创建一个从一号到第n号Python的数字列表

Question

How can I create a list one through nth that gives me a list, that is only one digit in length. So 100 would be 1,0,0 and 19 would be 1,9

>>> listGen(20)


>>> [1,2,3,4,5,6,7,8,9,1,0,1,1,1,2,1,3,1,4,1,5,1,6,1,7,1,8,1,9,2,0]

Answer 1

def listgen(n):
    return map(int, ''.join(map(str, range(1, n + 1))))

Breaking it apart:

n = 10
a = range(1, n + 1)   # [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
b = map(str, a)       # ['1', '2', '3', '4', '5', '6', '7', '8', '9', '10']
c = ''.join(b)        # '12345678910'
d = map(int, c)       # [1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 0]

The above code is written for python 2. For python 3 you could convert the map to list .

Answer 2

Or, without map()...

>>> [int(c) for c in ''.join([str(n) for n in range(1,21)])]

[1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 0, 1, 1, 1, 2, 1, 3, 1, 4, 1, 5, 1, 6, 1, 7, 1, 8, 1, 9, 2, 0]