False Iteration

Question

I want to write a for loop that tests two functions on a list. First one gives back an int, second a boolean. If boolean == False the function should remove the element from the list so in the end after calling the function L should have mutated. Here is an example.

def f(i):
    return i + 2

def g(i):
    return i > 5

L = [0, -10, 5, 6, -4]

def applyF_filterG(L, f, g):
    for i in L:
        if g(f(i)) == False:        
            L.remove(i)

L = [5, 6]

My problem is that with the function written above I get L = [-10, 5, 6] back because if i is removed, the i+1 turns into an i+2 because of the removed element. Does anybody know how to solve this? Big thanks!

`for i in L` instead of this use `for i in L[:]`, It will iterate though it's copy. — Rahul K P, Oct 04 '16 at 11:41
`[i for i in L if i+2 > 5]` it's also give the same result which you expect. — Rahul K P, Oct 04 '16 at 11:45

score 5 · Accepted Answer · answered Oct 04 '16 at 11:38

You should never remove elements from the list while iterating over it. just make a new list L1 equal to L and remove elements from L and iterate over L1.

def f(i):
    return i + 2

def g(i):
    return i > 5

L = [0, -10, 5, 6, -4]

def applyF_filterG(L, f, g):
    L1 = list(L)
    for i in L1:
        if g(f(i)) == False:        
            L.remove(i)


applyF_filterG( L, f, g )

print L

score 2 · Answer 2 · answered Oct 04 '16 at 11:39

2

Can you simplify this by just generating another list:

L = [0, -10, 5, 6, -4]
r = [a for a in L if a <= 5]

or something like that?

answered Oct 04 '16 at 11:39

Hannu

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The functions are just examples, they can change. That's why I can't be more specific in the loop – Mickey Mahoney Oct 04 '16 at 11:58
Then call a function r = [a for a in L if f(a) <= g()] It remains the same nevertheless. You need to create a new list instead of deleting items from the list you are looping through. After the loop you can then assign L=r. – Hannu Oct 04 '16 at 12:05

False Iteration

2 Answers2