Question

我正在开发一个涉及两个列表的程序：一个是随机顺序中的数字1-5，另一个是仅保存为不同元素的五个单词。基本上我希望能够以第一个列表的随机顺序打印出单词。我试过了：

order = ["2","5","4","1","3"]
words = ["Apple","Boat","Carrot","Dragonfly","Education"]
for i in range (0,5):
    print(words[order])

它只是说“TypeError: list indices must be integers, not list”。任何人都可以帮助我吗？

Answer 1

这里有两个问题：

您不会获取order的元素，只需将整个列表作为索引传递;和
如错误所示，您应该将字符串'2'转换为整数2。

我们可以使用 int(..) 。现在出现了一个新问题：列表具有从零开始的索引，列表中的索引是从一开始的。然而，我们可以从中减去一个。

这导致以下方法：

for i in order:
    print(words[int(i)-1])

Answer 2

更正您的代码将如下所示：

order = ["1","2","3","4","5"]
words = ["Apple","Boat","Carrot","Dragonfly","Education"]
for i in range(0,5):
    print(words[int(order[i])])

但这远不是一个干净的解决方案。你正在捣乱index es太多。

更好的方法是：

for x in order:
    print(words[int(x)-1])

所有这一切都说明，你所做的事情没有随机。考虑使用random.shuffle()。像这样：

from random import shuffle

order = ["1","2","3","4","5"]
words = ["Apple","Boat","Carrot","Dragonfly","Education"]
shuffle(order)  # the shuffling is done in-place

for i in [int(c)-1 for c in order]:
    print(words[i])

# prints
Carrot
Dragonfly
Education
Boat
Apple

Answer 3

words = ["Apple","Boat","Carrot","Dragonfly","Education"]
# 1 #
for _ in range(11):
    print(words[random.randrange(0, len(words))])
# 2 #
print("Before shuffle: ",words);random.shuffle(words);print("After shuffle: ",words)
# 3 #
print("Choose 1 random sample",random.sample(words, 1))

我不知道你的目标是什么，但这就是我想出的。此外，您不需要将它们存储为str in＆＃39; order＆＃39;名单。您可以拥有一个整数列表，您可以使用它而无需转换。如：order = [1,2,3,4,5]

我希望我的回答很有帮助。欢呼！

将列表的元素从str转换为int

3 个答案: