How to check for palindrome using Python logic

An alternative to the rather unintuitive [::-1] syntax is this:

>>> test = "abcba"
>>> test == ''.join(reversed(test))
True

The reversed function returns a reversed sequence of the characters in test.

''.join() joins those characters together again with nothing in between.

Just for the record, and for the ones looking for a more algorithmic way to validate if a given string is palindrome, two ways to achieve the same (using while and for loops):

def is_palindrome(word):

    letters = list(word)    
    is_palindrome = True
    i = 0

    while len(letters) > 0 and is_palindrome:       
        if letters[0] != letters[(len(letters) - 1)]:
            is_palindrome = False
        else:
            letters.pop(0)
            if len(letters) > 0:
                letters.pop((len(letters) - 1))

    return is_palindrome

And....the second one:

def is_palindrome(word):

    letters = list(word)
    is_palindrome = True

    for letter in letters:
        if letter == letters[-1]:
            letters.pop(-1)
        else:
            is_palindrome = False
            break

    return is_palindrome

The awesome part of python is the things you can do with it. You don't have to use indexes for strings.

The following will work (using slices)

def palindrome(n):
    return n == n[::-1]

What it does is simply reverses n, and checks if they are equal. n[::-1] reverses n (the -1 means to decrement)

"2) My for loop (in in range (999, 100, -1), is this a bad way to do it in Python?"

Regarding the above, you want to use xrange instead of range (because range will create an actual list, while xrange is a fast generator)

My opinions on question 3

I learned C before Python, and I just read the docs, and played around with it using the console. (and by doing Project Euler problems as well :)

Below the code will print 0 if it is Palindrome else it will print -1

Optimized Code

word = "nepalapen"
is_palindrome = word.find(word[::-1])
print is_palindrome

Output: 0

word = "nepalapend"
is_palindrome = word.find(word[::-1])
print is_palindrome

Output: -1

Explaination:

when searching the string the value that is returned is the value of the location that the string starts at.

So when you do word.find(word[::-1]) it finds nepalapen at location 0 and [::-1] reverses nepalapen and it still is nepalapen at location 0 so 0 is returned.

Now when we search for nepalapend and then reverse nepalapend to dnepalapen it renders a FALSE statement nepalapend was reversed to dnepalapen causing the search to fail to find nepalapend resulting in a value of -1 which indicates string not found.

Another method print true if palindrome else print false

word = "nepalapen"
print(word[::-1]==word[::1])

output: TRUE

There is also a functional way:

def is_palindrome(word):
  if len(word) == 1: return True
  if word[0] != word[-1]: return False
  return is_palindrome(word[1:-1])

I know that this question was answered a while ago and i appologize for the intrusion. However,I was working on a way of doing this in python as well and i just thought that i would share the way that i did it in is as follows,

word = 'aibohphobia'

word_rev = reversed(word)

def is_palindrome(word):
    if list(word) == list(word_rev):
        print'True, it is a palindrome'
    else:
        print'False, this is''t a plindrome'

is_palindrome(word)

The most pythonic way to do this is indeed using the slicing notation to reverse the string as mentioned already:

def is_palindrome(string: str) -> bool:
    return string == string[::-1]

In some other occasions though (like technical interviews), you may have to write a "proper" algorithm to find the palindrome. In this case, the following should do the trick:

def is_palindrome(string: str) -> bool:
    start = 0
    end = len(string) - 1
    
    while end >= start:
        if string[end] != string[start]:
            return False
        start += 1
        end -= 1
        
    return True

• Set pointers to the start and end of the string
• Iterate while end exceeds start
• If the character in end and start indices don't match then this is not a palindrome, otherwise keep comparing
• Increase start pointer by 1
• Decrease end pointer by 1

Test Cases:

import unittest

class Test(unittest.TestCase):

    palindromes = ['a', 'aa', 'aba', '12321']
    non_palindromes = ['ab', 'aab', 'cacacc']
    def test_is_palindrome(self):
        for case in self.palindromes:
            self.assertTrue(is_palindrome(case))

        for case in self.non_palindromes:
            self.assertFalse(is_palindrome(case))


if __name__ == '__main__':
    unittest.main()

Here a case insensitive function since all those solutions above are case sensitive.

def Palindrome(string): 

  return (string.upper() == string.upper()[::-1]) 

This function will return a boolean value.

doing the Watterloo course for python, the same questions is raised as a "Lesseon" find the info here:

http://cscircles.cemc.uwaterloo.ca/13-lists/

being a novice i solved the problem the following way:

def isPalindrome(S):
    pali = True
    for i in range (0, len(S) // 2):
        if S[i] == S[(i * -1) - 1] and pali is True:
            pali = True
        else:
            pali = False
    print(pali)
    return pali

The function is called isPalindrome(S) and requires a string "S". The return value is by default TRUE, to have the initial check on the first if statement.

After that, the for loop runs half the string length to check if the character from string "S" at the position "i" is the same at from the front and from the back. If once this is not the case, the function stops, prints out FALSE and returns false.

Cheers.kg

If the string has an uppercase or non-alphabetic character then the function converts all characters to lowercase and removes all non-alphabetic characters using regex finally it applies palindrome check recursively:

import re

rules = [
    lambda s: any(x.isupper() for x in s),
    lambda s: not s.isalpha()
]


def is_palindrome(s):
    if any(rule(s) for rule in rules):
        s = re.sub(r'[^\w]', '', s).lower()
    if len(s) < 2:
        return True
    if s[0] != s[-1]:
        return False
    return is_palindrome(s[1:-1])


string = 'Are we not drawn onward, we few, drawn onward to new era?'

print(is_palindrome(string))

the output is True for the input above.

