Question

我的问题是，当给出任何输入（格式正确）时，根据我的代码没有正确生成字典dancer_placings。输入1个舞者，1号，2个舞蹈，名为foo和bar，字典dancer_placings为{'1': {'bar': 0}}，当我希望它为{'1': {'foo': 0},{'bar': 0}}时

我显然犯了一个错误，那么如何修复我的代码以便它能够完成我打算做的事情呢？

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

我认为问题是你要为每个舞蹈覆盖dancer_placings的值。小调整应该返回您正在寻找的结果：

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

我通过快速划痕测试运行它，这就是结果：

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

Answer 2

您的代码存在一些问题，但我只是解决这些问题以使您的用例有效：

你每次都会覆盖舞者的位置：
dancer_placings.update（{舞者：{}}）

没有必要。这应该是一个缩进级别。

您正在创建一个不可变的元组。您需要使用列表。

试试这个：

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}]}

这似乎是你想要的答案！

Answer 3

这里讨论的代码是：

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

正如现在所写，每次通过舞者的迭代都会在每个舞蹈中循环。但是，语句dancer_placings.update({dancer:{}})将在内循环的每次迭代中“清除”当前舞者的值。因此，只有最后一次内部迭代“坚持”。

你想要的是这样的：

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

这将为外循环中的每个舞者创建一个空字典，并更新内循环中的舞蹈。

所以这会产生（注意字典键没有明确的顺序）：

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}}

从2个列表生成的字典没有做我想要的

3 个答案: