How to check for palindrome using Python logic [closed]

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How to check for palindrome using Python logic [closed]

I’m trying to check for a palindrome with Python. The code I have is very for-loop intensive.

And it seems to me the biggest mistake people do when going from C to Python is trying to implement C logic using Python, which makes things run slowly, and it’s just not making the most of the language.

I see on this website. Search for “C-style for”, that Python doesn’t have C-style for loops. Might be outdated, but I interpret it to mean Python has its own methods for this.

I’ve tried looking around, I can’t find much up to date (Python 3) advice for this. How can I solve a palindrome challenge in Python, without using the for loop?

I’ve done this in C in class, but I want to do it in Python, on a personal basis. The problem is from the Euler Project, great site By the way,.

def isPalindrome(n):
    lst = [int(n) for n in str(n)]
    if l==0 || l==1:
        return True
    elif len(lst)%2==0:
        for k in range (l)
        while (k<=((l-1)/2)):
            if (list[]):
for i in range (999, 100, -1):
    for j in range (999,100, -1):
        if isPalindrome(i*j):

I’m missing a lot of code here. The five hashes are just reminders for myself.

Concrete questions:

  1. In C, I would make a for loop comparing index 0 to index max, and then index 0+1 with max-1, until something something. How to best do this in Python?

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

  3. Does anybody have any good advice, or good websites, or resources for people in my position? I’m not a programmer, I don’t aspire to be one, I just want to learn enough so that when I write my bachelor’s degree thesis (electrical engineering), I don’t have to simultaneously LEARN an applicable programming language while trying to obtain good results in the project. “How to go from basic C to great application of Python”, that sort of thing.

  4. Any specific bits of code to make a great solution to this problem would also be appreciated, I need to learn good algorithms.. I am envisioning 3 situations. If the value is zero or single digit, if it is of odd length, and if it is of even length. I was planning to write for loops…

PS: The problem is: Find the highest value product of two 3 digit integers that is also a palindrome.

Asked By: DrOnline


Answer #1:

A pythonic way to determine if a given value is a palindrome:

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


  • We’re checking if the string representation of n equals the inverted string representation of n
  • The [::-1] slice takes care of inverting the string
  • After that, we compare for equality using ==
Answered By: Óscar López

Answer #2:

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

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

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

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

Answered By: RichieHindle

Answer #3:

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
            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]:
            is_palindrome = False
    return is_palindrome

Answer #4:

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 🙂

Answered By: jh314

Answer #5:

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


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



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"


Answered By: Ganesh Pandey

Answer #6:

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])
Answered By: Khozzy

Answer #7:

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'
        print'False, this is''t a plindrome'
Answered By: sanster9292

Answer #8:

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

is_palindrome = word.find(word[::-1])
Answered By: Inna Bohdanova

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