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implement rfind in Python

I am attempting to implement the rfind function in Python without using the built-in rfind method. It should work like the following:
Unlike the original method, takes an input string and a character as a parameter and returns the first index from the right where the character is found. Now I am getting stuck on what to do next..
# main function
inputString = input("Enter here: ")
inputChar = input("Which character would you like to find?")
print(Myrfind(inputString, inputChar))
def Myrfind(text,aChar):
reverseString = text[::-1]
for ch in reverseString:
if ch == aChar:
print()
else:
return -1

The simplest approach, IMHO, would be to just iterate over the string backwards and compare each character:
def myrfind(text, aChar):
for i in range(len(text) - 1, -1, -1):
if text[i] == aChar:
return i
return -1

Your function should be like (with minimal changes in current code):
def Myrfind(text,aChar):
reverseString = text[::-1]
for i, c in enumerate(reverseString): # enumerate() to iterate along with index
if c == aChar:
return len(text) - i - 1 # Return len(char) - i -1 since reverse string
else: # Return -1 if function is not
return -1 # exited by for loop
Sample run:
>>> Myrfind('Hello', 'o')
4
>>> Myrfind('Hello', 'l')
3
>>> Myrfind('Hello', 'e')
1
>>> Myrfind('Hello', 'a') # 'a' not in string
-1

So, in ch you have the character, not the index of the character, so even if you find it, you won't be able to know its index (you just know that the character is in your text string, nothing else)
Also, when you find a coincidence, you print... nothing (in print())
And since the return -1 (when not found) is within the for loop, you will stop the execution (returning -1) as soon as a character doesn't match the looked up character (you should be very familiar with the return statement)
You reverse the string, but that is probably not very... "adviseable", because then your indexes get reversed. For instance, let's say you want to find the last l in Hello. You are expecting:
H e l l o
Index 0 1 2 3 4
And that should be 3, but since you reverse it, and start walking it from left to right, you get:
o l l e H
Index 0 1 2 3 4
And the first match is in 1, now... You can still do it, though... You can substract that index to the total lenght, and you should get it.
Firstly, I would recommend you read what the enumerate built-in does (to not only get the character, but also the index)
A pretty close thing to what you have could be:
inputString = input("Enter here: ")
inputChar = input("Which character would you like to find?")
def Myrfind(text,aChar):
reverseString = text[::-1]
for i, ch in enumerate(reverseString):
if ch == aChar:
return len(text) - 1 - i
return -1
print(Myrfind(inputString, inputChar))
Or, instead of inverting the array, just start reading from the right (see what range does)
def Myrfind(text,aChar):
for i in range(len(text) - 1, -1, -1):
ch = text[i]
if ch == aChar:
return i
return -1

Getting the middle character in a odd length string

def get_middle_character(odd_string):
variable = len(odd_string)
x = str((variable/2))
middle_character = odd_string.find(x)
middle_character2 = odd_string[middle_character]
return middle_character2
def main():
print('Enter a odd length string: ')
odd_string = input()
print('The middle character is', get_middle_character(odd_string))
main()
I need to figure out how to print the middle character in a given odd length string. But when I run this code, I only get the last character. What is the problem?

You need to think more carefully about what your code is actually doing. Let's do this with an example:
def get_middle_character(odd_string):
Let's say that we call get_middle_character('hello'), so odd_string is 'hello':
variable = len(odd_string) # variable = 5
Everything is OK so far.
x = str((variable/2)) # x = '2'
This is the first thing that is obviously odd - why do you want the string '2'? That's the index of the middle character, don't you just want an integer? Also you only need one pair of parentheses there, the other set is redundant.
middle_character = odd_string.find(x) # middle_character = -1
Obviously you can't str.find the substring '2' in odd_string, because it was never there. str.find returns -1 if it cannot find the substring; you should use str.index instead, which gives you a nice clear ValueError when it can't find the substring.
Note that even if you were searching for the middle character, rather than the stringified index of the middle character, you would get into trouble as str.find gives the first index at which the substring appears, which may not be the one you're after (consider 'lolly'.find('l')...).
middle_character2 = odd_string[middle_character] # middle_character2 = 'o'
As Python allows negative indexing from the end of a sequence, -1 is the index of the last character.
return middle_character2 # return 'o'
You could actually have simplified to return odd_string[middle_character], and removed the superfluous assignment; you'd have still had the wrong answer, but from neater code (and without middle_character2, which is a terrible name).
Hopefully you can now see where you went wrong, and it's trivially obvious what you should do to fix it. Next time use e.g. Python Tutor to debug your code before asking a question here.

