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How to print sum of digits for a Python integer? [duplicate]

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Sum the digits of a number
(11 answers)
Closed last year.
Consider the following input prompt. I want to output the sum of digits that are inputted.
Example:
two_digit_number = input("Type a two digit number: ") 39
# program should print 12 (3 + 9) on the next line
If I input 39, I want to output 12 (3 + 9). How do I do this?

You can use sum(), transforming each digit to an integer:
num = input("Enter a two digit number: ")
print(sum(int(digit) for digit in num))

maybe like this:
numbers = list(input("Enter number: "))
print(sum(list(map(int, numbers))))
Read digits with input as a whole number of string type
split them with list() in characters in a list
use map() to convert type to int in a list
with sum() you're done.. just print()
Beware of invalid entries! "hundred" or "1 234 567"

you can achieve that like this:
a = input("Type a two digit number: ")
b = list(a)
print(int(b[0])+int(b[1]))

n = (int(input("Enter the two digit number\n")))
n = str(n)
p = int(n[0])+int(n[1])
print(p)

This is one of the many ways for it.
value = 39 # it is your input
value_string = str(value)
total = 0
for digit in value_string:
total += int(digit)
print(total)
new version according to need mentioned in comment:
value = 39
value_string = str(value)
total = 0
digit_list = []
for digit in value_string:
total += int(digit)
digit_list.append(digit)
print(" + ".join(digit_list))
print(total)

You can do that as follows:
two_digit_number = input("Type a two digit number: ")
digits = two_digit_number.split("+")
print("sum = ", int(digits[0])+int(digits[1]))
Explanation:
first read the input (ex: 1+1)
then use split() to separate the input statement into the strings of the input digits only
then use int() to cast the digit from string to int

How do I calculate the sum of the individual digits of an integer and also print the original integer at the end?

I wish to create a Function in Python to calculate the sum of the individual digits of a given number
passed to it as a parameter.
My code is as follows:
number = int(input("Enter a number: "))
sum = 0
while(number > 0):
remainder = number % 10
sum = sum + remainder
number = number //10
print("The sum of the digits of the number ",number," is: ", sum)
This code works up until the last print command where I need to print the original number + the statement + the sum. The problem is that the original number changes to 0 every time (the sum is correct).
How do I calculate this but also show the original number in the print command?

Keep another variable to store the original number.
number = int(input("Enter a number: "))
original = number
# rest of the code here
Another approach to solve it:
You don't have to parse the number into int, treat it as a str as returned from input() function. Then iterate each character (digit) and add them.
number = input("Enter a number: ")
total = sum(int(d) for d in number)
print(total)

You can do it completely without a conversion to int:
ORD0 = ord('0')
number = input("Enter a number: ")
nsum = sum(ord(ch) - ORD0 for ch in number)
It will compute garbage, it someone enters not a number

number = input("Enter a number: ")
total = sum((int(x) for x in list(number)))
print("The sum of the digits of the number ", number," is: ", total)
As someone else pointed out, the conversion to int isn't even required since we only operate a character at a time.

Only the last item is being appended to a list in a while loop (Python)

I'm trying to create a program that gets each digit of an inputted number into a list using a while loop. However, it only appends the last digit of the number to the list.
Code -
num = int(input("Enter a number: "))
numstr = str(num)
numlen = len(numstr)
x = 0
while x < numlen:
digits = []
a = numstr[x]
digits.append(a)
x = x + 1
print(digits)
So if I were to put in 372 as the number, the list would just simply be ['2'] with a length of 1.

Try this code:
digits = [i for i in str(num)]

You cannot do better than digits = list(str(num)). In fact, since input returns a string, even the conversion to a number is not necessary:
num = input("Enter a number: ")
digits = list(num)
(You still may want to ensure that what's typed is indeed a number, not a random string.)

Python: How to use a Variable Integer to Call a Specific Part of a String

Ok, so I've got a couple issues with a program (for school again) that I'm using to add up all the digits of a number. I've got some of the program down, except 2 things. First, how to use a variable (thelength below) in replacement of a number to call a specific digit of the input (I'm not sure if this is even possible, but it would be helpful). And second, how to add up different numbers in a string. Any ideas?
Here's what I have so far:
number = str(int(input("Please type a number to add up: ")))
length = len(number)
thelength = 0
total = 0
thenumbers = []
while thelength < length:
#The issue is me trying to use thelength in the next two lines, and the fact that number is now a string
total += number[thelength]
thenumbers.append(number[thelength])
thelength += 1
for num in thenumbers:
print(num[0])
print("+")
print("___")
print(total)
Thanks for any help I can get!

I don't know what "call a specific digit of the input" means, but the error in your code is here:
total += number[thelength]
total is an int, and you're trying to add a string to it, convert the digit to an integer first.
total += int(number[thelength])
Result:
1
2
3
+
___
6

import re
import sys
INPUT_VALIDATOR = re.compile("^[0-9]+$")
input_str = input("Please type a natural number to add up: ")
if INPUT_VALIDATOR.match(input_str) is None:
print ("Your input was not a natural number (a positive whole number greater or equal to zero)!")
print ("This displeases me, goodbye puny human.")
sys.exit(1)
total = 0
for digit_str in input_str:
print(digit_str)
total += int(digit_str)
print("+")
print("___")
print(total)
If you don't need to print the digits as you go, it's even easier:
# (Add the same code as above to get and validate the input string)
print(sum(int(digit_str) for digit_str in input_str))

number = int(input("Please type a number to add up: "))
total = 0
while number > 0:
total += number % 10
total /= 10
print(total)
num % 10 pretty much gets the last digit of a number
then we divide it by 10 to truncate the number by its last digit
we can loop through the number as long as it's above 0 and take the digital sum by using the method outlined above

Every thing that you need is convert the digits to int and sum them :
>>> s='1247'
>>> sum(map(int,s))
14
But as you get the number from input it could cause a ValueErorr , for refuse that you can use a try-except :
try :
print sum(map(int,s))
except ValueErorr :
print 'please write a valin number :'
Also if you are using python 2 use raw_input for get the number or if you are using python 3 just use input because the result of both is string !

ISBN final digit finder

I am working in python 3 and I am making a program that will take in a 10 digit ISBN Number and applying a method to it to find the 11th number.
Here is my current code
ISBN=input('Please enter the 10 digit number: ')
while len(ISBN)!= 10:
print('Please make sure you have entered a number which is exactly 10 characters long.')
ISBN=int(input('Please enter the 10 digit number: '))
continue
else:
Digit1=int(ISBN[0])*11
Digit2=int(ISBN[1])*10
Digit3=int(ISBN[2])*9
Digit4=int(ISBN[3])*8
Digit5=int(ISBN[4])*7
Digit6=int(ISBN[5])*6
Digit7=int(ISBN[6])*5
Digit8=int(ISBN[7])*4
Digit9=int(ISBN[8])*3
Digit10=int(ISBN[9])*2
Sum=(Digit1+Digit2+Digit3+Digit4+Digit5+Digit6+Digit7+Digit8+Digit9+Digit10)
Mod=Sum%11
Digit11=11-Mod
if Digit11==10:
Digit11='X'
ISBNNumber=str(ISBN)+str(Digit11)
print('Your 11 digit ISBN Number is ' + ISBNNumber)
I want to create some kind of loop so that the number after "Digit" for the variable name increases starting from 1 (or zero if it makes life easier), the number in the square brackets increases starting from 0 and the multiplication number to decrease from 11 to 2.
Is there any way of doing this code in a more efficient way?

I think this should do what you want.
def get_isbn_number(isbn):
digits = [(11 - i) * num for i, num in enumerate(map(int, list(isbn)))]
digit_11 = 11 - (sum(digits) % 11)
if digit_11 == 10:
digit_11 = 'X'
digits.append(digit_11)
isbn_number = "".join(map(str, digits))
return isbn_number
EXAMPLE
>>> print(get_isbn_number('2345432681'))
22303640281810242428
>>> print(get_isbn_number('2345432680'))
2230364028181024240X
Explanation of second line:
digits = [(11 - i) * num for i, num in enumerate(map(int, list(isbn)))]
Could be written out like:
isbn_letters = list(isbn) # turn a string into a list of characters
isbn_numbers = map(int, isbn_letters) # run the function int() on each of the items in the list
digits = [] # empty list to hold the digits
for i, num in enumerate(isbn_numbers): # loop over the numbers - i is a 0 based counter you get for free when using enumerate
digits.append((11 - i) * num) # If you notice the pattern, if you subtract the counter value (starting at 0) from 11 then you get your desired multiplier
Terms you should look up to understand the one line version of the code:
map,
enumerate,
list conprehension

ISBN=int(input('Please enter the 10 digit number: ')) # Ensuring ISBN is an integer
while len(ISBN)!= 10:
print('Please make sure you have entered a number which is exactly 10 characters long.')
ISBN=int(input('Please enter the 10 digit number: '))
continue
else:
Sum = 0
for i in range(len(ISBN)):
Sum += ISBN[i]
Mod=Sum%11
Digit11=11-Mod
if Digit11==10:
Digit11='X'
ISBNNumber=str(ISBN)+str(Digit11)
print('Your 11 digit ISBN Number is ' + ISBNNumber)