I am trying to reverse a list using a recursive function. Unfortunatley I am fairly new to recursion. Is this possible? That is my code thus far
def stringRev (word):
worLen = len(word)
if worLen == 1:
return word
return (word[-1]) + stringRev(word[:-1])
listWord = ["hey", "there", "jim"]
print(stringRev(listWord))
Your problem is that (word[-1]) is a string, not a list. So you are trying to add/concatenate a string and a list. I changed that expression to [word[-1]] to create a list.
>>> def stringRev (word):
... worLen = len(word)
... if worLen == 1:
... return word
... return [word[-1]] + stringRev(word[:-1])
...
>>> listWord = ["hey", "there", "jim"]
>>> print(stringRev(listWord))
['jim', 'there', 'hey']
>>>
PS. It would be helpful if you included the error you received when running your code: TypeError: Can't convert 'list' object to str implicitly
To reverse the order of the elements of the list, change:
return (word[-1]) + stringRev(word[:-1])
to
return [word[-1]] + stringRev(word[:-1])
(note the square brackets).
The problem is that you are trying to concatenate a string (word[-1]) with a list (word[:-1]).
The problem is that your function is expecting a single word, yet you're calling it with a list of words.
If you call it as follows, you'll see that it works just fine:
for word in ["hey", "there", "jim"]:
print(stringRev(word))
Or, if you wish to store the reversed strings in a list:
l = [stringRev(w) for w in ["hey", "there", "jim"]]
The one corner case where your function would fail is the empty string. I don't know whether that's a valid input, so it could be a non-issue (but trivial to fix nonetheless).
If you want it done in Python:
reversed(listWord)
assuming word is a list or a tuple
http://docs.python.org/2/library/functions.html#reversed
And to get a list:
list(reversed(listWord))
should work
But if you want an algorithm, I guess reversed is not your friend !
reversed doesn't return a string. It gives you an iterable object. For that reason, you'll often see s[::-1] to reverse a sequence as well.