Assign a dynamic default value for function variable?

Question

I have a function which creates a compound string based on how many times the user wants to input the string:

def string_maker(string_list, repeat_list):
  final_string = ''
  for i in range(len(string_list)):
    for _ in range(repeat_list[i]):
      final_string += string_list[i]
  return final_string

So, for example, string_maker(['a', 'b', 'c'], [1, 2, 1]) outputs abbc.

I want to make it so that if the user doesn't input a list for the repeat_list argument, it defaults to all 1s. This is how I'm trying to do it:

def string_maker(string_list, repeat_list = None):
  if repeat_list == None:
    repeat_list = [1] * len(string_list)
  final_string = ''
  for i in range(len(string_list)):
    for _ in repeat_list[i]:
      final_string += str
  return final_string

print(string_maker(['a', 'b', 'c']))

However, I receive an error saying 'int' object is not iterable, even though the new snippet I included returns the appropriate list [1, 1, 1]. Is there a better way to go about doing this?

Just to note that your inner loop isn't using range as you did originally... — Jon Clements
– Jon Clements, Commented Mar 25, 2020 at 22:07
The variable (?) str is undefined in the second code snippet (it's the name of a built-in class, so you shouldn't be using it a variable name in the first place). — martineau
– martineau, Commented Mar 25, 2020 at 22:10
The problem is this line: for _ in repeat_list[i]:... repeat_listp[i] will be an int object, and you cannot iterate over int object, as the error clearly states. — juanpa.arrivillaga
– juanpa.arrivillaga, Commented Mar 25, 2020 at 22:15
Thanks everything, this was a terrible example of my not being able to copy and paste my own code properly... — Oion Akif
– Oion Akif, Commented Mar 25, 2020 at 22:39

chepner · Accepted Answer · 2020-03-25 22:18:19Z

2

The join method is a much more efficient way to build this list.

from itertools import repeat


def string_maker(string_list, repeat_list=repeat(1)):
    return ''.join([x*n for x, n in zip(string_list, repeat_list)])

repeat(1) provides an infinite stream of 1s; zip only uses as many of them as it needs to pair with elements of string_list. x*n creates a string consisting of n repeated occurrences of x.

>>> string_maker(list("abc"))
'abc'
>>> string_maker(list("abc"), [1,2,1])
'abbc'

edited Mar 25, 2020 at 22:18

answered Mar 25, 2020 at 22:12

chepner

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