You need to simply access character based on index of string and string slicing. For example:
>>> s = '1234567'
>>> middle_index = len(s)/2
>>> first_half, middle, second_half = s[:middle_index], s[middle_index], s[middle_index+1:]
>>> first_half, middle, second_half
('123', '4', '567')
Explanation:
str[:n]: returns string from 0th index to n-1th index
str[n]: returns value at nth index
str[n:]: returns value from nth index till end of list

Should be like below:
def get_middle_character(odd_string):
variable = len(odd_string)/2
middle_character = odd_string[variable +1]
return middle_character

i know its too late but i post my solution
I hope it will be useful ;)
def get_middle_char(string):
if len(string) % 2 == 0:
return None
elif len(string) <= 1:
return None
str_len = int(len(string)/2))
return string[strlen]

reversedString = ''
print('What is your name')
str = input()
idx = len(str)
print(idx)
str_to_iterate = str
for char in str_to_iterate[::-1]:
print(char)
evenodd = len(str) % 2
if evenodd == 0:
print('even')
else:
print('odd')
l = str
if len(l) % 2 == 0:
x = len(l) // 2
y = len(l) // 2 - 1
print(l[x], l[y])
else:
n = len(l) // 2
print(l[n])

Remove Last instance of a character and rest of a string

If I have a string as follows:
foo_bar_one_two_three
Is there a clean way, with RegEx, to return: foo_bar_one_two?
I know I can use split, pop and join for this, but I'm looking for a cleaner solution.

result = my_string.rsplit('_', 1)[0]
Which behaves like this:
>>> my_string = 'foo_bar_one_two_three'
>>> print(my_string.rsplit('_', 1)[0])
foo_bar_one_two
See in the documentation entry for str.rsplit([sep[, maxsplit]]).

One way is to use rfind to get the index of the last _ character and then slice the string to extract the characters up to that point:
>>> s = "foo_bar_one_two_three"
>>> idx = s.rfind("_")
>>> if idx >= 0:
... s = s[:idx]
...
>>> print s
foo_bar_one_two
You need to check that the rfind call returns something greater than -1 before using it to get the substring otherwise it'll strip off the last character.
If you must use regular expressions (and I tend to prefer non-regex solutions for simple cases like this), you can do it thus:
>>> import re
>>> s = "foo_bar_one_two_three"
>>> re.sub('_[^_]*$','',s)
'foo_bar_one_two'

Similar the the rsplit solution, rpartition will also work:
result = my_string.rpartition("_")[0]
You'll need to watch out for the case where the separator character is not found. In that case the original string will be in index 2, not 0.
doc string:
rpartition(...)
S.rpartition(sep) -> (head, sep, tail)
Search for the separator sep in S, starting at the end of S, and return
the part before it, the separator itself, and the part after it. If the
separator is not found, return two empty strings and S.

Here is a generic function to remove everything after the last occurrence of any specified string. For extra credit, it also supports removing everything after the nth last occurrence.
def removeEverythingAfterLast (needle, haystack, n=1):
while n > 0:
idx = haystack.rfind(needle)
if idx >= 0:
haystack = haystack[:idx]
n -= 1
else:
break
return haystack
In your case, to remove everything after the last '_', you would simply call it like this:
updatedString = removeEverythingAfterLast('_', yourString)
If you wanted to remove everything after the 2nd last '_', you would call it like this:
updatedString = removeEverythingAfterLast('_', yourString, 2)

I know is python, and my answer may be a little bit wrong in syntax, but in java you would do:
String a = "foo_bar_one_two_three";
String[] b = a.split("_");
String c = "";
for(int i=0; i<b.length-1; a++){
c += b[i];
if(i != b.length-2){
c += "_";
}
}
//and at this point, c is "foo_bar_one_two"
Hope in python split function works same way. :)
EDIT:
Using the limit part of the function you can do:
String a = "foo_bar_one_two_three";
String[] b = a.split("_",StringUtils.countMatches(a,"_"));
//and at this point, b is the array = [foo,bar,one,two]

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?

Use .rfind():
>>> s = 'hello'
>>> s.rfind('l')
3
Also don't use str as variable name or you'll shadow the built-in str().

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

Use the str.rindex method.
>>> 'hello'.rindex('l')
3
>>> 'hello'.index('l')
2

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.

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'))

For this case both rfind() and rindex() string methods can be used, both will return the highest index in the string where the substring is found like below.
test_string = 'hello'
target = 'l'
print(test_string.rfind(target))
print(test_string.rindex(target))
But one thing should keep in mind while using rindex() method, rindex() method raises a ValueError [substring not found] if the target value is not found within the searched string, on the other hand rfind() will just return -1.

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

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

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

# Last Occurrence of a Character in a String without using inbuilt functions
str = input("Enter a string : ")
char = input("Enter a character to serach in string : ")
flag = 0
count = 0
for i in range(len(str)):
if str[i] == char:
flag = i
if flag == 0:
print("Entered character ",char," is not present in string")
else:
print("Character ",char," last occurred at index : ",flag)

you can use rindex() function to get the last occurrence of a character in string
s="hellloooloo"
b='l'
print(s.rindex(b))

str = "Hello, World"
target='l'
print(str.rfind(target) +1)
or
str = "Hello, World"
flag =0
target='l'
for i,j in enumerate(str[::-1]):
if target == j:
flag = 1
break;
if flag == 1:
print(len(str)-i)

How to get the position of a character in Python?

