计算w/o将整数转换为字符串的位数:

x=123
x=abs(x)
i = 0
while x >= 10**i:
    i +=1
# i is the number of digits

如果你想要一个整数的长度等于这个整数的位数，你总是可以把它转换成字符串，比如str(133)，然后像len(str(123))一样找到它的长度。

n = 3566002020360505
count = 0
while(n>0):
  count += 1
  n = n //10
print(f"The number of digits in the number are: {count}")

output: number中的位数为:16

正如亲爱的用户@Calvintwr提到的，函数数学。Log10在一个超出范围[-999999999999997,99999999999999997]的数字中有问题，我们会得到浮点数错误。我有这个问题与JavaScript(谷歌V8和NodeJS)和C (GNU GCC编译器)，所以一个“纯数学”的解决方案是不可能在这里。

基于这个要点和答案，亲爱的用户@Calvintwr

import math


def get_count_digits(number: int):
    """Return number of digits in a number."""

    if number == 0:
        return 1

    number = abs(number)

    if number <= 999999999999997:
        return math.floor(math.log10(number)) + 1

    count = 0
    while number:
        count += 1
        number //= 10
    return count

我在长度不超过20(包括20)的数字上进行了测试，没问题。它必须足够，因为64位系统上的最大整数长度是19 (len(str(sys.maxsize)) == 19)。

assert get_count_digits(-99999999999999999999) == 20
assert get_count_digits(-10000000000000000000) == 20
assert get_count_digits(-9999999999999999999) == 19
assert get_count_digits(-1000000000000000000) == 19
assert get_count_digits(-999999999999999999) == 18
assert get_count_digits(-100000000000000000) == 18
assert get_count_digits(-99999999999999999) == 17
assert get_count_digits(-10000000000000000) == 17
assert get_count_digits(-9999999999999999) == 16
assert get_count_digits(-1000000000000000) == 16
assert get_count_digits(-999999999999999) == 15
assert get_count_digits(-100000000000000) == 15
assert get_count_digits(-99999999999999) == 14
assert get_count_digits(-10000000000000) == 14
assert get_count_digits(-9999999999999) == 13
assert get_count_digits(-1000000000000) == 13
assert get_count_digits(-999999999999) == 12
assert get_count_digits(-100000000000) == 12
assert get_count_digits(-99999999999) == 11
assert get_count_digits(-10000000000) == 11
assert get_count_digits(-9999999999) == 10
assert get_count_digits(-1000000000) == 10
assert get_count_digits(-999999999) == 9
assert get_count_digits(-100000000) == 9
assert get_count_digits(-99999999) == 8
assert get_count_digits(-10000000) == 8
assert get_count_digits(-9999999) == 7
assert get_count_digits(-1000000) == 7
assert get_count_digits(-999999) == 6
assert get_count_digits(-100000) == 6
assert get_count_digits(-99999) == 5
assert get_count_digits(-10000) == 5
assert get_count_digits(-9999) == 4
assert get_count_digits(-1000) == 4
assert get_count_digits(-999) == 3
assert get_count_digits(-100) == 3
assert get_count_digits(-99) == 2
assert get_count_digits(-10) == 2
assert get_count_digits(-9) == 1
assert get_count_digits(-1) == 1
assert get_count_digits(0) == 1
assert get_count_digits(1) == 1
assert get_count_digits(9) == 1
assert get_count_digits(10) == 2
assert get_count_digits(99) == 2
assert get_count_digits(100) == 3
assert get_count_digits(999) == 3
assert get_count_digits(1000) == 4
assert get_count_digits(9999) == 4
assert get_count_digits(10000) == 5
assert get_count_digits(99999) == 5
assert get_count_digits(100000) == 6
assert get_count_digits(999999) == 6
assert get_count_digits(1000000) == 7
assert get_count_digits(9999999) == 7
assert get_count_digits(10000000) == 8
assert get_count_digits(99999999) == 8
assert get_count_digits(100000000) == 9
assert get_count_digits(999999999) == 9
assert get_count_digits(1000000000) == 10
assert get_count_digits(9999999999) == 10
assert get_count_digits(10000000000) == 11
assert get_count_digits(99999999999) == 11
assert get_count_digits(100000000000) == 12
assert get_count_digits(999999999999) == 12
assert get_count_digits(1000000000000) == 13
assert get_count_digits(9999999999999) == 13
assert get_count_digits(10000000000000) == 14
assert get_count_digits(99999999999999) == 14
assert get_count_digits(100000000000000) == 15
assert get_count_digits(999999999999999) == 15
assert get_count_digits(1000000000000000) == 16
assert get_count_digits(9999999999999999) == 16
assert get_count_digits(10000000000000000) == 17
assert get_count_digits(99999999999999999) == 17
assert get_count_digits(100000000000000000) == 18
assert get_count_digits(999999999999999999) == 18
assert get_count_digits(1000000000000000000) == 19
assert get_count_digits(9999999999999999999) == 19
assert get_count_digits(10000000000000000000) == 20
assert get_count_digits(99999999999999999999) == 20

所有使用Python 3.5测试的代码示例

如果您正在寻找一个不使用内置函数的解决方案。 唯一需要注意的是当你发送a = 000时。

def number_length(a: int) -> int:
    length = 0
    if a == 0:
        return length + 1
    else:
        while a > 0:
            a = a // 10
            length += 1
        return length
    

if __name__ == '__main__':
    print(number_length(123)
    assert number_length(10) == 2
    assert number_length(0) == 1
    assert number_length(256) == 3
    assert number_length(4444) == 4

设数字为n，则n中的位数为:

math.floor(math.log10(n))+1

注意，这将为+ve个整数< 10e15给出正确答案。除此之外，返回类型的数学的精度限制。Log10开始起作用，结果可能相差1。我可以简单地在后面用len(str(n));这需要O(log(n))时间，相当于10的幂次迭代。

感谢@SetiVolkylany让我注意到这个限制。令人惊讶的是，看似正确的解决方案在实现细节中有警告。