Remove odd number from list

Question

def purify(ls):
    y = ls
    for i in ls:
        if i % 2 != 0:
          y = y.remove(i)

    print y

The following code fails when I pass the list (4, 5, 5, 4). It returns (4, 5, 4) instead of (4, 4). What is wrong here ? I am not able to understand why changes in y list is affecting original list ls.

Rahul K P · Answer 1 · 2016-06-16T18:37:32.813

1

Do like this.

In [1]: a = (4, 5, 5, 4)
In [2]: result = [i for i in a if not i % 2]
In [3]: result
Out[1]: [4, 4]

In function.

def purify(ls):
   return [i for i in ls if not i % 2]

To understand more i expand my code.From this you can understand how it's work.

def purify(input_list):
    result = []
    for i in input_list:
        if not i % 2:
            result.append(i)
    return result

edited Jun 16 '16 at 18:37

answered Jun 16 '16 at 18:28

Rahul K P

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Moses Koledoye · Answer 2 · 2016-06-16T18:33:45.150

1

The indices change once you start removing items. It is not a good practice to iterate on a list while you're mutating its items in the loop. Iterate on a slice of the list ls[:] instead:

def purify(ls):
    for i in ls[:]:
        if i % 2 != 0:
            ls.remove(i)

Or just use a list comprehension:

[i for i in ls if i % 2 == 0]

edited Jun 16 '16 at 18:33

answered Jun 16 '16 at 18:28

Moses Koledoye

77,341
8
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But I am editing the list y and not original list ls. I am not able to understand why remove function on y list is affecting original list ls. – Mohan.W Jun 16 '16 at 21:03
The two variables are pointing to the same object. So if you do `x = y = ['a']` and then `x[0] = 4`, `y` gets updated also. – Moses Koledoye Jun 16 '16 at 21:05

score 0 · Answer 3 · answered Jun 16 '16 at 18:31

0

The following code is much cleaner and easier to understand:

l = [4, 5, 5, 4]

l = filter(lambda x: x % 2 == 0, l)

print(l)

[4, 4]

answered Jun 16 '16 at 18:31

Simon Kirsten

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score 0 · Answer 4 · answered Jun 16 '16 at 18:39

remove(...)
    L.remove(value) -- remove first occurrence of value.
    Raises ValueError if the value is not present.

correct version:

>>> filter(lambda x: x % 2 == 0, [4, 5, 5, 4])
[4, 4]

your correct version:

def purify(ls):
    y = ls[:]  # copy it instead of refering to it
    for i in ls:
        if i % 2 != 0:
            y.remove(i)  # remove will return none

    print y

Remove odd number from list

4 Answers4