Python: 从循环中创建两个列表并对两者求和

Question

我正在做一个CS50的信用项目，我在信用卡验证方面遇到了一些问题。

这里是我创建的函数：

def main():

while True :
    cardnumber = input("Please enter a credit card number: ")
    if cardnumber.isdecimal() and int(cardnumber) > 0 :
        break

count = len(cardnumber)

if count != 13 and count != 15 and count != 16:
    print("INVALID")
else:
    check(count, cardnumber)


def check(length, number):

lenght_max = 15

if length == 15 and int(number[0]) == 3 and (int(number[1]) == 4 or int(number[1]) == 7):
    if validator(number):
        print("AMEX")
elif length == 16 and int(number[0]) == 5 and int(number[1]) <= 5:
    if validator(number):
        print("MASTERCARD")
elif length == 16 or length == 13 and int(number[0]) == 4:
    if validator(number):
        print("VISA")
else:
    print("INVALID")
return number


def validator(num):

sum = 0
while num > 0:
    sum += num % 10
    num = num // 10

return sum
odd = [int(num[i]) * 2 for i in range(1, len(num), 2)]
even = [int(num[i]) for i in range(0, len(num), 2)]

new_sum = sum(validator(x) for x in odd) + sum(even)
if(new_sum % 10 == 0):
    return True
else:
    print("INVALID")

main()

我找到了打印偶数和赔率的方法（也乘以时间 2），但现在我必须对 booth 求和并检查余数是否为 0

这里是完整的说明： http://docs.cs50.net/problems/credit/credit.html

Answer 1

编写一个辅助函数来对您的数字求和。您需要广泛使用它。

def dig_sum(num):
   sum = 0
   while num > 0:
      sum += num % 10
      num = num // 10

   return sum

num = '378282246310005' # your credit card number
odd = [int(num[i]) * 2 for i in range(1, len(num), 2)] # these two remain the same
even = [int(num[i]) for i in range(0, len(num), 2)]

new_sum = sum(dig_sum(x) for x in odd) + sum(even)
if(new_sum % 10 == 0):
    print('Valid') #valid!

sum(dig_sum(x) for x in odd) 将获得您的 odd 列表中每个数字的数字总和，sum(...) 将找到结果总和。

输入：

'378282246310005'

输出：

Valid

Answer 2

您的函数的第一个问题是您没有将 even/odd 数字存储在某处：您每次构造一个包含一个元素的列表，然后打印该元素.

现在由于一个数字乘以两次，只能得到一个两位数，我们可以用：

def sum2(x):
    return (2*x)//10 + (2*x)%10

您可以构造一个 list all 奇数索引处的数字：

odd = [int(number[i]) for i in range(1,length,2)]

偶数索引的数字相同：

even = [int(number[i]) for i in range(0,length,2)]

现在我们可以简单地使用sum(..)内置函数来求和数字：

total = sum(sum2(oddi) for oddi in odd) + sum(even)

并检查它是否是 10 的倍数：

return total%10 == 0

或者把它们放在一起：

def validator(number, length):
    odd = [int(number[i]) for i in range(1,length,2)]
    even = [int(number[i]) for i in range(0,length,2)]
    total = sum(sum2(oddi) for oddi in odd) + sum(even)
    return total%10 == 0

或者我们可以使用下面的一行专家：

from itertools import zip_longest

def validator(number,length):
    numbi = iter(numbi)
    return sum(x+sum2(y) for x,y in zip_longest(numbi,numbi,fillvalue=0))%10 == 0

Python: 从循环中创建两个列表并对两者求和

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