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)我已經(jīng)嘗試了很多方法以“專業(yè)方式”編寫這段代碼，但這是我能讓它工作的唯一方法
2 回答
慕雪6442864
TA貢獻1812條經(jīng)驗 獲得超5個贊
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)
陪伴而非守候
TA貢獻1757條經(jīng)驗 獲得超8個贊
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沒有什么特別的錯誤。當(dāng)然，這里有一個你可以利用的模式：
for i in range(1, len(num), 2):
num[i] *= 2
但是，如果位置是任意的，“專業(yè)”的方法不是在函數(shù)中使用幻數(shù)，而是像這樣：
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)))
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