如何用Python查找字符串中相邻重叠子串'bob'的出现次数

Hey there! Let's break down why your code is returning 2 instead of the correct 3 when counting the substring 'bob' in s = 'xasdkbobobasdvobob'.

The Root Problem #

Chances are your original code was using a method that skips over overlapping occurrences of 'bob'. For example, if you used str.find() and jumped 3 characters ahead after each match, you'd miss the overlapping 'bob' in the bobob segment of your string. Let's look at that segment:

bobob contains two instances of 'bob' — starting at index 5 and index 7 (0-based). If you skip 3 characters after finding the first one, you'll jump past the second overlapping match entirely.

Solution 1: Iterate Through Possible Start Indices #

This straightforward approach checks every possible starting position for the 3-character substring 'bob', including overlapping ones:

s = 'xasdkbobobasdvobob'
bob_count = 0

# We only need to check up to len(s)-3 (since we need 3 characters for 'bob')
for i in range(len(s) - 2):
    if s[i:i+3] == 'bob':
        bob_count += 1

print(bob_count)  # Output: 3

Solution 2: Use Regular Expressions for Overlapping Matches #

If you prefer a more concise method, you can use Python's re module with a positive lookahead, which doesn't consume characters and allows overlapping matches:

import re

s = 'xasdkbobobasdvobob'
# (?=bob) matches positions where 'bob' follows, without consuming the string
bob_count = len(re.findall(r'(?=bob)', s))

print(bob_count)  # Output: 3

Both methods will correctly count all 3 occurrences of 'bob' in your string, including the overlapping one in the middle.

内容的提问来源于stack exchange，提问作者Mohamed AlNuaimi