Question

我正在快速制作游戏以获得乐趣，只是学习编码，而我似乎无法解决这个问题。

set1 = ["Horse", "Cow", "Pig", "Goat", "Chicken", "Sheep", "Donkey", "Duck", "Rabbit"]
set2 = ["Horse", "Cow", "Pig", "Goat", "Chicken", "Sheep", "Donkey", "Duck", "Rabbit"]
random.shuffle(set1,random.random)
random.shuffle(set2,random.random)
side1 = int(input("What number would you like to pick from side one?"))
side2 = int(input("What number would you like to pick from side two?"))
s1pick = set1.index(side1)
s2pick = set2.index(side2)
picks = []

我希望他们为side1和side2输入2个数字，然后存储来自set1和set2的相应值在s2pick和s1pick中{1}}。但我不断收到此错误消息：

Traceback (most recent call last):
    File "C:/Users/Me/PycharmProjects/nice memes/src/snap.py", line 26, in <module>
        s1pick = set1.index(side1)
ValueError: 4 is not in list

在pycharm社区版上使用python 3。谢谢！

Answer 1

list.index()返回您传递给它的元素值的索引。因此，对于列表l = ['a', 'b', 'c']，l.index('b')将返回1。

如果你想转向另一个方向，获取索引的值，那么你应该使用项目访问语法：

>>> l = ['a', 'b', 'c']
>>> l[1]
'b'

因此，在您的情况下，您希望使用set1[side1]来获取用户从set1中选择的值。

Answer 2

index() actually returns the index of the value that you pass to the method。您收到错误，因为值4不在您调用index()方法的列表中。如果您在从用户那里获得提示后已经将提示转换为int()，那么您可以指定您希望列表中的该索引处的元素如下：

s1pick = set1[side1]
s2pick = set2[side2]

但是，要小心。您获得的input()可能超出了您的列表范围。所以，我会把你的输入放到while循环中，以确保你得到一个有效的输入：

side1 = # input code
side2 = # input code

while side1 not in range(len(set1)) or side2 not in range(len(set2)):
    side1 = # input code
    side2 = # input code

Answer 3

比较两段代码：

list1 = ["Horse", "Cow", "Pig", "Goat", "Chicken", "Sheep", "Donkey", "Duck", "Rabbit"]
list2 = ["Horse", "Cow", "Pig", "Goat", "Chicken", "Sheep", "Donkey", "Duck", "Sheep"]

######################
index1 = 2
index2 = 3
element1 = list1[index1]
element2 = list2[index2]
print (element1) # output: Pig
print (element2) # output: Goat


######################
index1 = list1.index("Sheep")
index2 = list2.index("Donkey")
print (index1)   # output: 5
print (index2)   # output: 6

如何让用户输入一个数字，从当前列表中选择该项目并将其添加到新列表中？

3 个答案: