CODEWITHSUNDEEP
pythonlist
tintin
tintin

Reputation: 1529

delete a list of index from a list

So I've got into trouble when I want to delete a sequence of index from a list

    for i in linked_node:
        del l[i]

the linked_node is the index sequence this code simply wont work cause del changed the size of the list

using a for loop , I can do this by appending them to a new list

  l1 = []      
  for i in range(len(l)):
     if i is not in linked_node:
         l1.append(l[])

my question is , how to do this in an elegent way?without creating another list

Upvotes: 1

Views: 1051

Answers (4)

TerryA
TerryA

Reputation: 60024

Use a list comprehension instead:

>>> l = ['hi', 'hello', 'hey', 'hai']
>>> indexes = {1, 2}
>>> [j for i, j in enumerate(l) if i not in indexes]
['hi', 'hai']

Upvotes: 5

RickyA
RickyA

Reputation: 16029

Also using the reversed iteration trick, but somewhat faster because you don't have to sort your list:

l = ["a", "b", "c", "d"]

for idx in range(len(l)-1, -1, -1):
    i = l[idx]
    print(i)
    if i in ["b", "d", "c"]:
        del(l[idx])
print(l) 

>>> d
>>> c
>>> b
>>> a
>>> ['a']

Upvotes: 0

Sorin
Sorin

Reputation: 11968

you can do something like:

linked_node.sort(reversed=True)
for i in linked_node:
  del l[i]

Now this should work because the indices you are interested in stay the same. Another way to do it, if it's already sorted:

for (x, i) in enumerate(linked_node):
  del l[i - x]

Upvotes: 1

tobias_k
tobias_k

Reputation: 82949

If you do not want to create a new list, as in @Haidro's solution, but really delete the elements at the given indices from the existing list, you could also reverse the list of indices, so that indices at the end of the list are deleted first.

for i in sorted(linked_node, reverse=True):
    del l[i]

Otherwise, the elements you want to delete will shift to the left as you delete elements, thus changing their indices, and you end up deleting the wrong elements, or getting an IndexError.

Upvotes: 2

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