Python 在字符串中反向查找

发布于 2024-09-15 00:52:06 字数 460 浏览 3 评论 0原文

我有一个字符串和该字符串的任意索引。我想找到索引之前第一次出现的子字符串。

一个例子：我想通过使用索引和 str.rfind() 来查找第二个 I 的索引

s = "Hello, I am 12! I like plankton but I don't like Baseball."
index = 34 #points to the 't' in 'but'
index_of_2nd_I = s.rfind('I', index)
#returns = 36 and not 16

现在我希望 rfind() 返回第二个 I 的索引 (16) 但它返回 36。在文档中查找后，我发现 rfind 并不代表反向查找。

我对 Python 完全陌生，所以有内置的解决方案来反向查找吗？就像用一些 python [::-1] 魔法反转字符串并使用 find 等？或者我是否必须通过字符串反向迭代一个字符？

原文

I have a string and an arbitrary index into the string. I want find the first occurrence of a substring before the index.

An example: I want to find the index of the 2nd I by using the index and str.rfind()

s = "Hello, I am 12! I like plankton but I don't like Baseball."
index = 34 #points to the 't' in 'but'
index_of_2nd_I = s.rfind('I', index)
#returns = 36 and not 16

Now I would expect rfind() to return the index of the 2nd I (16) but it returns 36. after looking it up in the docs I found out rfind does not stand for reverse find.

I'm totally new to Python so is there a built in solution to reverse find? Like reversing the string with some python [::-1] magic and using find, etc? Or will I have to reverse iterate char by char through the string?

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孤君无依 2024-09-22 00:52:06

您的调用告诉 rfind 开始查找索引 34。您想要使用 rfind 重载，它需要一个字符串、一个开始和一个结束。告诉它从字符串的开头 (0) 开始，并停止查看 index：

>>> s = "Hello, I am 12! I like plankton but I don't like Baseball."
>>> index = 34 #points to the 't' in 'but'
>>> index_of_2nd_I = s.rfind('I', 0, index)
>>>
>>> index_of_2nd_I
16

Your call tell rfind to start looking at index 34. You want to use the rfind overload that takes a string, a start and an end. Tell it to start at the beginning of the string (0) and stop looking at index:

>>> s = "Hello, I am 12! I like plankton but I don't like Baseball."
>>> index = 34 #points to the 't' in 'but'
>>> index_of_2nd_I = s.rfind('I', 0, index)
>>>
>>> index_of_2nd_I
16

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满天都是小星星 2024-09-22 00:52:06

我很好奇如何实现通过 rpartition 从末尾查找字符串 n 次，并执行了第 n 个 rpartition 循环：

orig = s = "Hello, I am 12! I like plankton but I don't like Baseball."
found = tail = ''
nthlast = 2
lookfor = 'I'
for i in range(nthlast):
    tail = found+tail
    s,found,end = s.rpartition(lookfor)
    if not found:
        print "Only %i (less than %i) %r in \n%r" % (i, nthlast, lookfor, orig)
        break
    tail = end + tail
else:
    print(s,found,tail)

I became curious how to implement looking n times for string from end by rpartition and did this nth rpartition loop:

orig = s = "Hello, I am 12! I like plankton but I don't like Baseball."
found = tail = ''
nthlast = 2
lookfor = 'I'
for i in range(nthlast):
    tail = found+tail
    s,found,end = s.rpartition(lookfor)
    if not found:
        print "Only %i (less than %i) %r in \n%r" % (i, nthlast, lookfor, orig)
        break
    tail = end + tail
else:
    print(s,found,tail)

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