Question

所有

我想要定义一个 int（987654321）＆lt; =＆gt; [9,8,7,6,5,4,3,2,1] 转换器，如果int数的长度＆lt; 9 ，例如 10 该列表将 [0,0,0,0,0,0,0,1,0] ，如果长度> 9，例如9987654321，列表将是[9,9,8,7,6,5,4,3,2,1]

>>> i
987654321
>>> l
[9, 8, 7, 6, 5, 4, 3, 2, 1]
>>> z = [0]*(len(unit) - len(str(l)))
>>> z.extend(l)
>>> l = z
>>> unit
[100000000, 10000000, 1000000, 100000, 10000, 1000, 100, 10, 1]

>>> sum([x*y for x,y in zip(l, unit)])
987654321
>>> int("".join([str(x) for x in l]))
987654321


>>> l1 = [int(x) for x in str(i)]
>>> z = [0]*(len(unit) - len(str(l1)))
>>> z.extend(l1)
>>> l1 = z
>>> l1
[9, 8, 7, 6, 5, 4, 3, 2, 1]

>>> a = [i//x for x in unit]
>>> b = [a[x] - a[x-1]*10 for x in range(9)]
>>> if len(b) = len(a): b[0] = a[0]  # fix the a[-1] issue
>>> b 
[9, 8, 7, 6, 5, 4, 3, 2, 1]

我测试了以上解决方案，但发现那些可能不比我想要的更快/更简单，并且内部可能有长度相关的错误，任何人都可以与我分享这种转换的更好的解决方案？

谢谢！

Answer 1

也许我错过了一些东西，但这不应该足够（没有价值检查）？

def int_to_list(i):
    return [int(x) for x in str(i).zfill(9)]

def list_to_int(l):
    return int("".join(str(x) for x in l))

参考： str.zfill

Answer 2

那怎么样：

def int_to_list(num)
    return list ("%010d" % num)

Answer 3

def convert(number):
    stringified_number = '%s' % number
    if len(stringified_number) < 9:
        stringified_number = stringified_number.zfill(9)
    return [int(c) for c in stringified_number]

>>> convert(10)
[0, 0, 0, 0, 0, 0, 0, 1, 0]

>>> convert(987654321)
[9, 8, 7, 6, 5, 4, 3, 2, 1]

Answer 4

按整数位 -

将任意长度的整数按顺序放入列表中

a = 123456789123456789123456789123456789123456789123456789
j = len('{}'.format(a))
b = [0 for i in range(j)]
c = 0
while j > 0:
    b [c] = a % 10**j // 10**(j-1)
    j = j-1
    c = c + 1
print(b)

输出 -

[1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5， 6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,1,2,3， 4,5,6,7,8,9]

您可以将条件放在j上，以便将替代分配放到b。

将int转换为单个数字列表更快？

4 个答案: