delete items from list by indices

[blumenkraft]

Hi,

I have some list:
x = [8, 9, 1, 7]
and list of indices I want to delete from x:
indices_to_delete = [0, 3], so after deletion x must be equal to [9,
1].

What is the fastest way to do this? Is there any builtin?

Thanks.

 

[Chris Rebert]

Hi,

I have some list:
x = [8, 9, 1, 7]
and list of indices I want to delete from x:
indices_to_delete = [0, 3], so after deletion x must be equal to [9,
1].

What is the fastest way to do this? Is there any builtin?

#untested & unbenchmarked since it's 1am
offset = 0
for index in indices_to_delete:
del x[index-offset]
offset += 1

Cheers,
Chris

 

[Peter Otten]

I have some list:
x = [8, 9, 1, 7]
and list of indices I want to delete from x:
indices_to_delete = [0, 3], so after deletion x must be equal to [9,
1].

What is the fastest way to do this? Is there any builtin?

Why's that obsession with speed?

items = ["a", "b", "c", "d"]
delenda = [0, 3]
for i in sorted(delenda, reverse=True):

.... del items
....['b', 'c']

items = ["a", "b", "c", "d"]
delenda = set([0, 3])
items = [item for index, item in enumerate(items) if index not in delenda]
items

['b', 'c']

If you really need to modify the list in place change

items = [item for ...]

to

items[:] = [item for ...]

Try these and come back to complain if any of the above slows down your
script significantly...

Peter

 

[Ishwor]

Hi

Hi,

I have some list:
x = [8, 9, 1, 7]
and list of indices I want to delete from x:
indices_to_delete = [0, 3], so after deletion x must be equal to [9,
1].

What is the fastest way to do this? Is there any builtin?

Try this-

x = [8, 9, 1, 7]
[x.pop(i) for i in sorted(indices_to_delete,reverse=True)] [7, 8]
x

[9, 1]

Built-in used here is `sorted' and method on list used here is `pop'.
With regards to efficiency you may want to use the methods of list
which is more intuitive afaik and useful as its more reflective of
effect on the type list. It's a trivial choice here but later it might
help.

 

[blumenkraft]

I have some list:
x = [8, 9, 1, 7]
and list of indices I want to delete from x:
indices_to_delete = [0, 3], so after deletion x must be equal to [9,
1].

What is the fastest way to do this? Is there any builtin?

Why's that obsession with speed?

items = ["a", "b", "c", "d"]
delenda = [0, 3]
for i in sorted(delenda, reverse=True):

... š š del items
...>>> items

['b', 'c']

items = ["a", "b", "c", "d"]
delenda = set([0, 3])
items = [item for index, item in enumerate(items) if index not in delenda]
items

['b', 'c']

If you really need to modify the list in place change

items = [item for ...]

to

items[:] = [item for ...]

Try these and come back to complain if any of the above slows down your
script significantly...

Peter

Thanks, it helped (I didn't know about keyword argument "reverse" in
sorted function)