deleting items within a for loop - mutable indices

[SnuSnu]

Okay - here's a (probably) really easy question:

I can't do the following, so what's the best way to handle this
case of wanting to delete within a loop?

x = [0,1,2,3,4,5,6,7,8]
deletion_list = [2,7,8]

for i in deletion_list:
del x

This obviously doesn't work because the list keeps changing
each time I delete something, so the indices are no longer valid
after the first delete (so I get an IndexError on teh last delete).

My hack is to do this first (which makes sure that the higher indices
are deleted first), but this is not very elegant or quick:

deletion_list.sort()
deletion_list.reverse()
# BTW: It's a shame I can't do deletion_list.sort().reverse()

Is there a simple solution?

(It would be nice to do something like
del x[2,7,8] # or..
del [*deletion_list]
)

Thanks.

 

[Erik Max Francis]

Okay - here's a (probably) really easy question:

I can't do the following, so what's the best way to handle this
case of wanting to delete within a loop?

x = [0,1,2,3,4,5,6,7,8]
deletion_list = [2,7,8]

for i in deletion_list:
del x

This obviously doesn't work because the list keeps changing
each time I delete something, so the indices are no longer valid
after the first delete (so I get an IndexError on teh last delete).

The solution is usually to either iterate over an explicit copy of the
list, or use functional means to build the final list up from the
original list without mutating it at all.

In this case it's probably best to just iterate backwards over the
indices, since you know what you're getting and can control the order in
which you delete the items.

deletion_list.sort()
deletion_list.reverse()
# BTW: It's a shame I can't do deletion_list.sort().reverse()

This is a deliberate design decision. These methods are made to return
None so you're explicitly aware that they mutate the object, rather than
return a duplicate. You can always define your own functional version:

def mySort(seq):
result = seq[:]
result.sort()
return result

 

[Dan Bishop]

Okay - here's a (probably) really easy question:

I can't do the following, so what's the best way to handle this
case of wanting to delete within a loop?

x = [0,1,2,3,4,5,6,7,8]
deletion_list = [2,7,8]

for i in deletion_list:
del x

This obviously doesn't work because the list keeps changing
each time I delete something, so the indices are no longer valid
after the first delete (so I get an IndexError on teh last delete).

for i in deletion_list:
x = None
x = [num for num in x if num is not None]

 

[James Moughan]

Okay - here's a (probably) really easy question:

I can't do the following, so what's the best way to handle this
case of wanting to delete within a loop?

x = [0,1,2,3,4,5,6,7,8]
deletion_list = [2,7,8]

for i in deletion_list:
del x

This obviously doesn't work because the list keeps changing
each time I delete something, so the indices are no longer valid
after the first delete (so I get an IndexError on teh last delete).

My hack is to do this first (which makes sure that the higher indices
are deleted first), but this is not very elegant or quick:

deletion_list.sort()
deletion_list.reverse()
# BTW: It's a shame I can't do deletion_list.sort().reverse()

Is there a simple solution?

(It would be nice to do something like
del x[2,7,8] # or..
del [*deletion_list]
)

Thanks.

Eh, do you want to delete the elements in the deletion list from x, or
delete the elements in x which are at the location indicated in the
deletion list? Your example doesn't differentiate.

Anyhow, this is advice from a newbie but in the first case

x = [y for y in x if not y in deletion_list]

works well enough.

otherwise

tmp = [x for i in range(len(x)) if not i in deletion_list]
x = tmp

you could speed either up for largish lists with hashing;

dict = {}
for d in deletion_list:
dict[d] = 1
tmp = [x for i in range(len(x)) if not dict.has_key(i)]
x = tmp

alternatively you could make it linear in time for the number of
deletion elements, but the performance hit from creating a bunch of
lists hurts a bit too much, unless you don't care about order. Which
I'm guessing you do.

Jam

 

[Roger Binns]

are deleted first), but this is not very elegant or quick:

Here is a list comprehension that makes a new list
without the deletion_list items. It does have the virtue
of being a one liner

newx=[x for i in range(len(x)) if i not in deletion_list]

You will need to do timings to see what method is actually
the fastest. Don't forget to do so with list sizes the
same as you expect to be using in your real program. Search
archives for this group for 'timeit' to see examples of
how to do timings.

(I think that the size of the deletion_list as a proportion
of the original list is what best determines the relative
speeds of the various methods available).

Roger

 

[Henk-Jan de Jong]

Hello,

this should work:

for i in deletion_list:
x.remove(i)

Matuka

 

[Steven Rumbalski]

you could speed either up for largish lists with hashing;

dict = {}
for d in deletion_list:
dict[d] = 1
tmp = [x for i in range(len(x)) if not dict.has_key(i)]
x = tmp

If order doesn't matter and the items are unique you may wish to use
sets.Set from python 2.3 instead of a dict:

from sets import Set
set = Set(['a', 'b', 2, 42])
for item in deletion_list:

.... set.discard(item)
....Set(['b', 42])