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How to debug my Python program, which sums positive numbers based on their evenness

I'm trying to write program that asking the user for positive numbers, if it is an odd number, the software sums all of the odd digits in the number, same for even numbers. After that the software asking non stop for numbers and does the same thing as before, till the user type 0/negative number.
After that the software should print the number with the maximal sum. Sometimes it works and sometimes not.
Code:
def sum_Digits(n):
sum = 0
if n % 2 == 0: #For even numbers
while n>0:
if (n%10)%2 == 0:
sum += n%10
n = n//10
else:
n = n//10
print("sum: " , sum)
return sum
elif n % 2 != 0 : #For odd numbers
while n>0:
if (n%10)%2 != 0:
sum += n%10
n = n//10
else:
n = n//10
print("sum: " , sum)
return sum
def read_Numbers(N):
maX = 0
while N > 0: #while askNum Positive continue summing
suM = sum_Digits(N)
if suM > maX:
maX = N
N = int(input("Please eneter a Natural number: "))
if N <= 0:
return maX
def main():
num = int(input("Please enter a Natural number: ")) #asking the user to enter number
sum_Digits(num)
askNum = int(input("Please eneter a Natural number: "))
maxSum = read_Numbers(askNum)
print("Number with maximal sum: " , maxSum)
main()

Bug:
if user enter a negative number it will continue the loop and print the last entered number
because the program doesn't check for negative numbers
Fix:
I have added a condition if N <= 0:
return maX
this will return the last entered positive number
Bug:
sum_Digits(num)
I have removed this line because it is not needed
because we already call the function sum_Digits in the main function

Enter n different positive numbers. Calculate the average value of all prime numbers among them. [Python]

I need to do as the title suggests however I have ran into a problem. This is my code so far:
#Input
n = int(input('Enter n: '))
#Prime = 0
p = []
#Loopidty
for i in range(n):
#Ask user to input x
x = int(input('Enter a Number: '))
#Check if prime
if x < 2:
print('The number is not prime.')
else:
for n in range(2, x - 1):
if x % n == 0:
print('The number is not prime.')
break
else:
print('The number is prime.')
p.append(x)
#Answer
Answer = sum(p) / n
print('The answer is: ', Answer)
The issues is when I add the prime numbers to the list it only adds the first number and stops there, how do I combat this?
Many thanks

It's a good idea to break your code into smaller parts. We can create a function to check if a number is prime or not and check that it is correct. After that just add the rest of the code.
import math
def is_prime(x):
"""Check that a number `x``is prime"""
# You only need to go up to sqrt(x)
for n in range(2, int(math.sqrt(x) + 1)):
if x % n == 0:
return False
return True
# xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
sum = 0.0
num_entries = 0 # Number of prime numbers entered by the user
n = int(input('Enter n: '))
for i in range(n):
x = int(input('Enter a Number: '))
if is_prime(x):
sum += x
num_entries += 1
if num_entries == 0:
print("You didn't enter any prime number")
else:
print("The average of the entered prime numbers is ", sum / num_entries)

Problem in a python program to find 2 digits happy numbers ( stuck in loop )

I'm a beginner in python and I just wrote a program to find 2 digit happy numbers. we should give it a 2 digit number and then if it's a happy number it should print that it's a happy number and if it's not it should loop endlessly in a cycle which does not include 1.
this is my code :
input_number = int(input("Please Enter a positive 2 digits number\n"))
digits = [int(i) for i in str(input_number)]
while input_number != 1:
for i in range(len(digits)):
sum_of_squares = digits[i]**2 + digits[i-1]**2
input_number = sum_of_squares
print(input_number)
if input_number == 1:
print("Happy ;)")
break
my problem is that my program never leaves the loop, for example 49 is a happy number, but when i enter it, it prints 97 endlessly.
what's wrong with my code?

A slight variation to #joyal-mathew 's answer. This would work for any number of digits
input_number = int(input("Please Enter a positive number\n"))
while input_number != 1:
digits = [int(i) for i in str(input_number)]
digit_squares = [i*i for i in digits]
sum_of_squares = sum(digit_squares)
input_number = sum_of_squares
print(input_number)
iter_count+=1
if input_number == 1:
print("Happy ;)")
break

Inside the loop you need to change the value of digits. Also since it's for 2 digit numbers you can just use list indices of 0 and 1.
input_number = input()
digits = [int(i) for i in input_number]
while input_number != 1:
sum_of_squares = digits[0]**2 + digits[1]**2
digits = [int(i) for i in str(sum_of_squares)]
input_number = sum_of_squares
print(input_number)
if input_number == 1:
print("Happy ;)")
break

