The Shed is going Social! Join us on FaceBook and Twitter and chime in on the conversation.
|
 |
|
Dev Shed Forums
> Programming Languages
> Python Programming
|
Wrong result from recursive func. in Python
Discuss Wrong result from recursive func. in Python in the Python Programming forum on Dev Shed. Wrong result from recursive func. in Python Python Programming forum discussing coding techniques, tips and tricks, and Zope related information. Python was designed from the ground up to be a completely object-oriented programming language.
|
|
 |
|
|
|
|

Dev Shed Forums Sponsor:
|
|
|

December 15th, 2012, 10:48 AM
|
|
Registered User
|
|
Join Date: Dec 2012
Posts: 3
Time spent in forums: 57 m 54 sec
Reputation Power: 0
|
|
|
Wrong result from recursive func. in Python
Dear all,
I am quite new in Python. I want to write very simple function which gets an integer "c" and a list of integers "li". I would like to recursively find first two items where sum of them will be equal to "c". I have written this function in Python to find second option "n" and then I can easily find the first number by finding "c-n".
My code is:
Code:
def shop(c, li):
if ((c-li[0]) in li[1:]):
return li[0];
else:
shop(c, li[1:]);
The problem is when I call function as below:
Code:
c=100;
li=[5, 75, 25];
ii=shop(c, li);
print ii, "\n";
The result is always "None", however it should return "25". When I put a "print" statement before "return" I can see that the function has found the item correctly, but I am not sure why it is not returned.
Anybody know why this happens?
Thanks a lot!
|

December 15th, 2012, 11:02 AM
|
 |
Contributing User
|
|
|
|
|
Repair
Code:
def shop(c, li):
if ((c-li[0]) in li[1:]):
return li[0]
else:
return shop(c, li[1:])
Python functions, by default, return None. When the test failed (in other words the value is not in list) the else block evaluated, and then the function returned None.
__________________
[code] Code tags[/code] are essential for python code!
|

December 15th, 2012, 11:32 AM
|
|
Registered User
|
|
Join Date: Dec 2012
Posts: 3
Time spent in forums: 57 m 54 sec
Reputation Power: 0
|
|
Thanks a lot for your answer. But when I call the function as I mentioned before, at the first round the test if failed, since 100-5 is not in the list. But at the second round, when the function is called recursively, it should return a number since 100-75 exists in the list. Am i right or am I missing something?
Quote: | Originally Posted by b49P23TIvg
Code:
def shop(c, li):
if ((c-li[0]) in li[1:]):
return li[0]
else:
return shop(c, li[1:])
Python functions, by default, return None. When the test failed (in other words the value is not in list) the else block evaluated, and then the function returned None. |
|

December 15th, 2012, 12:59 PM
|
|
Registered User
|
|
Join Date: Dec 2012
Posts: 3
Time spent in forums: 57 m 54 sec
Reputation Power: 0
|
|
Hi again,
I solved the problem by rewriting the function as:
Code:
def shop(c, li):
return li[0] if (c-li[0] in li[1:]) else shop(c, li[1:])
|

December 15th, 2012, 05:41 PM
|
 |
Contributing User
|
|
|
|
|
Hmm. I see they added the ternary if expression to python2.
Your new statement demonstrates, in my opinion, good style because there's one return statement.
Do you still want an explanation of the failure in your first attempt?
|
Developer Shed Advertisers and Affiliates
| Thread Tools |
Search this Thread |
|
|
|
| Display Modes |
Rate This Thread |
Linear Mode
|
|
Posting Rules
|
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts
HTML code is Off
|
|
|
|
|