关于含range、len与切片的Python字符串统计代码运行原理的问询

First, let's look at the full code again for reference:

s = 'eljwboboblejr' # dont paste into grader
count = 0
for i in range (len(s)):
    if s[i:i+3]== 'bob':
        count+=1
print('Number of times bob occurs is: ' + str(count))

Let's break down the parts you're confused about one by one, using our example string to make it concrete.

What does len(s) do here? #

len(s) calculates the total number of characters in the string s. For our example, 'eljwboboblejr' has 12 characters, so len(s) returns 12.

The range(len(s)) part creates a sequence of numbers from 0 to 11 (since range() stops one number short of its input). We use this sequence to loop through every possible starting index in the string.

You might notice that when i is 10 or 11, we can't get a full 3-character chunk from the string — but that's okay! Python's slicing just returns whatever characters are left (1 or 2 characters in those cases), which can't match the 3-character 'bob', so those iterations don't affect our count. The code works perfectly fine this way.

How does s[i:i+3] == 'bob' work? #

This uses Python's string slicing to check if a 3-character chunk of s starting at index i exactly matches 'bob'.

Here's how slicing works: s[start:end] grabs all characters from the start index up to (but not including) the end index. So s[i:i+3] means:

• Start at position i
• Take the character at i, then i+1, then i+2 (since we stop at i+3)

Let's walk through our example:

• When i=4: s[4:7] pulls characters at positions 4, 5, 6 → 'b'+'o'+'b' → this matches 'bob', so count increases to 1.
• When i=6: s[6:9] pulls characters at positions 6,7,8 → 'b'+'o'+'b' → another match, so count jumps to 2.
• For every other i value, the 3-character slice won't match 'bob', so count stays the same.

By the end of the loop, count holds the total number of times 'bob' appears in the string — which is 2 for our example.

内容的提问来源于stack exchange，提问作者J. Zalewski