将递归结果转换为字符串，以便向后打印

Question

Here is my code so far:

import string  
def convert(num,base):
    if num==0:
        return 
    else:
        remainder = num%base
        num = num//base
        b=(str(remainder))
        print(b[::-1],end="")
        return convert(num,base)

Instead of printing this:

>>> convert(29,3)
2001

I need to print it backward like this (and it needs to be done by strings):

>>> convert(29,3)
1002

Seems like string does not work well with recursion:

>>> convert(29,3)
 ('2', ('0', ('0', ('1', None))))

Answer 1

Just change the order in which you print and recurse:

import string  
def convert(num,base):
    if num==0:
        return 
    remainder = num % base
    num = num // base
    convert(num, base)
    print(remainder, end="")

>>> convert(29,3)
1002

No idea what convertBase() is.

Answer 2

Throwing in my $0.02. I would opt for returning the entire string, instead of printing it as you go.

def convert(num, base):
    if num == 0:
        return ''
    num, remainder = divmod(num, base)
    return convert(num, base) + str(remainder)

EDIT might as well use the built-in divmod for computing the num and remainder.

Answer 3

Your recursive function is correct in the sense that it has a exit clause ( if num == 0: ), however due to the nature of recursive functions the order of execution will be the reverse of what you wish to print out. What needs to be done here is to keep adding the remainders to a list and then finally when you hit the exit clause, reverse that list and print it.

import string  
def convert(num, base, rem_list=None):
    if not rem_list:
        rem_list = []
    if num == 0:
        rem_list.reverse()
        print "".join(rem_list) 
    else:
        remainder = num % base
        num = num // base
        rem_list.append(str(remainder))
        return convert(num, base, rem_list)



>>> convert(29, 3)
1002
>>> convert(29, 3)
1002