我正在尝试通过删除索引来减少我的代码以使其更简单

Question

This is my recursive function named find(text,substring) that will return the substring if the substring is inside the text or return an empty string if substring is not inside the text. Another function named is_string_there(text, string) to test the find function. The function will return True if the string is inside the text or False if the string is not inside the text.

Am I able to cancel out text index and substring index to remove the two (0,0) at the print coding area to reduce my code to make it simpler and easier to understand so i can teach my students

def find(text, substring, text_index, substring_index): 
    if text_index == len(text) and substring_index != len(substring): 
        return False

    if substring_index == len(substring): 
        return True

    if text[text_index] == substring[substring_index]: 
        return find(text, substring, text_index+1, substring_index+1) 

    return False


def oo(text, substring, text_index, substring_index): 
    if text_index == len(text): 
        return False


    if text[text_index] == substring[substring_index]: 
        if find(text, substring, text_index, substring_index): 
            return True
        else: 
            return oo(text, substring, text_index+1, substring_index) 


    return oo(text , substring, text_index+1, substring_index)

print(oo("I love pie", "pie",0,0))
print(oo("I love pie1", "pie1",0,0))

Answer 1

In your code, you use oo() to check if a character matches, and then calls find() to check if the rest of the characters match. Instead of checking character-by-character, you could use the idea of checking string equality. In your oo() function, you can check if the substring is equal to a splice of the original text. This will reduce your code but will make your find() function obsolete, while still serving as a simple demonstration of recursion.

def oo(text, substring, index): 
    if index == len(text):
        return False  

    start = index
    end = start + len(substring)  

    if text[start:end] == substring: 
        return True
    else: 
        return oo(text, substring, index+1) 


print(oo("I love pie", "pie",0)) 
print(oo("I love pie1", "pie1",0))
print(oo("I love pie", "pie1",0))
print(oo("I love pie1", "pie",0))
print(oo("pie", "pie",0))
print(oo("pi", "pie1",0))
print(oo("pie1", "pie",1))

Output:

True
True
False
True
True
False
False

Answer 2

Well if you want to remove the indexes permanently from the print commands, you could just use a class. This way you can even create other functions to take user input etc and take advantage of class variables for readability. Not sure if if cases should in form if-elif-else though:

class test():    
    def __init__(self):
        self.reset()
        self.run()
    def reset(self):
        self.substring_index = 0 #instead of putting in function they are now class variables
        self.text_index = 0
    def run(self):#more stuff can be done here

        print(self.oo("I love pie","pie"))
        self.reset() #reset the class variables for each step
        print(self.oo("I love pie1","pie1"))

    def find(self,text, substring): #pretty much the same code just using class variables and class methods
        if self.text_index == len(text) and self.substring_index != len(substring): 
            return False

        elif self.substring_index == len(substring): 
            return True

        elif text[self.text_index] == substring[self.substring_index]: 
            self.text_index +=1
            self.substring_index +=1
            return self.find(text, substring) 

        else:
            return False


    def oo(self,text, substring): 
        if self.text_index == len(text): 
            return False


        elif text[self.text_index] == substring[self.substring_index]: 
            if self.find(text, substring): 
                return True
            else: 
                self.text_index +=1
                return oo(text, substring) 
        else:
            self.text_index +=1
        return self.oo(text , substring)

if __name__=='__main__':
    test()

Although just for reference, a super simple way for the problem is actually:

def oo(text,substring):
    if substring in text:
        return True
    else:
        return False

AS mentioned in comments:

def oo(text,substring):
    if text.find(substring) == 0:
        return True
    else:
        return False

Returns 0 if exists in text, -1 if not.

Interesting expansion possible when considering uppercase or lowercase diffrences. 'A' != 'a'