从2个列表中产生的字典不做我想要的

Question

我的问题是，当给出任何输入（正确格式化）时，字典dancer_placings无法根据我的代码正确生成。1个舞者的输入，有1，2舞，名为foo and bar，字典dancer_placings是 {'1': {'bar': 0}}，当我希望它为 {'1': {'foo': 0},{'bar': 0}}

时

我显然犯了一个错误，所以我该如何修复我的代码，以便它可以做我打算做的事情？

这是我的代码：

print("Welcome to Highland Scrutineer")
dancers = []
dances = []
dancer_placings = {}
dancers = []
dancer_num = int(input("How many dancers were there?: "))
while len(dancers) + 1 <= dancer_num:
    dancers.append(input("What was the number of the dancer? "))
print("The list of dancers is:")
for dancer in dancers:
    print(dancer)
dances = []
dance_num = int(input("How many dances did these dancers compete in? "))
while len(dances) + 1 <= dance_num:
    dances.append(input("What was the name of the dance? "))
print("The list of dances is:")
for dance in dances:
    print(dance)
for dancer in dancers:
    for dance in dances:
        dancer_placings.update({dancer:{}})
        dancer_placings[dancer].update({dance:0})
print(dancer_placings)

Answer 1

您的代码有一些问题，但我只是解决使您的用例工作的问题：

1. 您每次都在覆盖舞蹈置：
   dancer_placings.update（{{舞者：{}}）

这不是必需的。那应该是缩进水平。

2. 您目前正在创建一个不变的元组。您需要使用列表。

尝试以下操作：

print("Welcome to Highland Scrutineer")
dancers = []
dances = []
dancer_placings = {}
dancers = []
dancer_num = int(input("How many dancers were there?: "))
while len(dancers) + 1 <= dancer_num:
    dancers.append(input("What was the number of the dancer? "))
print("The list of dancers is:")
for dancer in dancers:
    print(dancer)
dances = []
dance_num = int(input("How many dances did these dancers compete in? "))
while len(dances) + 1 <= dance_num:
    dances.append(input("What was the name of the dance? "))
print("The list of dances is:")
for dance in dances:
    print(dance)
for dancer in dancers:
    dancer_placings[dancer] = []
    for dance in dances:
        dancer_placings[dancer].append({dance:0})
print(dancer_placings)

---

因此，这将导致以下输出：

user-macbookpro2:~ user$ python test.py
Welcome to Highland Scrutineer
How many dancers were there?: 1
What was the number of the dancer? 1
The list of dancers is:
1
How many dances did these dancers compete in? 2
What was the name of the dance? 'foo'
What was the name of the dance? 'bar'
The list of dances is:
foo
bar
{1: [{'foo': 0}, {'bar': 0}]}

---

这似乎是您所需的答案！

Answer 2

我认为您的问题是您正在为每种舞蹈的舞者_placings覆盖值。次要调整应返回您要寻找的结果：

for dancer in dancers:
    dancer_placings.update({dancer: {}})
for dance in dances:
    dancer_placings[dancer].update({dance: 0})

我通过快速的刮擦测试进行了运行，这就是结果：

{'1': {'bar': 0, 'foo': 0}}

Answer 3

所讨论的代码位是：

for dancer in dancers:
    for dance in dances:
        dancer_placings.update({dancer:{}})
        dancer_placings[dancer].update({dance:0})

正如现在所写的那样，每次舞者的每次迭代都会在每个舞蹈中循环。但是，语句dancer_placings.update({dancer:{}})将"清除"当前舞者在内部循环的每个迭代中的值。因此，只有最后一个内部迭代"粘贴"。

您想要的就是这样：

for dancer in dancers:
    dancer_placings.update({dancer:{}})
    for dance in dances:    
        dancer_placings[dancer].update({dance:0})

将为外圈中的每个舞者创建一个空词典，并更新内部循环中的舞蹈。

这样会产生（请注意，字典键没有明显的顺序）：

Welcome to Highland Scrutineer
How many dancers were there?: 3
What was the number of the dancer? 1
What was the number of the dancer? 2
What was the number of the dancer? 3
The list of dancers is:
1
2
3
How many dances did these dancers compete in? 4
What was the name of the dance? a
What was the name of the dance? b
What was the name of the dance? c
What was the name of the dance? d
The list of dances is:
a
b
c
d
{'1': {'a': 0, 'b': 0, 'd': 0, 'c': 0}, '3': {'a': 0, 'b': 0, 'd': 0, 'c': 0}, '2': {'a': 0, 'b': 0, 'd': 0, 'c': 0}}