A recursive and verbose solution:
def is_happy(int_tc, seen = None):
"""checks if the number is a happy number"""
if seen is None:
seen = [int_tc]
sum_squares = sum_of_squares_of_all_digits(seen[-1])
if sum_squares == 1:
return True
elif sum_squares in seen: # check if the numbers are repeating infinitely
return False
else:
seen.append(sum_squares)
return is_happy(int_tc, seen)
def sum_of_squares_of_all_digits(number):
"""return sum of square of all digits"""
return sum([digit ** 2 for digit in get_all_digits(number)])
def get_all_digits(number):
"""gets all the digits from the number right to left one bye one(yield)"""
while(number > 0):
yield number % 10
number = (number - number % 10) / 10

Python, largest odd integer

I am new to coding and have been learning a few days now. I wrote this program in Python while following along in some MIT OpenCourseware lectures and a few books. Are there anyways to more easily express the program?
Finger exercise: Write a program that asks the user to input 10 integers, and then prints the largest odd number that was entered. If no odd number was entered, it should print a message to that effect.
a = int(raw_input('Enter your first integer: '))
b = int(raw_input('Enter your second integer: '))
c = int(raw_input('Enter your third integer: '))
d = int(raw_input('Enter your fourth integer: '))
e = int(raw_input('Enter your fifth integer: '))
f = int(raw_input('Enter your sixth integer: '))
g = int(raw_input('Enter your seventh integer: '))
h = int(raw_input('Enter your eighth integer: '))
i = int(raw_input('Enter your ninth integer: '))
j = int(raw_input('Enter your tenth integer: '))
if a%2 ==0:
a = 0
else:
a = a
if b%2 ==0:
b = 0
else:
b = b
if c%2 ==0:
c = 0
else:
c = c
if d%2 ==0:
d = 0
else:
d = d
if e%2 ==0:
e = 0
else:
e = e
if f%2 ==0:
f = 0
else:
f = f
if g%2 ==0:
g = 0
else:
g = g
if h%2 ==0:
h = 0
else:
h = h
if i%2 ==0:
i = 0
else:
i = i
if j%2 ==0:
j = 0
else:
j = j
value = a, b, c, d, e, f, g, h, i, j
max = max(value)
if max ==0:
print 'There are no odd numbers.'
else:
print max, 'is the largest odd integer.'

A more compact form would be:
from __future__ import print_function
try: # Python 2
raw_input
except NameError: # Python 3 compatibility
raw_input = input
largest = None
for i in range(1, 11):
number = int(raw_input('Enter integer #%d: ' % i))
if number % 2 != 0 and (not largest or number > largest):
largest = number
if largest is None:
print("You didn't enter any odd numbers")
else:
print("Your largest odd number was:", largest)
This uses a simple loop to track how many integers were entered, but only stores the largest odd number encountered so far.

numbers = [input('Enter a number: ') for i in range(10)]
odds = [x for x in numbers if x % 2 == 1]
if odds:
print max(odds)
else:
print 'No odd numbers input'
Explanation:
numbers = [input('Enter a number: ') for i in range(10)]
This line is using a list comprehension to ask a user for 10 numbers. These numbers will be in the list object numbers
odds = [x for x in numbers if x % 2 == 1]
Next we are using another list comprehension to filter out all numbers in numbers that are not odd. Since odd numbers modulo 2 always equal 1, we are given a new list (odd) that only contains odd numbers.
if odds:
This is using python's truthy way of testing. Particularly, if a list is empty, this is False. If the list is not empty, it is True.
print max(odds)
Finally, if the above was True, we print the max value in the odds list
else:
print 'No odd numbers input'
If the if statement was False (there are no odds) we tell the user
A running copy looks like this:
Enter a number: 10
Enter a number: 12
Enter a number: 14
Enter a number: 15
Enter a number: 16
Enter a number: 17
Enter a number: 1
Enter a number: 2
Enter a number: 19
Enter a number: 2
19

Python has objects called list and tuple which represent a sequence of numbers—they serve many of the same purposes as "arrays" in other programming languages. An example of a list is [1,2,3,4,5]
Like most popular programming languages, Python also has the concept of a for loop.
myList = [1,2,3,4,5]
for x in myList:
print(x)
Python also has a somewhat unusual but very useful construct called a "list comprehension" which combines for loop, list, and an optional conditional in one neat syntax—check out these examples and see if you can understand how the result relates to the code
myNewList = [x+1 for x in myList]
myNewSelectiveList = [x+1 for x in myList if x >= 3]
and here's an example that's particularly useful in your exercise:
userInputs = [int(raw_input('Enter a number:')) for i in range(10)]
Finally, there's a function max which can take a list as its argument and which returns the largest item in the list. Once you have your 10 inputs in a list, you should be able to use these ingredients to find the highest odd number in one pretty short line (max over a list comprehension with an if conditional in it) .