How can I get the position of a character inside a string in Python?

There are two string methods for this, find() and index(). The difference between the two is what happens when the search string isn't found. find() returns -1 and index() raises a ValueError.
Using find()
>>> myString = 'Position of a character'
>>> myString.find('s')
2
>>> myString.find('x')
-1
Using index()
>>> myString = 'Position of a character'
>>> myString.index('s')
2
>>> myString.index('x')
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
ValueError: substring not found
From the Python manual
string.find(s, sub[, start[, end]])
Return the lowest index in s where the substring sub is found such that sub is wholly contained in s[start:end]. Return -1 on failure. Defaults for start and end and interpretation of negative values is the same as for slices.
And:
string.index(s, sub[, start[, end]])
Like find() but raise ValueError when the substring is not found.

Just for a sake of completeness, if you need to find all positions of a character in a string, you can do the following:
s = 'shak#spea#e'
c = '#'
print([pos for pos, char in enumerate(s) if char == c])
which will print: [4, 9]

>>> s="mystring"
>>> s.index("r")
4
>>> s.find("r")
4
"Long winded" way
>>> for i,c in enumerate(s):
... if "r"==c: print i
...
4
to get substring,
>>> s="mystring"
>>> s[4:10]
'ring'

Just for completion, in the case I want to find the extension in a file name in order to check it, I need to find the last '.', in this case use rfind:
path = 'toto.titi.tata..xls'
path.find('.')
4
path.rfind('.')
15
in my case, I use the following, which works whatever the complete file name is:
filename_without_extension = complete_name[:complete_name.rfind('.')]

What happens when the string contains a duplicate character?
from my experience with index() I saw that for duplicate you get back the same index.
For example:
s = 'abccde'
for c in s:
print('%s, %d' % (c, s.index(c)))
would return:
a, 0
b, 1
c, 2
c, 2
d, 4
In that case you can do something like that:
for i, character in enumerate(my_string):
# i is the position of the character in the string

string.find(character)
string.index(character)
Perhaps you'd like to have a look at the documentation to find out what the difference between the two is.

A character might appear multiple times in a string. For example in a string sentence, position of e is 1, 4, 7 (because indexing usually starts from zero). but what I find is both of the functions find() and index() returns first position of a character. So, this can be solved doing this:
def charposition(string, char):
pos = [] #list to store positions for each 'char' in 'string'
for n in range(len(string)):
if string[n] == char:
pos.append(n)
return pos
s = "sentence"
print(charposition(s, 'e'))
#Output: [1, 4, 7]

If you want to find the first match.
Python has a in-built string method that does the work: index().
string.index(value, start, end)
Where:
Value: (Required) The value to search for.
start: (Optional) Where to start the search. Default is 0.
end: (Optional) Where to end the search. Default is to the end of the string.
def character_index():
string = "Hello World! This is an example sentence with no meaning."
match = "i"
return string.index(match)
print(character_index())
> 15
If you want to find all the matches.
Let's say you need all the indexes where the character match is and not just the first one.
The pythonic way would be to use enumerate().
def character_indexes():
string = "Hello World! This is an example sentence with no meaning."
match = "i"
indexes_of_match = []
for index, character in enumerate(string):
if character == match:
indexes_of_match.append(index)
return indexes_of_match
print(character_indexes())
# [15, 18, 42, 53]
Or even better with a list comprehension:
def character_indexes_comprehension():
string = "Hello World! This is an example sentence with no meaning."
match = "i"
return [index for index, character in enumerate(string) if character == match]
print(character_indexes_comprehension())
# [15, 18, 42, 53]

more_itertools.locate is a third-party tool that finds all indicies of items that satisfy a condition.
Here we find all index locations of the letter "i".
Given
import more_itertools as mit
text = "supercalifragilisticexpialidocious"
search = lambda x: x == "i"
Code
list(mit.locate(text, search))
# [8, 13, 15, 18, 23, 26, 30]

Most methods I found refer to finding the first substring in a string. To find all the substrings, you need to work around.
For example:
Define the string
vars = 'iloveyoutosimidaandilikeyou'
Define the substring
key = 'you'
Define a function that can find the location for all the substrings within the string
def find_all_loc(vars, key):
pos = []
start = 0
end = len(vars)
while True:
loc = vars.find(key, start, end)
if loc is -1:
break
else:
pos.append(loc)
start = loc + len(key)
return pos
pos = find_all_loc(vars, key)
print(pos)
[5, 24]

A solution with numpy for quick access to all indexes:
string_array = np.array(list(my_string))
char_indexes = np.where(string_array == 'C')