为什么我在 Python 中遇到字符串索引问题

Question

I'm trying to understand why I'm having the same index again when I apply .index or .find why I'm getting the same index '2' again why not '3'? when a letter is repeated, and what is the alternative way to get an index 3 for the second 'l'

text = 'Hello'
for i in text:
    print(text.index(i))

the output is:

0
1
2
2
4

Answer 1

It's because .index() returns the lowest or first index of the substring within the string. Since the first occurrence of l in hello is at index 2, you'll always get 2 for "hello".index("l") .

So when you're iterating through the characters of hello , you get 2 twice and never 3 (for the second l ). Expanded into separate lines, it looks like this:

"hello".index("h")   # = 0
"hello".index("e")   # = 1
"hello".index("l")   # = 2
"hello".index("l")   # = 2
"hello".index("o")   # = 4

Edit: Alternative way to get all indices :

One way to print all the indices (although not sure how useful this is since it just prints consecutive numbers) is to remove the character you just read from the string:

removed = 0
string = "hello world"    # original string
for char in string:
  print("{} at index {}".format(char, string.index(char) + removed))  # index is index() + how many chars we've removed
  string = string[1:]    # remove the char we just read
  removed +=1            # increment removed count

Answer 2

text = 'Hello'
for idx, ch in enumerate(text):
    print(f'char {ch} at index {idx}')  

output

char H at index 0
char e at index 1
char l at index 2
char l at index 3
char o at index 4

Answer 3

If you want to find the second occurance, you should search in the substring after the first occurance

text = 'Hello'
first_index = text.index('l')
print('First index:', first_index)

second_index = text.index('l', first_index+1) # Search in the substring after the first occurance
print('Second index:', second_index)

The output is:

First index: 2
Second index: 3