如何解决“ TypeError：字符串索引必须为整数”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”（例如），并且有一个索引数组，querys = [1、3、2 [1, 3, 2] 。 我正在尝试在这些索引处找到最接近的字符。

例如： s[1] = b ，第二个b出现的最接近的索引是6。当我尝试运行上面的代码时，出现以下错误：

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

i已经是字符串中的字符了。 您只需要将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 2

您得到的错误是因为i是一个string但是索引必须是integers 。 （这应该从错误消息中清除）。 if s[i] ==c:更改为if i == c: ，将使您的代码运行，但是我认为这不会给您真正想要的答案。

既然你想最接近的索引中的字符是相同查询的指标，我认为你应该得到的结果应该是，只要名单len您的GTH queries 。 以下是一些可以帮助您实现该目标的代码。 我也扩大了您的榜样，以增进理解。

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')]

Answer 3

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索引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情况z ，以涵盖该字符没有其他实例的情况。 在这种情况下，过程将返回None 。