对于子孙后代来说，这无疑是迄今为止解决这个问题最慢的方法:

def num_digits(num, number_of_calls=1):
    "Returns the number of digits of an integer num."
    if num == 0 or num == -1:
        return 1 if number_of_calls == 1 else 0
    else:
        return 1 + num_digits(num/10, number_of_calls+1)

这里是最简单的方法，不需要将int转换为字符串:

假设给出的数字为15位，例如;n = 787878899999999;

n=787878899999999 
n=abs(n) // we are finding absolute value because if the number is negative int to string conversion will produce wrong output

count=0 //we have taken a counter variable which will increment itself till the last digit

while(n):
    n=n//10   /*Here we are removing the last digit of a number...it will remove until 0 digits will left...and we know that while(0) is False*/
    count+=1  /*this counter variable simply increase its value by 1 after deleting a digit from the original number
print(count)   /*when the while loop will become False because n=0, we will simply print the value of counter variable

输入:

n=787878899999999

输出:

15

我的代码相同如下，我已经使用了log10方法:

from math import *

def digit_count(数量):

if number>1 and round(log10(number))>=log10(number) and number%10!=0 :
    return round(log10(number))
elif  number>1 and round(log10(number))<log10(number) and number%10!=0:
    return round(log10(number))+1
elif number%10==0 and number!=0:
    return int(log10(number)+1)
elif number==1 or number==0:
    return 1

我必须在1和0的情况下指定，因为log10(1)=0和log10(0)=ND，因此上面提到的条件不满足。但是，此代码仅适用于整数。

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。

假设您要求的是可以存储在整数中的最大数字，则该值与实现有关。我建议你在使用python时不要这样想。在任何情况下，相当大的值都可以存储在python 'integer'中。记住，Python使用鸭子类型!

编辑: 我在澄清提问者想要数字数之前给出了我的答案。就此而言，我同意公认答案所建议的方法。没什么可补充的了!