Python recursion not returning value

Question

For some reason x in this code is not updating within the recursion. Shouldn't x update as I'm calling b(c) within a(y)? When x updates in b(c) but doesn't return to

global nested
def extract(nested,depth):
    y = depth[0]
    depth = depth[1:]
    extract = nested[y]
    newlist(extract)
    return depth
def newlist(x):
    nested = x
    return nested
def recursiveRef(nested,depth):
    """Return element from nested list
    list ->int
    """
    if len(depth) == 0:
        return nested
    else:
        return recursiveRef(nested,extract(nested,depth))

In some languages, you'd need x = b(c) not just b(c). Not sure on python — Adam
– Adam, Commented May 14, 2011 at 10:23
You haven't defined depth anywhere (not initialized it, nor are you incrementing it). Can you please update the code? — csl
– csl, Commented May 14, 2011 at 10:27
This code should not even execute, giving you NameError in your recursion(). If you use global variables, please specify so. — Daniel Kluev
– Daniel Kluev, Commented May 14, 2011 at 10:28
your code is quite obscure, can you provide a simpler (non) working version? — vrde
– vrde, Commented May 14, 2011 at 10:29
Why you need this, or just a few samples of input and expected output, would really help understand what you're doing. — Fred Nurk
– Fred Nurk, Commented May 14, 2011 at 11:27

Mark Tolonen · Accepted Answer · 2011-05-14 15:31:35Z

3

Is this what you are trying to do?

def recursiveRef(nested,depth):
    """Return element from nested list
    list ->int
    """
    if len(depth) == 0:
        return nested
    else:
        return recursiveRef(nested[depth[0]],depth[1:])

print recursiveRef([[1,2,3],[4,[5,6],7]],[1])
print recursiveRef([[1,2,3],[4,[5,6],7]],[1,1])
print recursiveRef([[1,2,3],[4,[5,6],7]],[1,1,1])

Output

[4, [5, 6], 7]
[5, 6]
6

answered May 14, 2011 at 15:31

Mark Tolonen

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1 Comment

Steven Over a year ago

Yes that is what i want to do

Petar Minchev · Accepted Answer · 2011-05-14 10:25:03Z

1

I am not a Python master but I think the problem is that x is local to the recursion function. You are changing an other global x in your b(c). Please, correct me if I am wrong.

answered May 14, 2011 at 10:25

Petar Minchev

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Comments

Gurgeh · Accepted Answer · 2011-05-14 11:24:07Z

1

Your code looks suspicious. When you think you want a global variable, usually you want a class. Consider wrapping your functions in a class and use self.x instead of x.

That said, "global" should not be written in the top of your program. Instead, you need it in every function that modifies your global variable, but not those that only reads it.

def newlist(x):
    global nested
    nested = x
    return nested

answered May 14, 2011 at 11:24

Gurgeh

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Comments

vrde · Accepted Answer · 2011-05-14 10:32:51Z

0

The x variable in the b function is not binded to the one in the recursion function.

I really don't understand what you are tring to do -- I can be wrong but I suggest you to consider to put the two functions a and b into recursion, creating a closure. In that way the nested functions can see and modify all the variables defined in the outer scope.

answered May 14, 2011 at 10:32

vrde

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Comments

Hugh Bothwell · Accepted Answer · 2011-05-14 18:46:48Z

0

An even simpler non-recursive version:

def inTree(tree, ref):
    for offs in ref:
        tree = tree[offs]
    return tree

answered May 14, 2011 at 18:46

Hugh Bothwell

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