Filtering a list of strings in python

Question

For an assignment I have to filter a list without using the built-in function in Python. Here's my code :

def satisfiesF(L):
    result = 0
    i = 0
    L1 = []
    while i < len(L):
        s = L[i]
        if f(s) == True :
            result += 1
            L1.append(L[i])
        i += 1
    L = L1
    print L
    return result
def f(s):
    return 'a' in s
L = ['a', 'b', 'a']
print satisfiesF(L)
print L

It prints :

['a', 'a']
2
['a', 'b', 'a']

Why is the second L not the same as the first? I have searched for hours but I can't understand. It seems to me that here L is passed by reference so why doesn't the value change?

I've tried changing the line L = L1 in L = list(L1) but the result is the same.

Answer 1

It's a simple mistake of the scope of local and global variables.

Inside your function satisfiesF(L) the line

L = L1

changes the L which is declared inside the function which you passed to the function satisfiesF() as an argument. You could have passed L' and tried printing L' , it would work just fine. But print L would give variable out of scope error.

But when you do

print L

in the last line, it will print the L declared just two lines above it as L = ['a', 'b', 'a']

Answer 2

As Himanshu points out, the scope is the issue. That is why this works, since it sets self.L

class my_test:
    def __init__(self, values, check):
        self.L = values
        self.check = check

    def run_test(self):
        print self.satisfiesF()
        print self.L

    def f(self, s):
        return self.check in s

    def satisfiesF(self):
        result = 0
        i = 0
        L1 = []
        while i < len(self.L):
            s = self.L[i]
            if self.f(s) == True :
                result += 1
                L1.append(self.L[i])
            i += 1
        self.L = L1
        print self.L
        return result

x = my_test(['a', 'b', 'a'], 'a')
x.run_test()

Output is

['a', 'a']
2
['a', 'a']

Answer 3

L = L1

This makes L reference to the list that L1 references. Add print(id(L)) before and after this line, you could see its id has changed.

One way to keep L the same id while getting the content of L1 is to clear it first and then append:

del L[:]
L.append(L1)