using
Python 2.3.4 (#53, May 25 2004, 21:17:02) [MSC v.1200 32 bit (Intel)] on
win32
OK, I have a recursive function that should return a list, but doesn't
<start session>
def test(word):
if type(word) == str:
print "it's a word"
test([word])
if type(word) == list:
print "The conditional worked, see ->", word
return word a = test('foobity')
it's a word
The conditional worked, see -> ['foobity'] print a
None
</end session>
What am I missing?
-derek. 4 3147
"Derek Rhodes" <rh****@worldpa th.net> writes: if type(word) == str: print "it's a word" test([word])
The last line tests [word] and throws away the value. YOu have to say
"return test([word])".
Derek Rhodes <rhoder <at> worldpath.net> writes: OK, I have a recursive function that should return a list, but doesn't
def test(word): if type(word) == str: print "it's a word" test([word])
if type(word) == list: print "The conditional worked, see ->", word return word
By default, if a Python function does not hit a return statement before the
end of the function, it returns the None value. Notice that if word is a str,
your function executes the first if-block, including the recursive call and
then skips the second if-block. So in this case, you never hit a return
statement and so Python returns None. You probably meant to write:
def test(word):
if type(word) == str:
return test([word])
if type(word) == list:
return word
If you run into these kind of mistakes frequenly, it might be worth having
only one return point in each function. You would then write your code
something like:
def test(word):
if isinstance(word , str):
result = test([word])
elif isinstance(word , list):
result = word
else:
raise TypeError('unsu pported type %r' % type(word))
return result
Of course, this particular example probably doesn't merit a recursive function
anyway, but you get the idea...
Steve
Derek Rhodes wrote: OK, I have a recursive function that should return a list, but doesn't
<start session>
def test(word): if type(word) == str: print "it's a word" test([word])
if type(word) == list: print "The conditional worked, see ->", word return word
What am I missing?
You are forgetting to return the value.
change this part ::
if type(word) == str:
print "it's a word"
test([word])
to
if type(word) == str:
print "it's a word"
return test([word]) # return the result of test([word])
George
"Steven Bethard" <st************ @gmail.com> wrote in message
news:ma******** *************** *************** @python.org... Derek Rhodes <rhoder <at> worldpath.net> writes: OK, I have a recursive function that should return a list, but doesn't
def test(word): if type(word) == str: print "it's a word" test([word])
if type(word) == list: print "The conditional worked, see ->", word return word
By default, if a Python function does not hit a return statement before the end of the function, it returns the None value. Notice that if word is a str, your function executes the first if-block, including the recursive call and then skips the second if-block. So in this case, you never hit a return statement and so Python returns None. You probably meant to write:
def test(word): if type(word) == str: return test([word]) if type(word) == list: return word
If you run into these kind of mistakes frequenly, it might be worth having only one return point in each function. You would then write your code something like:
def test(word): if isinstance(word , str): result = test([word]) elif isinstance(word , list): result = word else: raise TypeError('unsu pported type %r' % type(word)) return result
Of course, this particular example probably doesn't merit a recursive function anyway, but you get the idea...
Steve
WOW, thanks everyone for the quick reply!
-Derek. This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics |
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