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How can I retain zero in python for calculation?

So I supposed to verify if the input number is a UPC number or not. I have to allowed leading zeros and accounted it in the calculation.
Here is my current code, it works for all number except number has leading zeros:
Condition for UPC code valid:
Calculate the sum of multiplying odd digit index by 3 and even digit index by 1 of the input number.
Calculate the sum we just did modulo 10, get result digit.
If the resulting digit is between 1 and 9 then subtract result digit from 10. If the result digit is 0, add 0 to the to the base number to get the completed number.
def UPC_code(num):
sum_digit = 0
index = 0
num_temp = str(num)[:-1]
len_nt = len(num_temp)
for digit in num_temp:
if (index + 1) % 2 != 0: # If number position is odd
sum_digit += int(digit) * 3 # Sum = digit * 3
if index < len_nt: # Increase index till end
index += 1
elif (index + 1) % 2 == 0: # If number position is even
sum_digit += int(digit) * 1 # Sum = digit * 1
if index < len_nt:
index += 1
# print(sum_digit)
res_digit = sum_digit % 10
if 1 <= res_digit <= 9:
res_digit = 10 - res_digit # Res digit meet condition = 10 - res digit
if res_digit == num % 10:
return True
elif res_digit != num % 10:
return False
else:
print("Something went wrong")
# End UPC_code()
Call UPC_code()
import code_check.py as cc
num = str(input())
num_int = int(num)
if cc.UPC_code(num_int) is True and num_int != 0:
print(num, "valid UPC code.")
else:
print("Not valid")
Expected input:
042100005264
Expected output:
042100005264 valid UPC code
Actual output:
Not valid

it works for all number except number has leading zeros
As you have doubtless discovered, python does not allow you to write 0700. Historically that would have meant 0o700, or 448, which is likely not what you want anyhow...
In this case the solution is simple. If you need to handle numbers like 00007878979345, handle strings.
Thus refactor your code to take a string. As a bonus, int("000008") is 8, so when you need the number as a number you don't even have to do anything.

Python: Finding Average of Numbers

My task is to:
"Write a program that will keep asking the user for some numbers.
If the user hits enter/return without typing anything, the program stops and prints the average of all the numbers that were given. The average should be given to 2 decimal places.
If at any point a 0 is entered, that should not be included in the calculation of the average"
I've been trying for a while, but I can't figure out how to make the programs act on anything I instruct when the user hits 'enter' or for it to ignore the 0.
This is my current code:
count = 0
sum = 0
number = 1
while number >= 0:
number = int(input())
if number == '\n':
print ('hey')
break
if number > 0:
sum = sum + number
count= count + 1
elif number == 0:
count= count + 1
number += 1
avg = str((sum/count))
print('Average is {:.2f}'.format(avg))

You're very close! Almost all of it is perfect!
Here is some more pythonic code, that works.
I've put comments explaining changes:
count = 0
sum = 0
# no longer need to say number = 1
while True: # no need to check for input number >= 0 here
number = input()
if number = '': # user just hit enter key, input left blank
print('hey')
break
if number != 0:
sum += int(number) # same as sum = sum + number
count += 1 # same as count = count + 1
# if number is 0, we don't do anything!
print(f'Average is {count/sum:.2f}') # same as '... {:.2f} ...'.format(count/sum)
Why your code didn't work:
When a user just presses enter instead of typing a number, the input() function doesn't return '\n', rather it returns ''.
I really hope this helps you learn!

Try this:
amount = 0 # Number of non-zero numbers input
nums = 0 # Sum of numbers input
while True:
number = input()
if not number: # Breaks out if nothing is entered
break
if int(number) != 0: # Only add to the variables if the number input is not 0
nums+=int(number)
amount += 1
print(round(nums/amount,2)) # Print out the average rounded to 2 digits
Input:
1
2
3
4
Output:
2.5
Or you can use numpy:
import numpy as np
n = []
while True:
number = input()
if not number: # Breaks out if nothing is entered
break
if int(number) != 0: # Only add to the variables if the number input is not 0
n.append(int(number))
print(round(np.average(n),2)) # Print out the average rounded to 2 digits

A list can store information of the values, number of values and the order of the values.
Try this:
numbers = []
while True:
num = input('Enter number:')
if num == '':
print('Average is', round(sum(numbers)/len(numbers), 2)) # print
numbers = [] # reset
if num != '0' and num != '': numbers.append(int(num)) # add to list
Benefit of this code, it does not break out and runs continuously.

