Question

How does string.join resolve? I tried using it as below:

import string 
list_of_str = ['a','b','c'] 
string.join(list_of_str.append('d'))

But got this error instead (exactly the same error in 2.7.2):

Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/usr/local/lib/python2.6/string.py", line 318, in join
    return sep.join(words)
TypeError

The append does happen, as you can see if you try to join list_of_string again:

print string.join(list_of_string)
-->'a b c d'

---

here’s the code from string.py (couldn’t find the code for the builtin str.join() for sep):

def join(words, sep = ' '):
    """join(list [,sep]) -> string

    Return a string composed of the words in list, with
    intervening occurrences of sep.  The default separator is a
    single space.

    (joinfields and join are synonymous)

    """
    return sep.join(words)

What’s going on here? Is this a bug? If it’s expected behavior, how does it resolve/why does it happen? I feel like I’m either about to learn something interesting about the order in which python executes its functions/methods OR I’ve just hit a historical quirk of Python.

---

Sidenote: of course it works to just do the append beforehand:

list_of_string.append('d')
print string.join(list_of_string)
-->'a b c d'

Answer 1

list_of_str.append('d')

does not return the new list_of_str.

The method append has no return value and so returns None.

To make it work you can do this:

>>> import string
>>> list_of_str = ['a','b','c']
>>> string.join(list_of_str + ['d'])

Although that is not very Pythonic and there is no need to import string… this way is better:

>>> list_of_str = ['a','b','c']
>>> ''.join(list_of_str + ['d'])