Bob', 2], 4: ['Bill', 1], 5: ['Bill', 2]}
这里我使用Counter来跟踪输入字典中名称的出现。希望这可以帮助。也许不是最优雅的代码,但它的工作原理。
答案 2 :(得分:4)
不使用任何模块,这是我提出的代码。也许不是那么短暂,但我害怕模块。
def new_dict(d):
check = [] #List for checking against
new_dict = {} #The new dictionary to be returned
for i in sorted(d.keys()): #Loop through all the dictionary items
val = d[i] #Store the dictionary item value in a variable just for clarity
check.append(val) #Add the current item to the array
new_dict[i] = [d[i], check.count(val)] #See how many of the items there are in the array
return new_dict
像这样使用:
d = {1:'Bob', 2:'Joe', 3:'Bob', 4:'Bill', 5:'Bill'}
d = new_dict(d)
print d
输出:
{1: ['Bob', 1], 2: ['Joe', 1], 3: ['Bob', 2], 4: ['Bill', 1], 5: ['Bill', 2]}