python - 卢恩公式的实现

Question

我试图在 Python中实现Luhn 公式。这是我的代码：

import sys


def luhn_check(number):
    if number.isdigit():
        last_digit = int(str(number)[-1])
        reverse_sequence = list(int(d) for d in str(int(number[-2::-1])))

        for i in range(0, len(reverse_sequence), 2):
            reverse_sequence[i] *= 2

        for i in range(len(reverse_sequence)):
            if reverse_sequence[i] > 9:
                reverse_sequence[i] -= 9

        sum_of_digits = 0
        for i in range(len(reverse_sequence)):
            sum_of_digits += reverse_sequence[i]

        result = divmod(sum_of_digits, 10)

        if result == last_digit:
            print("[VALID] %s" % number)
        else:
            print("[INVALID] %s" % number)
        quit()

    print("[ERROR] \" %s \" is not a valid sequence." % number)
    quit()


def main():
    if len(sys.argv) < 2:
        print("Usage: python TLF.py <sequence>")
        quit()

    luhn_check(sys.argv[1])

if __name__ == '__main__':
    main()

但它不能正常工作：

[INVALID] 4532015112830366
[INVALID] 6011514433546201
[INVALID] 6771549495586802

等等...

但是代码的逻辑对我来说似乎没问题。我遵循了这个工作流程：

卢恩公式

删除数字的最后一位数字。最后一位数字是我们要检查的反转数字
将奇数位（1、3、5 等）的数字乘以 2，然后将所有大于 9 的结果减去 9
将所有数字加在一起
校验位（卡的最后一个数字）是您需要添加以获得 10 的倍数（模 10）的金额

score 13 · Accepted Answer

我认为算法不正确。

第二步，您需要对产品的数字求和，而不是减去 9。参考：维基百科。

在维基百科中有这个例子：

def luhn_checksum(card_number):
    def digits_of(n):
        return [int(d) for d in str(n)]
    digits = digits_of(card_number)
    odd_digits = digits[-1::-2]
    even_digits = digits[-2::-2]
    checksum = 0
    checksum += sum(odd_digits)
    for d in even_digits:
        checksum += sum(digits_of(d*2))
    return checksum % 10

def is_luhn_valid(card_number):
    return luhn_checksum(card_number) == 0


result = is_luhn_valid(4532015112830366)
print 'Correct:' + str(result)
result = is_luhn_valid(6011514433546201)
print 'Correct:' + str(result)
result = is_luhn_valid(6771549495586802)
print 'Correct:' + str(result)

结果：

>>>Correct:True
>>>Correct:True
>>>Correct:True

score 4 · Accepted Answer

这是我发现的 Luhn 测试最简洁的 python 公式：

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

上述函数和其他 Luhn 实现（使用不同的编程语言）可在https://www.rosettacode.org/wiki/Luhn_test_of_credit_card_numbers中找到。

score 3 · Accepted Answer

我会保持简单易读，如下所示：

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

但是有很多方法可以做同样的事情。显然，您必须以不同的方式查看实际输出，这只是将总数相加以确定该值是否有效。

最好的！

score 2 · Accepted Answer

看到这个Python 食谱

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 )

score 2 · Accepted Answer

您的代码中有一些错误：

result = divmod(sum_of_digits, 10)

返回一个元组，你只需要取模，即使用

result = sum_of_digits % 10

其次，要检查有效性，您不要省略最后一位数字（即校验和），而是将其包含在计算中。利用

reverse_sequence = list(int(d) for d in str(int(number[::-1])))

并检查结果是否为零：

if not result:
    print("[VALID] %s" % number)

或者，如果您坚持保持这种不需要的复杂性，请检查最后一位是否与校验和模 10取反：保持

reverse_sequence = list(int(d) for d in str(int(number[-2::-1])))

但使用

if (result + last_digit) % 10 == 0:
    print("[VALID] %s" % number)

对于更简单和更短的代码，我可以给你一个参考我的旧答案。

score 2 · Accepted Answer

以下内容可能会帮助一些人从 python 中的 Luhn 算法入手。

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!')

score 1 · Accepted Answer

在一行代码中实现 Luhn 公式（如果没有换行符）

输入参数必须是字符串

def is_card_valid(card_number: str) -> bool:
    return (sum(
        (element + (index % 2 == 0) * (element - 9 * (element > 4))
         for index, element in enumerate(map(int, card_number[:-1])))
    ) + int(card_number[-1])) % 10 == 0

有地图功能

def is_card_valid(card_number: str) -> bool:
    return (sum(
        map(lambda n: n[1] + (n[0] % 2 == 0) * (n[1] - 9 * (n[1] > 4)),
            enumerate(map(int, card_number[:-1])))
    ) + int(card_number[-1])) % 10 == 0

score 0 · Accepted 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])

score 0 · Accepted Answer

您的逻辑的主要问题是您正在应用一些仅用于奇数到偶数的计算。

我有一些建议给你：

不是接收一个 int 作为输入，而是接收一个字符串，这样您就可以处理带有尾随零的数字。
尝试应用列表推导式或map在处理列表时使用
您可以定义一些辅助函数来提高代码的可修改性。

考虑到我的建议，这就是我解决这个问题的方法。请注意，我使用的是 Python 的类型注释。我还将用注释突出显示 Luhn 算法的步骤。

def compute_sum_of_remainig_digits(remaining_digits: List[int]) -> int:
    # use a list comprehension to extract digits in even positions
    even_positions = remaining_digits[-2::-2]
    # use a list comprehension to extract digits in odd positions
    odd_positions = remaining_digits[-1::-2]
    # 2. Multiply the digits in odd positions (1, 3, 5, etc.) by 2
    # and subtract 9 to all any result higher than 9
    computed_odd = [2 * d - 9 if 2 * d > 9 else 2 * d for d in odd_positions]
    # 3. Add all the numbers together
    return sum(even_positions + computed_odd)


def compute_checksum(number: str) -> int:
    # the following line convert a string of digits into list of digits using map
    # this way you do not need to explicitly declare a for loop
    digits = list(map(int, number))
    # 1. Drop the last digit from the number...
    last_digit = digits[-1]
    total = compute_sum_of_remainig_digits(digits[:-1 ])

    # 4. The check digit (the last number of the card) is
    # the amount that you would need to add...
    return (total + last_digit)


def check_valid_number(number: str) -> bool:
    # 4. ...to get a multiple of 10 (Modulo 10)
    return (compute_checksum(number) * 9) % 10 == 0

现在让我们验证前面的代码：

>>> valid_number = "4929355681587216"
>>> invalid_number = "5183814022326227"
>>> check_valid_number(valid_number)
True
>>> check_valid_number(invalid_number)
False

我希望这个答案可以帮助您或帮助其他有麻烦的人了解如何计算 Luhn's checker。

python - 卢恩公式的实现

9 回答 9

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