Iterate a list using recursion

Question

How can I write this same function using recursion?

def replace(thelist,a,b):
"""Returns: a COPY of thelist but with all occurrences of a replaced by b. 

    Example: replace([1,2,3,1], 1, 4) = [4,2,3,4].

Precondition: thelist is a list of ints
              a and b are ints"""

I believe I'm supposed to make a copy of the original, and then from there make a = b in the list, but I do not know how to implement that. Please help me solve this, and thank you!

newlist = thelist[:] if newlist == []: return 0 return ( a = b if newlist[0] == a else a) + replace(newlist[1:],a,b) — steve12, Oct 28 '14 at 03:34
I know some of it is wrong but I'm not sure where to go from there — steve12, Oct 28 '14 at 03:34

score 0 · Accepted Answer · answered Oct 28 '14 at 03:31

Try this:

def replace(thelist, a, b):
    if not thelist:
        return []
    elif thelist[0] == a:
        return [b] + replace(thelist[1:], a, b)
    else:
        return [thelist[0]] + replace(thelist[1:], a, b)

It's a recursive solution, and works as expected:

replace([1, 2, 3, 1], 1, 4)
=> [4, 2, 3, 4]

Iterate a list using recursion

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