Recursive function in removing occurence in list python

Question

def remove_all1(x,s):
    def loop(x,s,ss):
        while s!=[]:
            if x == s[0]:
                return loop(x,s[2:],ss+[s[1]])
            else:
                return loop(x,s[1:],ss+[s[0]])
        return ss
    return loop(x,s,[])

This is the one I made by recursive tail function to remove recursive value x and if :

print(remove_all1(3,[4,3,5,6,3,2,1]))

I would get:

[4,5,6,2,1]

The exact result I want. But for the recursive function here:

def remove_all0(x,s):
    while s!=[]:
        if x == s[0]:
            ss = [s[1]] + remove_all0(x,s[2:])
            return ss 
        else:
            s1 = [s[0]] + [remove_all0(x, s[1:])]
            return s1 
    if s==[]:
        return s
print(remove_all0(3,[4,3,5,6,3,2,1]))

I would not get the same result. Instead, I got:

[4, [5, 6, [2, 1, []]]]

Why does this happen?

remove_all0 returns a list, which you're wrapping in another list when you call it recursively — jonrsharpe
– jonrsharpe, Commented Oct 11, 2015 at 12:24
I think this post might also be of interest stackoverflow.com/questions/1157106/… it shows how to efficiently remove all occurrences of an item from a list in Python. — Anthony Blackshaw
– Anthony Blackshaw, Commented Oct 11, 2015 at 13:00

Eric Renouf · Accepted Answer · 2015-10-11 12:56:59Z

2

The problem is in your else clause when you wrap the returned value from remove_all0 which is a list, and make it a list containing that list. Just remove those square brackets and you'll get the results you want:

def remove_all0(x,s):
    while s!=[]:
        if x == s[0]:
            ss = [s[1]] + remove_all0(x,s[2:])
            return ss
        else:
            s1 = [s[0]] + remove_all0(x, s[1:])
            return s1
    if s==[]:
        return s
print(remove_all0(3,[4,3,5,6,3,2,1]))

gives

[4, 5, 6, 2, 1]

answered Oct 11, 2015 at 12:56

Eric Renouf

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