如何更改列表中特定位置的数字

Question

def id(id):    num = [int(x) for x in str(id)]    num[1] = num[1]*2    num[3] = num[3]*2    num[5] = num[5]*2    num[7] = num[7]*2    print(num)x = id(123456789)我已经尝试了很多方法以“专业方式”编写这段代码，但这是我能让它工作的唯一方法

Answer 1

def multiply_even_indexes(number):

    # Going for each digit, and multiply by 2 if it's index is even

    # int(d)*2**(i % 2) means that we:

    # 1. Convert x to number

    # 2. Multiply x with 2 in the power of either 0 or 1 (depends if `i` is even)

    #    That means:

    #    For i = 1: We get i%2==1 (reminder of 1) so it's multiply by 2^1=2 (so we multiply by two the second element)

    #    For i = 2: We get i%2==0 (mo reminder) so it's multiply by 2^0=1 (so we don't change the third element)

    digits_result = [int(d)*2**(i % 2) for i, d in enumerate(str(number))]

    return digits_result 

x = multiply_even_indexes(123456789)

# [1, 4, 3, 8, 5, 12, 7, 16, 9]

print(x)

Answer 2

如果您只是想更改某些特定位置的值，那么您做事的方式并没有什么特别的错误。当然，这里有一个你可以利用的模式：

for i in range(1, len(num), 2):

    num[i] *= 2

但是，如果位置是任意的，“专业”的方法不是在函数中使用幻数，而是像这样：

POSITIONS = [1, 4, 5, 7]

def make_id(input_id): # don't shadow built-in names

    num = map(int, str(input_id))

    for i in POSITIONS:

        num[i] *= 2

    # return some value, don't just print it

    return int(''.join(map(str, num)))