Using multiples of 5 as a variable to use modulo more easily

I'm relatively new to Python, and coding in general (first-semester, intro computer science, I'm not a comp sci major.), and a homework assignment has me trying to figure out how many numbers between two numbers contain "5" (Such as between 105 and 168 there are 16 numbers that contain 5.) I started my code with:
def give_me_five(start, end):
ctr = start
max = end
numberoffives = 0
while ctr <= end:
print (ctr)
if ctr % 5 == 0 and (ctr % 10 != 0):
elif ctr % 5 == 0 and (ctr % 10 == (somethingsomethingmultiplesof5)
numberoffives += 1
if ctr // 10 == 5:
numberoffives += 1
ctr += 1
return numberoffives
The problem I'm encountering is that I can't use this to check for numbers in the 50s, 150s, etc... I tried using the ctr % 5 == 0 and (ctr % 10 == ) bit there to try and include them but now I can't figure out a way to proceed at all, and frankly I feel like I'm overcomplicating things. Does anyone have any advice for how to make my code actually work?

Since the digit 5 can be anywhere in a number it's easiest to convert each integer between the two numbers to a string instead so that you can use the in operator to check if 5 is a substring:
def give_me_five(start, end):
return sum('5' in str(i) for i in range(start, end + 1))
or if you prefer to do it with math, you can keep dividing a given number by 10 while checking if the remainder is 5 until the quotient becomes 0:
def give_me_five(start, end):
count = 0
for i in range(start, end + 1):
while i > 0:
if i % 10 == 5:
count += 1
break
i //= 10
return count
so that:
give_me_five(105, 168)
returns: 16

If I am understanding correctly, we want to do: for each number x in range, check to see if any of its digits digit is equal to 5.
How do we do this?
One thing we could do is cast the digit => string, and then check to see if the substring '5' exists. But this is a naive solution, requiring more memory and time to store all the digits into a string (rather than the few bytes necessary to represent the existing int).
So a smarter way would be to continually extract the least significant digit (digit = x % 10), check if digit == 5, and then divide out the least significant digit to repeat (x //= 10 WHILE x > 0).

How can I Keep the following program running until it becomes a single digit number?

I want to write a program that can calculate the sum of an integer as well as count its digits . It will keep doing this until it becomes a one digit number.
For example, if I input 453 then its sum will be 12 and digit 3.
Then it will calculate the sum of 12=1+2=3 it will keep doing this until it becomes one digit. I did the first part but i could not able to run it continuously using While . any help will be appreciated.
def main():
Sum = 0
m = 0
n = input("Please enter an interger: ")
numList = list(n)
count = len(numList)
for i in numList:
m = int(i)
Sum = m+Sum
print(Sum)
print(count)
main()

It is not the most efficient way, but it doesn't matter much here; to me, this is a problem to elegantly solve by recursion :)
def sum_digits(n):
n = str(n)
if int(n) < 10:
return n
else:
count = 0
for c in n:
count += int(c)
return sum_digits(count)
print sum_digits(123456789) # --> 9 # a string
A little harder to read:
def sum_digits2(n):
if n < 10:
return n
else:
return sum_digits2(sum(int(c) for c in str(n))) # this one returns an int

There are a couple of tricky things to watch out for. Hopefully this code gets you going in the right direction. You need to have a conditional for while on the number of digits remaining in your sum. The other thing is that you need to covert back and forth between strings and ints. I have fixed the while loop here, but the string <-> int problem remains. Good luck!
def main():
count = 9999
Sum = 0
m = 0
n = input("Please enter an integer: ")
numList = list(n)
while count > 1:
count = len(numList)
for i in numList:
m = int(i)
Sum = m+Sum
print(Sum)
print(count)
# The following needs to be filled in.
numlist = ???
main()

You can do this without repeated string parsing:
import math
x = 105 # or get from int(input(...))
count = 1 + int(math.log10(x))
while x >= 10:
sum = 0
for i in xrange(count):
sum += x % 10
x /= 10
x = sum
At the end, x will be a single-digit number as described, and count is the number of original digits.