I'm also studying Guttag's book from scratch. I came with the following solution:
list = []
odds = False
print('You will be asked to enter 10 integer numbers, one at a time.')
for i in range(1,11):
i = int(input('Number: '))
list.append(i)
list = sorted(list, reverse = True)
for j in list:
if j%2 == 1:
print('The largest odd number is', j, 'from', list)
odds = True
break
if odds == False:
print('There is no odd number from', list)
I went through a long(er) version similar to the OP, but since the reading list for MIT 6.00x explicitly suggested studying topic 3.2 alongside chapter 2, I thought that lists would be an acceptable answer.
The code above should allow for negatives and zero.

I'm also working through Guttag's book, this solution uses some of the above code but might have a different spin on things. I filtered out the user input right away to include only odd integer input. If all of the input is even, then the list is empty, the code checks for an empty list, then sorts whatever is left (odd integers) and returns the last one. Please let me know if there are any hidden problems in this (I'm rather new to writing algorithms as well).
arr = []
max = 0
while max < 10:
userNum = int(input('enter an int: '))
if userNum %2 != 0:
arr.append(userNum)
max = max + 1
if len(arr) == 0:
print('all are even')
oddArr = sorted(arr)
print(oddArr[-1])

I just find this question while looking for alternative answers here is my code :
Note:I changed the code a bit so I can decide how many numbers I want to enter
alist=[]
olist=[]
num=int(input("how many numbers ? : "))
for n in range(num):
numbers=int(input())
alist.append(numbers)
for n in range(len(alist)):
while alist[n]%2 != 0 :
olist.append(alist[n])
break
else:
n +=1
olist.sort()
if len(olist) != 0:
print("biggest odd number: ",olist[-1])
else:
print("there is no odd number ")

The question is very early in the book and assumes no knowledge of lists, or indeed the Nonetype (it has mentioned it but not explained how it is used). Also, some solutions here will not work if the highest odd is negative because they initialise largest = 0 before the loop.
This works:
iters = 10
largest = "spam"
while iters != 0:
user_num = int(input("Enter an integer: "))
if user_num % 2 != 0:
if largest == "spam" or user_num > largest:
largest = user_num
iters -= 1
if largest == "spam":
print("You did not enter an odd integer")
else:
print("The largest odd integer was", largest)

itersLeft = 10 #define number of integers
x=0 #creating a variable for storing values
max=0 #creating a variable for defining max
while itersLeft!=0:
x=int(input())
itersLeft = itersLeft-1
if x%2!=0 and x>max:
max=x
if max!=0:
print(max)
elif max==0:
print("No odd number was entered")
*Note: works only for non-negative numbers

