Help- Simple recursive function to build a list

[actuary77]

I am trying to write simple recursive function to build a list:


def rec(n,alist=[]):
_nl=alist[:]
print n,_nl
if n == 0:
print n,_nl
return _nl
else:
_nl=_nl+[n]
rec(n-1,_nl)

_nl=[]
_nl=rec(4)
print _nl

### shouldn't this work?

_nl=rec(4)




The output is:
4 []
3 [4]
2 [4, 3]
1 [4, 3, 2]
0 [4, 3, 2, 1]
0 [4, 3, 2, 1]
None
None


Question:
============
Why isn't the function returning a list?
Why is function returning None?

Any guidance from one of you pro's would be greatly appreciated.

Thanks,
DKoch
(e-mail address removed)

 

[Roy Smith]

I am trying to write simple recursive function to build a list:


def rec(n,alist=[]):
_nl=alist[:]
print n,_nl
if n == 0:
print n,_nl
return _nl
else:
_nl=_nl+[n]
rec(n-1,_nl)

_nl=[]
_nl=rec(4)
print _nl

### shouldn't this work?

_nl=rec(4)




The output is:
4 []
3 [4]
2 [4, 3]
1 [4, 3, 2]
0 [4, 3, 2, 1]
0 [4, 3, 2, 1]
None
None


Question:
============
Why isn't the function returning a list?
Why is function returning None?

I'm not 100% sure what you're trying to do here, but I suspect you left a
"return" out of the else block, i.e. "return rec(n-1, _nl)". That would
certainly explain why it's returning None.

What problem are you trying to solve, or are you just experimenting with
recursion?

 

[Kent Johnson]

I am trying to write simple recursive function to build a list:


def rec(n,alist=[]):
_nl=alist[:]
print n,_nl
if n == 0:
print n,_nl
return _nl
else:
_nl=_nl+[n]
rec(n-1,_nl)

should be
return rec(n-1,_nl)

>>> def rec(n,alist=[]):

... _nl=alist[:]
... print n,_nl
... if n == 0:
... print n,_nl
... return _nl
... else:
... _nl=_nl+[n]
... return rec(n-1,_nl)
...4 []
3 [4]
2 [4, 3]
1 [4, 3, 2]
0 [4, 3, 2, 1]
0 [4, 3, 2, 1][4, 3, 2, 1]

Kent