计算数字中的位数

Question

我写了一个名为

count_digit

的函数：

# Write a function which takes a number as an input
# It should count the number of digits in the number
# And check if the number is a 1 or 2-digit number then return True 
# Return False for any other case
def count_digit(num):
    if (num/10 == 0):
        return 1
    else:
        return 1 + count_digit(num / 10);


print(count_digit(23))

我得到

325

作为输出。这是为什么？我该如何纠正？

Answer 1

这是一种

Python3

行为。

/

返回

float

而不是

integer

除法。

将代码更改为：

def count_digit(num):
    if (num//10 == 0):
        return 1
    else:
        return 1 + count_digit(num // 10)

print(count_digit(23))

Answer 2

将整数转换为字符串，然后对转换后的字符串使用 len() 方法。除非您也考虑将浮点数作为输入，而不仅仅是整数。

Answer 3

假设您总是向函数发送整数并且您要求数学答案，这可能是您的解决方案：

import math


def count_digits(number):
    return int(math.log10(abs(number))) + 1 if number else 1


if __name__ == '__main__':
    print(count_digits(15712))
    # prints: 5

Answer 4

递归

def count_digit(n):
    if n == 0:
        return 0
    return count_digit(n // 10) + 1

简单

def count_digit(n):
    return len(str(abs(n)))

Answer 5

这是一个简单的Python解决方案；

num=23
temp=num #make a copy of integer
count=0

#This while loop will run unless temp is not zero.,so we need to do something to make this temp ==0

while(temp): 
  temp=temp//10  # "//10" this floor division by 10,remove last digit of the number.,means we are removing last digit and assigning back to the number;unless we make the number 0
  count+=1 #after removing last digit from the number;we are counting it;until the number becomes Zero and while loop becomes False
print(count)

Answer 6

Greg_Kan 建议：

len(str(num))