There is much easier way I just found. It's only 1 line.

is_palindrome = word.find(word[::-1])

You are asking palindrome in python. palindrome can be performed on strings, numbers and lists. However, I just posted a simple code to check palindrome of a string.

# Palindrome of string
str=raw_input("Enter the string\n")
ln=len(str)
for i in range(ln/2) :
    if(str[ln-i-1]!=str[i]):
        break
if(i==(ln/2)-1):
    print "Palindrome"
else:
    print "Not Palindrome"

Assuming a string 's'

palin = lambda s: s[:(len(s)/2 + (0 if len(s)%2==0 else 1)):1] == s[:len(s)/2-1:-1]  
# Test
palin('654456')  # True
palin('malma')   # False
palin('ab1ba')   # True

#compare 1st half with reversed second half
# i.e. 'abba' -> 'ab' == 'ba'[::-1]

def is_palindrome( s ):
   return True if len( s ) < 2 else s[ :len( s ) // 2 ] == s[ -( len( s ) // 2 ):][::-1]

You can use Deques in python to check palindrome

def palindrome(a_string): ch_dequeu = Deque() for ch in a_string: ch_dequeu.add_rear(ch) still_ok = True while ch_dequeu.size() > 1 and still_ok: first = ch_dequeu.remove_front() last = ch_dequeu.remove_rear() if first != last: still_ok = False return still_ok

class Deque: def __init__(self): self.items = [] def is_empty(self): return self.items == [] def add_rear(self, item): self.items.insert(0, item) def add_front(self, item): self.items.append(item) def size(self): return len(self.items) def remove_front(self): return self.items.pop() def remove_rear(self): return self.items.pop(0)

import string

word = input('Please select a word to test \n')
word = word.lower()
num = len(word)

x = round((len(word)-1)/2)
#defines first half of string
first = word[:x]

#reverse second half of string
def reverse_odd(text):
    lst = []
    count = 1
    for i in range(x+1, len(text)):

        lst.append(text[len(text)-count])
        count += 1
    lst = ''.join(lst)
    return lst

#reverse second half of string
def reverse_even(text):
    lst = []
    count = 1
    for i in range(x, len(text)):
        lst.append(text[len(text)-count])
        count += 1
    lst = ''.join(lst)
    return lst


if reverse_odd(word) == first or reverse_even(word) == first:
    print(string.capwords(word), 'is a palindrome')
else:
    print(string.capwords(word), 'is not a palindrome')

the "algorithmic" way:

import math

def isPalindrome(inputString):
    if inputString == None:
        return False

    strLength = len(inputString)
    for i in range(math.floor(strLength)):
        if inputString[i] != inputString[strLength - 1 - i]:
            return False
    return True

There is another way by using functions, if you don't want to use reverse

#!/usr/bin/python

A = 'kayak'

def palin(A):

    i = 0
    while (i<=(A.__len__()-1)):
        if (A[A.__len__()-i-1] == A[i]):
            i +=1
        else:
         return False

if palin(A) == False:

    print("Not a Palindrome")

else :

    print ("Palindrome")

It looks prettier with recursion!

def isPalindrome(x):
z = numToList(x)
length = math.floor(len(z) / 2)
if length < 2:
    if z[0] == z[-1]:
        return True
    else:
        return False
else:
    if z[0] == z[-1]:
        del z[0]
        del z[-1]
        return isPalindrome(z)
    else:
        return False

def is_palindrome(string):
   return string == ''.join([letter for letter in reversed(string)])

print ["Not a palindrome","Is a palindrome"][s == ''.join([s[len(s)-i-1] for i in range(len(s))])]

This is the typical way of writing single line code

def pali(str1):
    l=list(str1)
    l1=l[::-1]
    if l1==l:
        print("yess")
    else:
        print("noo")
str1="abc"
a=pali(str1)
print(a)

I tried using this:

def palindrome_numer(num):
num_str = str(num)
str_list = list(num_str)
if str_list[0] == str_list[-1]:
    return True
return False

and it worked for a number but I don't know if a string

def isPalin(checkWord):
    Hsize = len(lst)/2
    seed = 1
    palind=True
    while seed<Hsize+1:
        #print seed,lst[seed-1], lst [-(seed)]
        if(lst[seed-1] != lst [-seed]):
            palind = False
            break
        seed = seed+1
    return palind

lst = 'testset'
print lst, isPalin(lst)    
lst = 'testsest'
print lst, isPalin(lst) 

Output

testset True
testsest False

You could use this one-liner that returns a bool value:

str(x)==str(x)[::-1]

This works both for words and numbers thanks to the type casting...

This method can solve the issue:

def is_palindrome(word:str) -> bool:
    word_lst = list(word)
    if not word_lst and (len(word_lst)<=1) and (len(word_lst)>=10**5):
        return
    reversed_word_lst = word_lst[::-1]
    return word_lst == reversed_word_lst

Best!

def palindrome_or_not(val):
    length = len(val)
    for i in range(length // 2):
        if val[i] != val[length - 1 - i]:
            return 'Not Palindrome'
    return 'Palindrome'

print(palindrome_or_not('malayalam'))

#!/usr/bin/python

str = raw_input("Enter a string ")
print "String entered above is %s" %str
strlist = [x for x in str ]
print "Strlist is %s" %strlist
strrev = list(reversed(strlist)) 
print "Strrev is %s" %strrev
if strlist == strrev :
   print "String is palindrome"
else :
   print "String is not palindrome"

maybe you can try this one:

list=input('enter a string:')

if (list==list[::-1]):
    print ("It is a palindrome")
else:
   print("it is not palindrome")