Question

amount_of_names = int(input("Input amount of players: "))
names = []

for counter in range(amount_of_names):
  name = str(input("Input players first name: "))
  names.append(name)
#print (names)
selected_name = str(input("Input your name: "))

while name in names:
  import random 
  names = random.choice(names)
  break
print (names)

我正在尝试将列表中的名称与＆＃34;您的姓名＆＃34;匹配。并循环直到名称不匹配，例如Bob：Sam（是的）Sam：Sam（不） - 提前致谢

Answer 1

制作list，然后使用append为其添加输入。

AmountOfNames = int(input("Input amount of players: "))
Names = []

for counter in range (0, AmountOfNames):
    RegisteredName = input("Input player {}'s first name: ".format(counter+1))
    Names.append(RegisteredName)

print(Names)

示例：

Input amount of players: 3
Input player 1's first name: Simon
Input player 2's first name: James
Input player 3's first name: Jacob
['Simon', 'James', 'Jacob']

或者你可以这样做：

>>> names = input("Enter the first names of the players: ").split(", ")
Enter the first names of the players: Simon, James, Jacob
>>> names
['Simon', 'James', 'Jacob']

使用上述方法，如果他们每次都不用逗号和空格分隔它们，它会出现故障，所以要非常小心。

Answer 2

试试这个：

double

现在，amount_of_names = int(input("Input amount of players: ")) names = [] for counter in range(amount_of_names): name = str(input("Input players first names: ")) names.append(name) print names列表包含names名称

Answer 3

在循环外面创建一个空数组。例如。 NameList = []然后，在循环内，对于每次迭代，将名称添加到使用NameList.append（// name）的列表中

循环直到（py）

3 个答案: