Python：在字符串中查找子串并返回子串的索引

Question

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!

Answer 1

There's a builtin method find on string objects.

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 :)

find returns -1 if the string cannot be found.

Answer 2

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

Answer 3

There is one other option in regular expression , the search method

import re

string = 'Happy Birthday'
pattern = 'py'
print(re.search(pattern, string).span()) ## this prints starting and end indices
print(re.search(pattern, string).span()[0]) ## this does what you wanted

By the way, if you would like to find all the occurrence of a pattern, instead of just the first one, you can use finditer method

import re

string = 'i think that that that that student wrote there is not that right'
pattern = 'that'

print([match.start() for match in re.finditer(pattern, string)])

which will print all the starting positions of the matches.

Answer 4

Adding onto @demented hedgehog answer on using find()

In terms of efficiency

It may be worth first checking to see if s1 is in s2 before calling find() .
This can be more efficient if you know that most of the times s1 won't be a substring of s2

Since the in operator is very efficient

 s1 in s2

It can be more efficient to convert:

index = s2.find(s1)

to

index = -1
if s1 in s2:
   index = s2.find(s1)

This is useful for when find() is going to be returning -1 a lot.

I found it substantially faster since find() was being called many times in my algorithm, so I thought it was worth mentioning

Answer 5

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.

Answer 6

Here is a simple approach:

my_string = 'abcdefg'
print(text.find('def'))

Output:

3

I the substring is not there, you will get -1 . For example:

my_string = 'abcdefg'
print(text.find('xyz'))

Output:

-1

Sometimes, you might want to throw exception if substring is not there:

my_string = 'abcdefg'
print(text.index('xyz')) # It returns an index only if it's present

Output:

Traceback (most recent call last):

File "test.py", line 6, in print(text.index('xyz'))

ValueError: substring not found

Answer 7

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.