The other answers and comments suggesting lists and loops are much nicer, but they're not the only way to change and shorten your code.
In your tests, the else: a = a sections are doing nothing, assigning a to itself is no change, so they can all be removed, and the if tests brought onto one line each:
a = int(raw_input('Enter your first integer: '))
b = int(raw_input('Enter your second integer: '))
c = int(raw_input('Enter your third integer: '))
d = int(raw_input('Enter your fourth integer: '))
e = int(raw_input('Enter your fifth integer: '))
f = int(raw_input('Enter your sixth integer: '))
g = int(raw_input('Enter your seventh integer: '))
h = int(raw_input('Enter your eighth integer: '))
i = int(raw_input('Enter your ninth integer: '))
j = int(raw_input('Enter your tenth integer: '))
if a%2 == 0: a = 0
if b%2 == 0: b = 0
if c%2 == 0: c = 0
if d%2 == 0: d = 0
if e%2 == 0: e = 0
if f%2 == 0: f = 0
if g%2 == 0: g = 0
if h%2 == 0: h = 0
if i%2 == 0: i = 0
if j%2 == 0: j = 0
value = a, b, c, d, e, f, g, h, i, j
max = max(value)
if max ==0:
print 'There are no odd numbers.'
That's the most obvious change that makes it easier to follow, without fundamentally changing the pattern of what you are doing.
After that, there are ways you could rewrite it - for example, staying with math only, even numbers divide by 2 with remainder 0 and odd numbers have remainder 1. So doing (x % 2) * x will change even numbers into 0, but keep odd numbers the same.
So you could replace all the if tests, with no test, just an assignment:
a = (a % 2) * a
b = (b % 2) * b
c = (c % 2) * c
...
if e%2 == 0: e = 0
if f%2 == 0: f = 0
The lines get slightly shorter, and if you're OK with how that works, you could merge those into the value line, and put that straight into max to get:
a = int(raw_input('Enter your first integer: '))
b = int(raw_input('Enter your second integer: '))
c = int(raw_input('Enter your third integer: '))
d = int(raw_input('Enter your fourth integer: '))
e = int(raw_input('Enter your fifth integer: '))
f = int(raw_input('Enter your sixth integer: '))
g = int(raw_input('Enter your seventh integer: '))
h = int(raw_input('Enter your eighth integer: '))
i = int(raw_input('Enter your ninth integer: '))
j = int(raw_input('Enter your tenth integer: '))
largest = max(a%2*a, b%2*b, c%2*c, d%2*d, e%2*e, f%2*f, g%2*g, h%2*h, i%2*i, j%2*j)
if largest == 0:
print 'There are no odd numbers.'
else:
print largest, 'is the largest odd integer.'
There isn't really a way to shorten assigning ten variables without some kind of loop, and it's arguable whether any of this is 'more easily expressing the program', but it does take 58 lines down to 17, remove 10 conditional tests, 10 else/no-op assignments and 1 variable while keeping approximately the same structure / workings.
PS. I changed max = max() because calling your variable by the same name as the function is a bad idea - you can't use the function again, and it's confusing to other Python programmers reading your code who already know what 'max' does, if you've reused that name for something else.
Edit: A commentor suggests negative numbers matter. The above writing answers "here's my code, how could I express it more easily?" without introducing any new Python or changing the behaviour, but it cannot handle negative odd numbers; max() will always choose a zero over a negative odd number, and the program will wrongly answer that "there are no odd numbers".
I don't think that's fixable without introducing any new concepts at all. And if introducing new concepts is happening, use lists and loops. Andy's suggestion to build a list that includes only the odd numbers, and then take the max value of that, for example.
But, doing something to handle them without lists -- there is another approach that hardly changes the shape of the code at all, introducing the boolean OR operation, which compares two true/false values and returns false if they are both false, otherwise true.
Python does a lot of automagic behind-the-scenes conversion to true/false to make logical operators work well. Variables with no value (zero, empty containers, empty strings) are all "false" and variables with some value are all "true".
From earlier, we have one bit that knocks even numbers to zero (a%2*a) and now we want to knock zero off the numberline completely:
-3 or None -> -3
-1 or None -> -1
0 or None -> None
1 or None -> 1
3 or None -> 3
5 or None -> 5
Introducing: a%2*a or None. It's magical, ugly, hard to follow, but valid Python -- and I'm chuffed because it's like solving a puzzle and it works, you know? Change the max line and the test to:
largest = max(a%2*a or None, b%2*b or None, c%2*c or None, d%2*d or None, e%2*e or None,
f%2*f or None, g%2*g or None, h%2*h or None, i%2*i or None, j%2*j or None)
if largest == None:
Evens get squished to zeros, zeros get squished to nothing, odds come through unchanged. Max now only has odd numbers to work with so it can now pick a negative odd number as the answer. Case closed. btw. use lists.

Compare 10 inputs and print the highest odd number
y = 0
for counter in range(10):
x = int(input("Enter a number: "))
if (x%2 == 1 and x > y):
y = x
if (y == 0):
print("All are even")
else:
print(y, "is the largest odd number")

Try the following -
def largest_odd():
q = int(input('Please enter a number: '))
w = int(input('Please enter a number: '))
e = int(input('Please enter a number: '))
lis = []
if q%2 != 0:
lis.insert (q,q)
if w%2 != 0:
lis.insert (w,w)
if e%2 != 0:
lis.insert (e,e)
Great = max(lis)
print(Great)

