使用递归在字符串中显示多个子字符串索引

Question

def subStringMatchExact(target, key):

    if (target.find(key) == -1):
        return []
    else:
        foundStringAt = [target.find(key)]
        target = target[foundStringAt[0] + len(key):]
        return foundStringAt + subStringMatchExact(target, key)

string = subStringMatchExact("your code works with wrongly correlated coefficients which incorporates more costs", "co") 

print(string)

当前输出错误:

[5, 22, 9, 19, 14]

我在上一个递归步骤中总结子字符串的长度时遇到了麻烦.就像列表的第二个元素应该29而不是22像len(previousSubstring) + len(key) - 1 + len(currentSubstring).

有什么想法来改进我的代码和/或修复我的错误吗？

Answer 1

快捷的方式

您不必实施自己的解决方案,它已经完成了!使用模块中的finditer功能re:

>>> import re
>>> s = 'your code works with wrongly correlated coefficients which incorporates more costs'
>>> matches = re.finditer('co', s)
>>> positions = [ match.start() for match in matches ]
>>> positions
[5, 29, 40, 61, 77]

你自己的方式

如果你想自己实现(使用递归),你可以利用str.find函数的额外参数.让我们看看help(str.find)它的内容:

S.find(sub [,start [,end]]) -> int

    Return the lowest index in S where substring sub is found,
    such that sub is contained within s[start:end].  Optional
    arguments start and end are interpreted as in slice notation.

    Return -1 on failure.

有一个额外的参数叫做start告诉从str.find哪里开始搜索子字符串.这正是我们所需要的!

因此,修改您的实现,我们可以获得一个简单,快速和美观的解决方案:

def substring_match_exact(pattern, string, where_should_I_start=0):
    # Save the result in a variable to avoid doing the same thing twice
    pos = string.find(pattern, where_should_I_start)
    if pos == -1:
        # Not found!
        return []
    # No need for an else statement
    return [pos] + substring_match_exact(pattern, string, pos + len(key))

这里的递归是什么？

• 您首先在位置0开始搜索字符串中的子字符串.
• 如果未找到子字符串,则返回空列表[].
• 如果找到子字符串,它将返回[pos]加上子字符串将出现在从位置开始的字符串中的所有位置pos + len(key).

使用我们全新的功能

>>> s = 'your code works with wrongly correlated coefficients which incorporates more costs'
>>> substring_match_exact('co', s)
[5, 29, 40, 61, 77]