如何解决“TypeError：字符串索引必须是整数”错误？

Question

def closest(s, queries):
    for query in queries:
        c = s[query]
        for i in s[query:]:
            if s[i] == c:
                return i
            else:
                return -1

在上面的代码中，我有一个字符串 s = "abcaccba"（比如说），我有一个索引数组，查询 = [1, 3, 2]。我试图在这些索引处找到最接近的字符。

例如：s[1] = b 和 b 第二次出现的最接近的索引是 6。当我尝试运行上面的代码时，我得到这个错误：

File "solution.py", line 23, in closest
    if s[i] == c:
TypeError: string indices must be integers

我在这里做错了什么？？

Answer 1

您可以使用find() 字符串方法来查找字符串中的字符位置。

查看每个query-indexed 字符前后的字符串段。然后使用它们的相对位置来确定它们在s中的真实索引。

s = "abcaccbaz"
queries = [1, 3, 2, 0, 5, 6, 8]
closest = []

for q in queries:
    char = s[q]
    pre = s[:q][::-1] # reversed, so nearest character to char is first
    post = s[q+1:]

    pre_dist = pre.find(char) 
    pre_found = pre_dist >= 0

    post_dist = post.find(char) 
    post_found = post_dist >= 0

    if post_found and (post_dist <= pre_dist or not pre_found):
        closest.append(post_dist + len(pre) + 1)
    elif pre_found and (pre_dist <= post_dist or not post_found):
        closest.append(q - pre_dist - 1)
    else:
        closest.append(None)

closest
# [6, 0, 4, 3, 4, 1, None]

我添加了一个额外的边缘案例z，以涵盖该角色没有其他实例的情况。在这种情况下，该过程返回None。

Answer 2

i 已经是字符串中的一个字符。只需将s[i] 更改为i 并计算字符的位置即可。

def closest(s, queries):
    for query in queries:
        c = s[query]
        for i in s[query:]:
            if i == c: # i is a char , i = s[position]
                print( i,(s[query+1:].index(i)+query+1) )
                break
            else:
                pass

s = "abcaccba"
queries = [1, 3, 2]
closest(s, queries)

Answer 3

您得到的错误是因为i 是string，但索引必须是integers。 （这应该从错误消息中清楚地看出）。将 if s[i] ==c: 更改为 if i == c: 将使您的代码运行，但我认为它不会为您提供您真正想要的答案。

由于您想要字符与查询索引相同的最接近的索引，我认为您应该得到一个结果列表，该列表应该与您的 @987654327 的 length 一样长@。下面是一些可以帮助您实现该目标的代码。我还扩展了您的示例以增强理解。

s = "abcaccba" 
queries = [1,3,2,0, 5]
def closest(s, queries):
    print('String: ', s)
    print('Query: ', queries)
    results = []
    for query in queries:
        c = s[query]
        if s[query] in s[query+1 :]:
            results.append(('query = {}'.format(query), 'character = {}'.format(c), 'index of next occurrence = {}'.format(1+query+s[query+1 :].index(c))))
            #results.append((query, c, 1+query+s[query+1 :].index(c)))
        else:
            results.append(('query = {}'.format(query), 'character = {}'.format(c), 'index of next occurrence = {}'.format(-1)))
            #results.append((query, c, -1))
    return results
closest(s  = s, queries =  queries) 

这是输出

String:  'abcaccba'
Query:  [1, 3, 2, 0, 5]

[('query = 1', 'character = b', 'index of next occurrence = 6'),
 ('query = 3', 'character = a', 'index of next occurrence = 7'),
 ('query = 2', 'character = c', 'index of next occurrence = 4'),
 ('query = 0', 'character = a', 'index of next occurrence = 3'),
 ('query = 5', 'character = c', 'index of next occurrence = -1')]