How to check for palindrome using Python logic [closed]

Answer 1

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

Answer 2

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

Answer 3

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')

Answer 4

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

Answer 5

word = "<insert palindrome/string>"
reverse = word[::-1] 
is_palindrome = word.find(reverse)
print is_palindrome

This was a question in Udacity comp 101, chapter 1. Gives a 0 for palindrome gives a -1 for not. Its simple, and does not use loops.

Answer 6

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"