根据位数将 integer 向上舍入到下一个 99

Question

How do I use math.ceil() to round up a number to the maximum of it's base. For example, if the number is between

0 - 9 -> 9
10 - 99 -> 99
100 - 999 -> 999
1000 - 9999 -> 9999

so forth. I've done it by counting the number of digits but I'm looking for a pythonic way

Answer 1

Here is one way to do this using the ceiling along with logarithms:

def round_by_tens(inp):
    z = math.ceil(math.log10(inp+1))
    q = (10 ** z) - 1
    return q

nums = [5, 50, 500, 5000]
for num in nums:
    print(round_by_tens(num))

This prints:

9.0
99.0
999.0
9999.0

The logic here is to first compute the ceiling power of 10 which would be required to generate the tens factor which is the upper bound of the input. Then, we simply take that tens upper bound, and subtract one from it to generate the output you expect.

Answer 2

You can use the following also:

def repeat_to_length(string_to_expand, length):
    return int(string_to_expand * length)


num = 0
print(repeat_to_length("9", len(str(num))))
# 9
num = 45
print(repeat_to_length("9", len(str(num))))
# 99
num = 123
print(repeat_to_length("9", len(str(num))))
# 999

Answer 3

I would probably use this:

def round9(x):
    r = 10
    while x >= 10: 
        r *= 10
        x //= 10
    return r-1