I am on the same exercise just now, and came up with the following solution, which seems to work just fine and is in line with the topics taught in the book so far(variable assignments, conditionals, and while loop):
EXERCISE:
Write a program that asks the user to input 10 integers, and
then prints the largest odd number that was entered. If no odd number was entered, it should print a message to that effect.
from __future__ import print_function
try: # Python 2
raw_input
except NameError: # Python 3 compatibility
raw_input = input
numbers_count = 0
next_input = 0
max_odd_number = None
while numbers_count < 10:
numbers_count += 1
next_input = raw_input("Please enter a number: " + str(numbers_count) +
" of 10\n")
if int(next_input)%2 != 0:
# on the entry of first number, we check max_odd_number - if it is of
# type None, it means no value has been assigned to it so far thus the
# first odd number entry becomes the first maximum odd number, be it
# positive or negative.
if max_odd_number == None:
max_odd_number = int(next_input)
if int(next_input) > max_odd_number:
max_odd_number = int(next_input)
if max_odd_number == None:
print ("None of the numbers entered were odd!")
else:
print ("Maximum odd number you entered is: " + str(max_odd_number))
Any comments would be appreciated.
Thanks,
A

A simple answer is :
x = 0
result = None;
while(x < 10):
inputx = raw_input('Enter integer #%d: ' % x)
inputx = int(inputx)
if (inputx % 2 == 1):
if(inputx > result):
result = inputx
x += 1
if result is None:
print 'no odd number was entered'
else:
print result
Note: if enter a String like '3f',it will throw a ValueError:
invalid literal for int() with base 10: '3f'
So finally ，the best anwser is
result = None
x = 0
while(x < 10):
inputx = raw_input('Enter integer #%d: ' % x)
try:
inputx = int(inputx)
except ValueError:
print'you enter value ',inputx,' is not a Integer. please try again!'
continue
if (inputx % 2 == 1):
if(inputx > result):
result = inputx
x+=1
if result is None:
print 'no odd number was entered'
else:
print 'the largest odd number is: ',result

The question if introduced in the book just after giving knowledge to if condition and while iteration statement.
Though there are lot many datatypes that could be used in python to get easy solution, we need to use only primitives that too the very basics.
The code below takes 10 user inputs(only odd) and outputs the largest of the 10 numbers.
Answer to the question:(Code)
a1= int(input("Enter the number1: "))
while a1%2 ==0:
print("Entered number is not an odd number.")
a1= int(input("Enter the number1: "))
a2= int(input("Enter the number2: "))
while a2%2 ==0:
print("Entered number is not an odd number.")
a2= int(input("Enter the number2: "))
a3= int(input("Enter the number3: "))
while a3%2 ==0:
print("Entered number is not an odd number.")
a3= int(input("Enter the number3: "))
a4= int(input("Enter the number4: "))
while a4%2 ==0:
print("Entered number is not an odd number.")
a4= int(input("Enter the number4: "))
a5= int(input("Enter the number5: "))
while a5%2 ==0:
print("Entered number is not an odd number.")
a5= int(input("Enter the number5: "))
a6= int(input("Enter the number6: "))
while a6%2 ==0:
print("Entered number is not an odd number.")
a6= int(input("Enter the number6: "))
a7= int(input("Enter the number7: "))
while a7%2 ==0:
print("Entered number is not an odd number.")
a7= int(input("Enter the number7: "))
a8= int(input("Enter the number8: "))
while a8%2 ==0:
print("Entered number is not an odd number.")
a8= int(input("Enter the number8: "))
a9= int(input("Enter the number9: "))
while a9%2 ==0:
print("Entered number is not an odd number.")
a9= int(input("Enter the number9: "))
a10= int(input("Enter the number10: "))
while a10%2 ==0:
print("Entered number is not an odd number.")
a10= int(input("Enter the number10: "))
if a1>a2 and a1>a3 and a1>a4 and a1>a5 and a1>a6 and a1>a7 and a1>a8 and a1>a9 and a1>a10:
print(str(a1) +" is the largest odd number.")
elif a2>a1 and a2>a3 and a2>a4 and a2>a5 and a2>a6 and a2>a7 and a2>a8 and a2>a9 and a2>a10:
print(str(a2) +" is the largest odd number.")
elif a3>a1 and a3>a2 and a3>a4 and a3>a5 and a3>a6 and a3>a7 and a3>a8 and a3>a9 and a3>a10:
print(str(a3) +" is the largest odd number.")
elif a4>a1 and a4>a2 and a4>a3 and a4>a5 and a4>a6 and a4>a7 and a4>a8 and a4>a9 and a4>a10:
print(str(a4) +" is the largest odd number.")
elif a5>a1 and a5>a2 and a5>a3 and a5>a4 and a5>a6 and a5>a7 and a5>a8 and a5>a9 and a5>a10:
print(str(a5) +" is the largest odd number.")
elif a6>a1 and a6>a2 and a6>a3 and a6>a4 and a6>a5 and a6>a7 and a6>a8 and a6>a9 and a6>a10:
print(str(a6) +" is the largest odd number.")
elif a7>a1 and a7>a2 and a7>a3 and a7>a4 and a7>a5 and a7>a6 and a7>a8 and a7>a9 and a7>a10:
print(str(a7) +" is the largest odd number.")
elif a8>a1 and a8>a2 and a8>a3 and a8>a4 and a8>a5 and a8>a6 and a8>a7 and a8>a9 and a8>a10:
print(str(a8) +" is the largest odd number.")
elif a9>a1 and a9>a2 and a9>a3 and a9>a4 and a9>a5 and a9>a6 and a9>a7 and a9>a8 and a9>a10:
print(str(a9) +" is the largest odd number.")
else:
print(str(a10) +" is the largest odd number.")
Hope this helps.