I would like to give credit to this stackoverflow question for a succinct way to sum up digits of a number, and the answers above for giving you some insight to the solution.
Here is the code I wrote, with functions and all. Ideally you should be able to reuse functions, and here the function digit_sum(input_number) is being used over and over until the size of the return value (ie: length, if sum_of_digits is read as a string) is 1. Now you can use the while loop to keep checking till the size is what you want, and then abort.
def digit_sum(input_number):
return sum(int(digit) for digit in str(input_number))
input_number = input("Please enter a number: ")
sum_of_digits = digit_sum(input_number)
while(len(str(sum_of_digits)) > 1):
sum_of_digits = digit_sum(input_number)
output = 'Sum of digits of ' + str(input_number) + ' is ' + str(sum_of_digits)
print output
input_number = sum_of_digits

this is using recursive functions
def sumo(n):
sumof = 0
while n > 0:
r = n%10 #last digit
n = n/10 # quotient
sumof += r #add to sum
if sumof/10 == 0: # if no of digits in sum is only 1, then return
return sumof
elif sumof/10 > 0: #else call the function on the sumof
sumo(sumof)

Probably the first temptation would be to write
while x > 9:
x = sum(map(int, str(x)))
that literally means "until there is only one digit replace x by the sum of its digits".
From a performance point of view however one should note that computing the digits of a number is a complex operation because Python (and computers in general) store numbers in binary form and each digit in theory requires a modulo 10 operation to be extracted.
Thus if the input is not a string to begin with you can reduce the number of computations noting that if we're interested in the final sum (and not in the result of intermediate passes) it doesn't really matter the order in which the digits are summed, therefore one could compute the result directly, without converting the number to string first and at each "pass"
while x > 9:
x = x // 10 + x % 10
this costs, from a mathematical point of view, about the same of just converting a number to string.
Moreover instead of working out just one digit however one could also works in bigger chunks, still using maths and not doing the conversion to string, for example with
while x > 99999999:
x = x // 100000000 + x % 100000000
while x > 9999:
x = x // 10000 + x % 10000
while x > 99:
x = x // 100 + x % 100
while x > 9:
x = x // 10 + x % 10
The first loop works 8 digits at a time, the second 4 at a time, the third two and the last works one digit at a time. Also it could make sense to convert the intermediate levels to if instead of while because most often after processing n digits at a time the result will have n or less digits, leaving while loops only for first and last phases.
Note that however the computation at this point is so fast that Python general overhead becomes the most important part and thus not much more can be gained.

You could define a function to find the sum and keep updating the argument to be the most recent sum until you hit one digit.
def splitSum(num):
Sum = 0
for i in str(num):
m = int(i)
Sum = Sum + m
return str(Sum)
n = input("Please enter an integer: ")
count = 0
while count != 1:
Sum = splitSum(n)
count = len(Sum)
print(Sum)
print(count)
n = Sum

Python: display the number of one's in any user given integer number

How do I display the number of one's in any given integer number?
I am very new to python so keep this in mind.
I need the user to input a whole number.
Then I want it to spit out how many one's are in the number they input.
i am using python 3.3.4
How would I be able to do this with this following code?
num = int(input ("Input a number containing more than 2 digits"))
count = 0
for i in range (0):
if num(i) == "1":
count = count + 1
print (count)
I don't know what i'm doing wrong
it gives me 'int' object is not callable error

Something like this:
Int is not iterable so you may need to convert into string:
>>> num = 1231112
>>> str(num).count('1')
4
>>>

str(num).count('1') works just fine, but if you're just learning python and want to code your own program to do it, I'd use something like this, but you're on the right track with the code you supplied:
count = 0
for i in str(num):
if i == "1":
count = count + 1 # or count += 1
print(count)

Just to help you, here is a function that will print out each digit right to left.
def count_ones(num):
while num:
print(num % 10) # last digit
num //= 10 # get rid of last digit

num = 1112111
answer = str(num).count("1")

num = int(input (" Input a number to have the number of ones be counted "))
count = 0
for i in str(num):
if i == "1":
count = count + 1 # or count += 1
print (' The number of ones you have is ' + str(count))
So i took the user input and added it to the correct answer since when i tried the answer from crclayton it didn't work. So this works for me.