How do I print the digits of a number from left to right in Python without using string index or lis-CodePudding

num = int(input("Enter a number: "))

while num > 0:
    num2 = num
    dig = 0

    while num2 > 0:
        num2 = num2 //10
        dig = dig   1

    x = num // 10**(dig-1)
    if dig == 1:
        print (x, end = " ")
    else:
        print (x, end = ", ")
    num = num % 10**(dig-1)

I wrote this code which outputs the number from left to right unless there is a zero at the middle or at the last of the number

For example: If input = 12345; output = 1, 2, 3, 4, 5

However, if input = 120 OR 102; output = 1, 2, OR 1, 2

Is there any workaround for this? I have been told not to use string indexing or lists to solve this.

CodePudding user response：

We can recursively iterate from right to left and then return the value we saw:

def get_digits(a: int):
    if a == 0:
        return ""
    if a > 9:
        return get_digits(a//10)   f", {a}" 
    return f"{a}"


    
print("100:", get_digits(100))
print("12345:", get_digits(12345))
print("120:", get_digits(120))
print("102:", get_digits(102))

The output will be:

100: 1, 0, 0
12345: 1, 2, 3, 4, 5
120: 1, 2, 0
102: 1, 0, 2

CodePudding user response：

Assuming strings are not strictly forbidden, here are some possible alternative solutions.

def get_digits(num):
    return [int(i) for i in str(num)]

Another solution [1]:

import math

def get_digits(num):
    return [(num//(10**i)) for i in range(math.ceil(math.log(num, 10))-1, -1, -1)]

This would be optimal if you just needed to print the digits directly.

print(", ".join([i for i in str(num)]))

If you're in a beginning Python class, your professor likely only wants to see something like this which doesn't use lists or string indexes:

num = 1203

for i, n in enumerate(str(num)):
    if i == 0:
        print(n, end="")
    else:
        print(', '   n, end='')

[Out]: 1, 2, 0, 3