Question

为什么这会说列表索引超出范围？这可能不是最好的方法，但只是想知道为什么会这样！感谢

key = 13    

alphabet = ["A", "B", "C", "D", "E", "F", "G", "H", "I", "J", "K", "L", "M", "N", "O", "P", "Q", "R", "S", "T", "U", "V", "W", "X", "Y", "Z"]

replacement_set = ["A", "B", "C", "D", "E", "F", "G", "H", "I", "J", "K", "L", "M", "N", "O", "P", "Q", "R", "S", "T", "U", "V", "W", "X", "Y", "Z"]

count = -1

while count <= 25:
    count += 1
    if count < int(key):
        index_value = int(count + int(key))
        replacement_set[index_value] = alphabet[count]
        print replacement_set
    elif count >= int(key):     
        index_value = int(count - int(key))
        replacement_set[index_value] = alphabet[count]
        print replacement_set
    else:
            break

print replacement_set

Answer 1

Simeon Visser的回答回答了你的问题，但我认为值得指出Pythonic解决你的问题：

from string import ascii_uppercase as uppercase

replacement_set = uppercase[13:] + uppercase[:13]

uppercase[:13]是切片表示法的一个示例。您可以从the Python documentation了解更多相关信息。

Answer 2

IndexError发生在这里：

elif count >= int(key):     
    index_value = int(count - int(key))
    replacement_set[index_value] = alphabet[count] # <<<<<

因为count的值为26而发生这种情况。换句话说，您正在尝试访问超出alphabet范围的元素。请注意，您的循环不会阻止此操作，因为您在此处添加了1：

while count <= 25:
    count += 1

换句话说：当count为25时，您仍然会通过添加一个来26。您可以将while行转换为：

来修复代码

while count < 25:

Answer 3

发生错误的原因：

while count <= 25:
    count += 1
    print count

结束输出：

24
25
26

...而字母[26]超出范围，因为密钥必须低于len（字母）。

使用应该使用类似的东西：

key = 13
for count, letter in enumerate(alphabet):
    # calculate index_value, so it has to be in correct range
    index_value = (count - key) % len(alphabet)
    replacement_set[index_value] = letter
    print replacement_set

（更多pythonic使用for - 循环和enumerate）

为什么这个迷你程序说列表索引超出范围？

3 个答案: