I was trying to implement the Luhn Formula in Python. Here is my code:
import sys
def luhn_check(number):
if number.isdigit():
last_digit = int(
I'd keep it simple and easy to read, something like this:
def luhn(value):
digits = list(map(int,str(value))
oddSum = sum(digits[-1::-2])
evnSum = sum([sum(divmod(2 * d, 10)) for d in digits[-2::-2]])
return (oddSum + evnSum) % 10 == 0
But there's tons of ways to do the same thing. Obviously, you'd have to do it differently to see the actual output, this just sums up the total to determine if the value is valid.
Best!
There are some errors in your code:
result = divmod(sum_of_digits, 10)
returns a tuple, you need only modulo, that is use
result = sum_of_digits % 10
Second, to check for validity, you don't omit last digit (that is checksum), but include it in computations. Use
reverse_sequence = list(int(d) for d in str(int(number[::-1])))
And check for result being zero:
if not result:
print("[VALID] %s" % number)
Or if you insist on keeping this not needed complexity, check for last digit to be inverse of checksum modulo 10: keep
reverse_sequence = list(int(d) for d in str(int(number[-2::-1])))
but use
if (result + last_digit) % 10 == 0:
print("[VALID] %s" % number)
For a simplier and shorter code, I can give you a reference to my older answer.
check_numbers = ['49927398716', '4847352989263095', '79927398713', '5543352315777720']
def Luhn_Check(number):
"""
this function checks a number by using the Luhn algorithm.
Notes (aka - How to Luhn) :
Luhn algorithm works in a 1 2 1 2 ... order.
Therefore, in computer speak, 1 0 1 0 ... order
step 1:
-> reverse the # so we are not working from right to left (unless you want too)
-> double every second number
step 2:
-> if the doubled number is greater then 9, add the individual digits of the number to get a single digit number
(eg. 12, 1 + 2 = 3)
step 3:
-> sum all the digits, if the total modulo is equal to 0 (or ends in zero) then the number is valid...
"""
reverse_numbers = [int(x) for x in number[::-1]] # convert args to int, reverse the numbers, put into a list
dbl_digits = list() # create empty list
digits = list(enumerate(reverse_numbers, start=1)) # enumerate numbers starting with an index of 1
for index, digit in digits:
if index % 2 == 0: # double every second (other) digit.
doub_digit = digit * 2
dbl_digits.append(doub_digit - 9) if doub_digit > 9 else dbl_digits.append(doub_digit)
else:
# if not '0' append to list (this would be the 1 in Luhn algo sequence (1 2 1 2 ...)
dbl_digits.append(digit)
return sum(dbl_digits) % 10
if (__name__ == "__main__"):
print("Valid Numbers: %s " % [x for x in check_numbers if Luhn_Check(x) == 0])
This is the most concise python formula for the Luhn test I have found:
def luhn(n):
r = [int(ch) for ch in str(n)][::-1]
return (sum(r[0::2]) + sum(sum(divmod(d*2,10)) for d in r[1::2])) % 10 == 0
The above function and other Luhn implementations (in different programming languages) are available in https://www.rosettacode.org/wiki/Luhn_test_of_credit_card_numbers .
see this Python recipe
def cardLuhnChecksumIsValid(card_number):
""" checks to make sure that the card passes a luhn mod-10 checksum """
sum = 0
num_digits = len(card_number)
oddeven = num_digits & 1
for count in range(0, num_digits):
digit = int(card_number[count])
if not (( count & 1 ) ^ oddeven ):
digit = digit * 2
if digit > 9:
digit = digit - 9
sum = sum + digit
return ( (sum % 10) == 0 )
The following might help some people to start with the Luhn algorithm in python.
num = list(input("Please enter the number to test (no space, no symbols, only \
numbers): "))
num = list(map(int, num))[::-1] #let's transform string into int and reverse it
for index in range(1,len(num),2):
if num[index]<5:
num[index] = num[index] *2
else: #doubling number>=5 will give a 2 digit number
num[index] = ((num[index]*2)//10) + ((num[index]*2)%10)
checksum=sum(num)
print("checksum= {}".format(checksum))
if checksum%10 !=0:
print('the number is not valid')
else:
print('the number is valid!')