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

In the code above, I have a string s = "abcaccba" (say) and I have an array of indices, queries = [1, 3, 2] . I am trying to find the closest occurrence of the character at these indices.

For example: s[1] = b and the closest index at which the second occurrence of b is, is at 6. When I try to run the code above, I get this error:

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

What am I doing wrong here??

Answer 1

i already a character in string. You just need to change s[i] to i and calculate the position of character.

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

The error you are getting is because i is a string but indexes must be integers . (This should be clear from the error message). Changing the piece if s[i] ==c: to if i == c: would make your code run but I do not think it will give you the answer you are really looking for.

Since you want the closest index in which the character is the same as that of the index of the query, I think you should get as a list of results which should be as long as the len gth of your queries . Below is some code that can help you achieve that goal. I have also extended your example to enhance understanding.

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) 

Here is the output

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

You can use the find() string method to look up character positions in a string.

Look at the string segments ahead and behind of each query -indexed character. Then use their relative position to determine their true index in s .

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]

I added an additional edge case, z , to cover the case where there are no additional instances of that character. In that case, the procedure returns None .