Python. list.pop() based on index

Question

I was working with list in python and I need to remove non-true values.

Can someone explain why here I get index out of range error:

    for n in range(len(lst)-1):                  #index outside the range
        if not bool(lst[n]):
            lst.pop(n)
    return lst

It is kind of work with while loop

def compact(lst):
    while n < len(lst):
        if not bool(lst[n]):
            lst.pop(n)
        n+=1
        print(n)

    return lst

But in this case loop will skip some items.

function is called like:
compact([0, 1, 2, '', [], False, (), None, 'All done'])

Asked By: iGRiK

||

Source

Answer 1

When you delete an element, all the elements after it renumber: the n+1th element becomes nth element, etc. But you progress to the next n anyway. This is why you are skipping some elements.

In the first snippet, you pre-construct the list of indices to iterate over; but as the list shortens, some of the later indices will not exist any more.

In the second snippet, you compare with the actual length of the list on each iteration, so you never get an invalid index.

Also note that bool is not needed whenever you are evaluating a condition, as it is implicitly applied in such a context.

In order to do this correctly, you have two choices:

iterate from end of the list backwards. If you delete an element, the elements in front of it do not get renumbered.
```
for n in range(len(lst) - 1, -1, -1):
    if not lst[n]:
        lst.pop(n)
```
using the while method, make sure to either increment n (moving n closer to the end of the list), or delete an element (moving the end of the list closer to n); never both at the same time. This will ensure no skipping.
```
n = 0
while n < len(lst):
    if not lst[n]:
        lst.pop(n)
    else:
        n += 1
```

The third option is to avoid the index loop altogether, and do it more pythonically, generating a new list using a comprehension with a filtering condition, or filter.

new_lst = list(filter(bool, lst))

or

new_lst = [e for e in lst if e]

or, if you really want to change the original list, replace its content like this:

lst[:] = filter(bool, lst)

Answered By: Amadan

Answer 2

Index out of range error:
lst.pop(n) This will pop element from the list but for n in range(len(lst)-1) will still loop through the lst assuming same length as original.

To avoid this start loop from last index in reverse order so that even after pop values will still be present for indexes yet to be read.

   def compact(lst):
    for n in range(len(lst)-1,-1,-1):                  
        if not bool(lst[n]):
            lst.pop(n)
    return lst

    ls = [0, 1, 2, '', '',[], False, (), None, 'All done','']
    print(compact(ls))

    # Output:
    # [1, 2, 'All done']

Answered By: Shriya Jain

Python. list.pop() based on index

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