I'd like to know how do I check if an input is a palindrome with a while loop, using Python.
Thanks:
i tried this
i = 0
n = len(msg_list)
while i < n:
palindrome = msg_list[i]
if palindrome == msg_list[-1]:
print("Palindrome? True")
msg_list.pop(-1)
else:
print("Palindrome? False")
i=i+1
but at the end I receive an error message that the list index is out of range
You don't need to iterate till the end, but only till the middle character. And compare every character to the character at the same index when counted in reverse:
s = "abcca"
length = len(s)
i = 0
while i < length / 2 + 1:
if s[i] != s[-i - 1]:
print "Not Palindrome"
break
i += 1
else:
print "Palidrome"
else part of the while loop is executed, when the loop completes its iteration without any break.
Alternatively, if you can use anything else than a while loop, then this task is just of single line in Python:
if s == s[::-1]:
print "Palindrome"
Oh, it became two lines.
With a while loop
import string
palin = 'a man, a plan, a canal, panama'
def testPalindrome(in_val):
in_val = in_val.lower()
left, right = 0, len(in_val) - 1
while left < right:
char_left, char_right = '#', '#'
while char_left not in string.lowercase:
char_left = in_val[left]
left += 1
while char_right not in string.lowercase:
char_right = in_val[right]
right -= 1
if char_left != char_right:
return False
return True
print testPalindrome(palin)
Without
>>> palindrome = 'a man, a plan, a canal, panama'
>>> palindrome = palindrome.replace(',', '').replace(' ', '')
>>> palindrome
'amanaplanacanalpanama'
>>> d[::-1] == d
True
A short solution using reversed:
for c, cr in s, reversed(s):
if c != cr:
print("Palindrome? False")
break
else:
print("Palindrome? True")
Another way using a while loop. Once two characters don't match, the while loop stops, so it's quite efficient but of course not the best way to do it in Python.
def palindrome(word):
chars_fw = list(word)
chars_bw = list(reversed(word))
chars_num = len(word)
is_palindrome = True
while chars_num:
if chars_fw[chars_num-1] != chars_bw[chars_num-1]:
is_palindrome = False
break
chars_num -= 1
return is_palindrome
Figured I would add another alternative for people still viewing this question. It uses a while loop, and is reasonably concise (although, I still prefer the if word = word[::-1] approach.
def is_palindrome(word):
word = list(word.replace(' ', '')) # remove spaces and convert to list
# Check input
if len(word) == 1:
return True
elif len(word) == 0:
return False
# is it a palindrome....
while word[0] == word[-1]:
word.pop(0)
word.pop(-1)
if len(word) <= 1:
return True
return False
word = "quiniuq"
pairs = zip(word,reversed(word))
a,b = next(pairs)
try:
while a == b:
a,b = next(pairs)
return False # we got here before exhausting pairs
except StopIteration:
return True # a == b was true for every pair
The use of a while loop here is contrived, but it will consume the whole list and perform the test.
If a while loop wasn't a requirement, one would do: all(a == b for a,b in zip(word,reversed(word)))
