如何在不使用起始索引或范围作为参数的情况下，找到句子中第二次或第三次出现的索引数？(python)

Question

我有一句话-

Sammy likes to swim in the ocean, likes to spin servers, and likes to smile.

我想找到单词likes第二次出现的起始索引号。

但我不想使用起始索引号或范围作为str.find()函数的参数。

如果可能的话，我希望避免正则表达式，因为对于初学者来说很难理解。

注意:使用str.find()函数并不是强制性的。我只想知道，如果不给开始索引或范围作为参数，这是否可能。

Answer 1

法向循环法：

string = "Sammy likes to swim in the ocean, likes to spin servers, and likes to smile."
word = "likes"
indexes = []
for i in range(len(string)):
    if string[i:i+len(word)] == word:
        indexes.append(i)
print(indexes[1]) # Second location, outputs: 34

列表理解：

[i for i in range(len(string)) if string[i:i+len(word)] == word][1] # Outputs: 34

Answer 2

使用第二个捕获组的正则表达式、start和end：

import re
s = 'Sammy likes to swim in the ocean, likes to spin servers, and likes to smile.'
m = re.match(r'.*(likes).*(likes).*', s)
m.start(2)  # start of second capturing group
# 61
m.end(2)  # end of second capturing group
# 66
s[m.start(2), m.end(2)]
# 'likes'

Answer 3

使用regex怎么样？

import re

string = 'Sammy likes to swim in the ocean, likes to spin servers, and likes to smile.'
pattern = 'likes'
likes = [(m.start(0), m.end(0)) for m in re.finditer(pattern, string)]
# [(6, 11), (34, 39), (61, 66)]

likes1有你想要的