[python] Find index of last occurrence of a substring in a string

I want to find the position (or index) of the last occurrence of a certain substring in given input string str.

For example, suppose the input string is str = 'hello' and the substring is target = 'l', then it should output 3.

How can I do this?

If you don't wanna use rfind then this will do the trick/

def find_last(s, t):
    last_pos = -1
    while True:
        pos = s.find(t, last_pos + 1)
        if pos == -1:
            return last_pos
        else:
            last_pos = pos

Use the str.rindex method.

>>> 'hello'.rindex('l')
3
>>> 'hello'.index('l')
2

you can use rindex() function to get the last occurrence of a character in string

s="hellloooloo"
b='l'
print(s.rindex(b))

Python String rindex() Method

Description
Python string method rindex() returns the last index where the substring str is found, or raises an exception if no such index exists, optionally restricting the search to string[beg:end].

Syntax
Following is the syntax for rindex() method -

str.rindex(str, beg=0 end=len(string))

Parameters
str - This specifies the string to be searched.

beg - This is the starting index, by default its 0

len - This is ending index, by default its equal to the length of the string.

Return Value
This method returns last index if found otherwise raises an exception if str is not found.

Example
The following example shows the usage of rindex() method.

Live Demo

!/usr/bin/python

str1 = "this is string example....wow!!!";
str2 = "is";

print str1.rindex(str2)
print str1.index(str2)

When we run above program, it produces following result -

5
2

Ref: Python String rindex() Method - Tutorialspoint

Not trying to resurrect an inactive post, but since this hasn't been posted yet...

(This is how I did it before finding this question)

s = "hello"
target = "l"
last_pos = len(s) - 1 - s[::-1].index(target)

Explanation: When you're searching for the last occurrence, really you're searching for the first occurrence in the reversed string. Knowing this, I did s[::-1] (which returns a reversed string), and then indexed the target from there. Then I did len(s) - 1 - the index found because we want the index in the unreversed (i.e. original) string.

Watch out, though! If target is more than one character, you probably won't find it in the reversed string. To fix this, use last_pos = len(s) - 1 - s[::-1].index(target[::-1]), which searches for a reversed version of target.

You can use rfind() or rindex()
^{Python2 links: rfind() rindex()}

>>> s = 'Hello StackOverflow Hi everybody'

>>> print( s.rfind('H') )
20

>>> print( s.rindex('H') )
20

>>> print( s.rfind('other') )
-1

>>> print( s.rindex('other') )
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ValueError: substring not found

The difference is when the substring is not found, rfind() returns -1 while rindex() raises an exception ValueError (Python2 link: ValueError).

If you do not want to check the rfind() return code -1, you may prefer rindex() that will provide an understandable error message. Else you may search for minutes where the unexpected value -1 is coming from within your code...

Example: Search of last newline character

>>> txt = '''first line
... second line
... third line'''

>>> txt.rfind('\n')
22

>>> txt.rindex('\n')
22

Try this:

s = 'hello plombier pantin'
print (s.find('p'))
6
print (s.index('p'))
6
print (s.rindex('p'))
15
print (s.rfind('p'))

The more_itertools library offers tools for finding indices of all characters or all substrings.

Given

import more_itertools as mit


s = "hello"
pred = lambda x: x == "l"

Code

Characters

Now there is the rlocate tool available:

next(mit.rlocate(s, pred))
# 3

A complementary tool is locate:

list(mit.locate(s, pred))[-1]
# 3

mit.last(mit.locate(s, pred))
# 3

Substrings

There is also a window_size parameter available for locating the leading item of several items:

s = "How much wood would a woodchuck chuck if a woodchuck could chuck wood?"
substring = "chuck"
pred = lambda *args: args == tuple(substring)

next(mit.rlocate(s, pred=pred, window_size=len(substring)))
# 59