Question

我试图创建一个获取用户输入的python字典，将其拆分为两个索引的列表，将第二个索引作为字典键，将第一个索引作为字典值 - 但作为元组进行转换。希望有道理！我有它工作，但当我输入具有相同键的另一个输入时，我希望新值附加到字典中的元组。我知道元组是不可变的并且没有追加（或者我认为）所以我需要使用什么魔法才能将它添加到字典中的元组中？

到目前为止我的代码是：

desserts = {}

name_vote = input ('Name:vote ')

while name_vote != '':
  no_colon_name_vote = name_vote.replace(":", " ")
  name, vote = no_colon_name_vote.split()
  name = tuple([name])
  if vote not in desserts:
    desserts[vote] = name
  else:
    desserts[vote].append(name)   #this is where I'm hitting a brick wall
  name_vote = input ('Name:vote ')

print(desserts)

我希望两个输入的输出应该是

Name:vote Luke:icecream
Name:vote Bob:icecream
Name:vote 
{'icecream': ('Luke', 'Bob')}

Answer 1

我想我可能有它！

desserts = {}

name_vote = input ('Name:vote ')

while name_vote != '':
  no_colon_name_vote = name_vote.replace(":", " ")
  name, vote = no_colon_name_vote.split()
  name = tuple([name])
  if vote not in desserts:
    desserts[vote] = name
  else:
    original = desserts[vote]
    desserts[vote] = (original + name)
  name_vote = input ('Name:vote ')

print(desserts)

Answer 2

使用list存储值和defaultdict会更容易，如果您要使用可变容器一直添加名称会更有意义：

from collections import defaultdict
desserts = defaultdict(list)
name_vote = input ('Name:vote ')
while name_vote != '':
    no_colon_name_vote = name_vote.replace(":", " ")
    name, vote = no_colon_name_vote.split()
    desserts[vote].append(name)  
    name_vote = input ('Name:vote ')
print(desserts)

如果你想要元组，你可以将列表转换为元组：

for k,v in desserts.iteritems():
    desserts[k] = tuple(v)

在Python中附加作为元组的字典值

2 个答案: