Python: Find a substring in a string and returning

问题:

I have:

• a function: def find_str(s, char)

• and a string: "Happy Birthday",

I essentially want to input "py" and return 3 but I keep getting 2 to return instead.

Code:

def find_str(s, char):
    index = 0           
    if char in s:
        char = char[0]
        for ch in s:
            if ch in s:
                index += 1
            if ch == char:
                return index

    else:
        return -1

print(find_str("Happy birthday", "py"))

Not sure what's wrong!

回答1:

Ideally you would use str.find or str.index like demented hedgehog said. But you said you can't ...

Your problem is your code searches only for the first character of your search string which(the first one) is at index 2.

You are basically saying if char[0] is in s, increment index until ch == char[0] which returned 3 when I tested it but it was still wrong. Here's a way to do it.

def find_str(s, char):
    index = 0

    if char in s:
        c = char[0]
        for ch in s:
            if ch == c:
                if s[index:index+len(char)] == char:
                    return index

            index += 1

    return -1

print(find_str("Happy birthday", "py"))
print(find_str("Happy birthday", "rth"))
print(find_str("Happy birthday", "rh"))

It produced the following output:

3
8
-1

回答2:

There's a builtin method on string objects to do this in python you know?

s = "Happy Birthday"
s2 = "py"

print s.find(s2)

Python is a "batteries included language" there's code written to do most of what you want already (whatever you want).. unless this is homework :)

Edit: find returns -1 if the string cannot be found.

回答3:

late to the party, was searching for same, as "in" is not valid, I had just created following.

def find_str(full, sub):
    index = 0
    sub_index = 0
    position = -1
    for ch_i,ch_f in enumerate(full) :
        if ch_f.lower() != sub[sub_index].lower():
            position = -1
            sub_index = 0
        if ch_f.lower() == sub[sub_index].lower():
            if sub_index == 0 :
                position = ch_i

            if (len(sub) - 1) <= sub_index :
                break
            else:
                sub_index += 1

    return position

print(find_str("Happy birthday", "py"))
print(find_str("Happy birthday", "rth"))
print(find_str("Happy birthday", "rh"))

which produces

3
8
-1

remove lower() in case case insensitive find not needed.

回答4:

Not directly answering the question but I got a similar question recently where I was asked to count the number of times a sub-string is repeated in a given string. Here is the function I wrote:

def count_substring(string, sub_string):
    cnt = 0
    len_ss = len(sub_string)
    for i in range(len(string) - len_ss + 1):
        if string[i:i+len_ss] == sub_string:
            cnt += 1
    return cnt

The find() function probably returns the index of the fist occurrence only. Storing the index in place of just counting, can give us the distinct set of indices the sub-string gets repeated within the string.

Disclaimer: I am 'extremly' new to Python programming.