Python：获取最大的 n 位数

Question

Very stupid question but how can I quickly get the largest n-digit number with python? It's clear that number will always be n-occurrences of 9. So n=1 -> 9, n=2 -> 99 etc.

What I have is this:

max_str = ''
# there should be a better way to do this, right?
for i in range(self.number_length):
    max_str += '9'
max_value = int(max_str)

This works. But as the comment says, there must be a better way, right?

EDIT:

I made a quick and dirty performance comparison of the different options:

%%timeit
10**5 - 1

8.97 ns ± 0.0952 ns per loop (mean ± std. dev. of 7 runs, 100000000 loops each)

%%timeit
pow(10,5)-1

305 ns ± 1.13 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

%%timeit
int("9" * 5)

137 ns ± 0.321 ns per loop (mean ± std. dev. of 7 runs, 10000000 loops each)

%%timeit
max_str = ''
for i in range(5):
    max_str += '9'

401 ns ± 1.63 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

10**n - 1 for some reason is by far the fastest solution while pow(10,n) -1 isn't much faster than my loop approach.

Answer 1

All you need to do is:

10**n - 1

where n is the desired number of digits.

Answer 2

n_largest_number=pow(10,n)-1

Explanation: pow(a,b) function outputs a^b.