In my experience the way to more easily express the program is with a function.
This would have been my answer in Syntax tested for Python 3.7.6:
'''
Finger exercise:
Write a program that asks the user to input 10 integers,
and then prints the largest odd number that was entered.
If no odd number was entered, it should print a message to that effect.
'''
def LargestOdd(numbers=[]):
'''
Parameters
----------
numbers : list, whcih should contain 10 integers.
DESCRIPTION. The default is [].
Returns
-------
The largest odd integer in the list number.
'''
odd_numbers=[]
if len(numbers)==10:
for n in numbers:
if n%2 != 0:
odd_numbers.append(n)
max_numb=max(odd_numbers)
print('The largest odd number is '+str(max_numb))
else:
print('Please, enter 10 numbers')
LargestOdd([1,2,3,7,45,8,9,10,30,33])
Output: The largest odd number is 45

n = int(input("Enter the no of integers:"))
lst = []
count = 1
while count <=n:
no = int(input("Enter an integer:"))
count = count +1
if (no%2!=0):
lst.append(no)
print ("The list of odd numbers",lst)
print("The maximum number from the list of odd number is:",max(lst))

Here is my solution:
def max_odd():
"""Returns largest odd number from given 10 numbers by the use.
if the input is not valid, the message is displayed to enter valid numbers"""
x = [input('Enter a value: ') for i in range(10)]
x = [int(i) for i in x if i]
if x:
try:
x = [i for i in x if i%2 != 0]
return(max(x))
except:
return('All even numbers provided.')
else:
return('Please enter a valid input')

I started learning coding from Guttag's book. And since this is in the 2nd chapter, the solution follows only basic if condition and while loop
#input 10 integers
n1 = int(input('Enter 1st integer: '))
n2 = int(input('Enter 2nd integer: '))
n3 = int(input('Enter 3rd integer: '))
n4 = int(input('Enter 4th integer: '))
n5 = int(input('Enter 5th integer: '))
n6 = int(input('Enter 6th integer: '))
n7 = int(input('Enter 7th integer: '))
n8 = int(input('Enter 8th integer: '))
n9 = int(input('Enter 9th integer: '))
n10 = int(input('Enter 10th integer: '))
#create list from input
list = [n1,n2,n3,n4,n5,n6,n7,n8,n9,n10]
largest = list[0] #Assign largest to the first integer
x = 1 #index starts at 1
while x < len(list):
if list[x]%2 != 0:
if list[x] > largest:
largest = list[x]
x += 1
if largest%2 == 0:
print('All numbers are even')
else:
print('Largest odd number is', largest)

#This is the simplest program for this question
a=[input('Enter a number: ') for i in range(10)]
#This gets 10 inputs from the user and stores it as a string in a list
b=[int(a[i]) for i in range(10)]
#Here the string values were converted into integer values
for i in range(10):
if max(b)%2==0:
b.remove(max(b))
#Now the loop checks for the max number and if it's even it deletes it.
c=max((b),default='Nil')
if c=="Nil":
print("Please enter an odd number")
else:
print(c,"Is the Largest Odd Number")
#Now the largest number left is an odd number and we finally print it!!!!

This code should work work as well. Syntax tested for python 2.7
def tenX(): #define function
ten = [] #empty list for user input
odds = [] #empty list for odd numbers only
counter = 10
ui = 0
while counter > 0 :
ui = raw_input('Enter a number: ')
ten.append(int(ui)) #add every user input to list after int conversion
counter -= 1
for i in ten:
if i % 2 != 0:
odds.append(i)
print "The highest number is", max(odds) #max() returns highest value in a list
>>> tenX() #call function

The Next Palindrome number

I am beginner in programming, So can you please tell me what's wrong with my code?
I want to print next palindrome number if the number entered by the user (n) is not palindrome
n = int(input("Enter any number :- "))
reverse = 0
temp = n
while (n!=0):
reverse = reverse * 10
reverse = reverse + n%10
n=n//10
if(temp==reverse):
print ("Already palindrome:: ")
if(temp != reverse):
new_temp = temp
new_reverse = 0
for i in range(new_temp,new_temp+10):
while(temp != 0):
new_reverse = new_reverse * 10
new_reverse = new_reverse + temp%10
temp = temp//10
if(new_temp==new_reverse):
print ("Next pallindrome is :- ",new_temp)
break
if(new_temp != new_reverse):
temp = new_temp+1

There are two problems with your code.
1) Your "for i in range" loop calculates the reverse of the temp variable, but you don't change the temp variable's value.
You do
new_temp = temp
for i in range(new_temp,new_temp+10):
[SNIP]
if(new_temp != new_reverse):
temp = new_temp+1 #this value never changes.
So you're making 10 iterations with one and the same value.
2) Ten iterations might not be enough to find a palindrome. Keep going until you find a palindrome.
Working code:
def reverse(num):
reverse= 0
while num:
reverse= reverse*10 + num%10
num= num//10
return reverse
num= int(input("Enter any number :- "))
if num==reverse(num):
print ("Already palindrome.")
else:
while True:
num+= 1
if num==reverse(num):
print ("Next palindrome is : %s"%num)
break

To check if a number is a palindrome, you don't need to convert it to a number. In fact, its a lot simpler if you just check the string equivalent of your number.
>>> i = '212'
>>> i == i[::-1]
True
>>> i = '210'
>>> i == i[::-1]
False
Use this to your advantage, and create a function:
def is_palindrome(foo):
return str(foo) == str(foo)[::-1]
Next, to find the next palindrome, simply increment the number till your palindrome check is true.
Combine all that, and you have:
def is_palindrome(n):
return str(n) == str(n)[::-1]
n = raw_input('Enter a number: ')
if is_palindrome(n):
print('Congratulations! {0} is a palindrome.'.format(n))
else:
n1 = n
while not is_palindrome(n1):
n1 = int(n1)+1
print('You entered {0}, but the next palindrome is {1}'.format(n, n1))
Here is how it works:
$ python t.py
Enter a number: 123
You entered 123, but the next palindrome is 131
$ python t.py
Enter a number: 121
Congratulations! 121 is a palindrome.

If it helps, I believe it's possible to solve this problem with n/2 iterations where n is the length of the input number. Here's my solution in Python:
def next_palin_number(number):
number+=1
# Convert the number to a list of its digits.
number = list(str(number))
# Initialize two indices for comparing symmetric digits.
i = 0
j = len(number) - 1
while i < j:
# If the digits are different:
if number[i] != number[j]:
# If the lower-power digit is greater than the higher-power digit:
if int(number[j]) > int(number[i]):
if number[j-1]!='9':
number[j - 1] = str(int(number[j - 1]) + 1)
number[j] = number[i]
else:
number = list(str(int(''.join(number[:j]))+1))+number[j:]
else:
number[j] = number[i]
i += 1
j -= 1
# Concatenate and return the result.
return "".join(number)

This problem has a wonderful number of ways to solve them.
One of them is
def nearest_palindrome(number):
#start writitng your code here
while True:
number+=1
if str(number) == str(number)[::-1]:
return number
number=12300
print(nearest_palindrome(number))
Thanks for your time to read my answer : )

I have written this for finding next pallindrome number given a pallindrome number.
def palindrome(num):
bol=False
#x=len(str(num))
num=num+1
while(bol==False):
if(check_palindrome(num)):
bol=True
else:
num=num+1
return num
def check_palindrome(n):
temp=n
rev=0
while(n>0):
dig=n%10
rev=rev*10+dig
n=n//10
if(temp==rev):
return True
b=palindrome(8)
print(b)

def next_palin_drome(n):
while True:
n+=1
if str(n) == str(n)[::-1]:
return n
n=12231
print(next_palin_drome(n))
output:12321

def nearest_palindrome(number):
for i in range(1,number):
number=number+1
tem=str(number)
tem1=tem[-1::-1]
if(tem==tem1):
return number
else:
continue
number=12997979797979797
print(nearest_palindrome(number))

def nearest_palindrome(number):
n = len(str(number))//2
if(len(str(number)) % 2 == 0):
#number like 1221
number_1 = int((str(number))[:n]) #12
number_2 = int((str(number))[n:]) #21
if(number_1 < number_2):
number_1 += 1
number_2 = int(str(number_1)[::-1])
else:
number_2 = int(str(number_1)[::-1])
# if last half part is zero then just reverse the first number
if number_2 == 0:
number_2 = str(number_1)[::-1]
#combining the both parts
ans = int(str(number_1) + str(number_2))
return ans
else:
#numer like 12510 n=2
nu = int((str(number))[:n+1]) #add in this number
number_1 = int((str(number))[:n]) # 12
number_2 = int((str(number))[n+1:]) # 21
if (number_1 < number_2):
nu += 1
number_2 = int((str(nu))[::-1][1:])
else:
number_2 = int((str(nu))[::-1][1:])
#if last half part is zero then just reverse the first number
if number_2 == 0:
number_2 = str(nu)[::-1]
number_2 = number_2[1:]
#combinning both parts
ans = int(str(nu) + str(number_2))
return ans
number=12331
print(nearest_palindrome(number))

If a definite range is given:
# function to check if the number is a palindrome
def palin(x):
s=str(x)
if s==s[::-1]:
return True
else:
return False
n=int(input("Enter the number"))
# Putting up range from the next number till 15 digits
for i in range(n+1,int(10e14)):
if palin(i) is True:
print(i)
break

A brute force method:
def math(n):
while not var:
n += 1
if str(n) == str(n)[::-1] : f = 'but next is : '+str(n); return f
n = int(input()); t = math(n); print('Yes',t) if str(n) == str(n)[::-1] else print('No',t); global var; var = False

This is a good fast solution. I saw that the other solutions were iterating and checking through every +1 they did, but this is really slow for big numbers.
This solution has O(n) time if you look at the length of the number
beginNumber = 123456789101112131415161718 #insert number here for next palidrome
string = str(beginNumber + 1)
length = len(string)
number= [int(x) for x in list(string)]
for i in range(length//2):
if (number[i] != number[length-1-i]):
if (number[i]<number[length-1-i]):
number[length-2-i] += 1
number[length-1-i] = number[i]
print("".join([str(x) for x in number]))

I have written this for finding next pallindrome number given a pallindrome number..
#given a pallindrome number ..find next pallindrome number
input=999
inputstr=str(input)
inputstr=inputstr
#append 0 in beginning and end of string ..in case like 99 or 9999
inputstr='0'+inputstr+'0'
length=len(inputstr)
halflength=length/2;
#if even length
if(length%2==0):
#take left part and reverse it(which is equal as the right part )
temp=inputstr[:length/2]
temp=temp[::-1]
#take right part of the string ,move towards lsb from msb..If msb is 9 turn it to zero and move ahead
for j,i in enumerate(temp):
#if number is not 9 then increment it and end loop
if(i!="9"):
substi=int(i)+1
temp=temp[:j]+str(substi)+temp[j+1:]
break;
else:
temp=temp[:j]+"0"+temp[j+1:]
#now you have right hand side...mirror it and append left and right part
output=temp[::-1]+temp
#if the length is odd
if(length%2!=0 ):
#take the left part with the mid number(if length is 5 take 3 digits
temp=inputstr[:halflength+1]
#reverse it
temp=temp[::-1]
#apply same algoritm as in above
#if 9 then make it 0 and move on
#else increment number and break the loop
for j,i in enumerate(temp):
if(i!="9"):
substi=int(i)+1
temp=temp[:j]+str(substi)+temp[j+1:]
break;
else:
temp=temp[:j]+"0"+temp[j+1:]
#now the msb is the middle element so skip it and copy the rest
temp2=temp[1:]
#this is the right part mirror it to get left part then left+middle+right isoutput
temp2=temp2[::-1]
output=temp2+temp
print(output)
similarly for this problem take the left part of given number ...reverse it..store it in temp
inputstr=str(number)
if(inputstr==inputstr[::-1])
print("Pallindrome")
else:
temp=inputstr[:length/2]
temp=temp[::-1]
for j,i in enumerate(temp):
if(i!="9"):
substi=int(i)+1
temp=temp[:j]+str(substi)+temp[j+1:]
break;
else:
temp=temp[:j]+"0"+temp[j+1:]
now depending on length of your number odd or even generate the output..as in the code
if even then output=temp[::-1]+temp
if odd then temp2=temp1[1:]
output=temp2[::-1]+temp
I am not sure about this solution..but hope it helps