delete items from a set while iterating over it

Question

I have a set myset, and I have a function which iterates over it to perform some operation on its items and this operation ultimately deletes the item from the

Answer 1

Another way could be :

s=set()
s.add(1)
s.add(2)
s.add(3)
s.add(4)
while len(s)>0:
    v=next(iter(s))
    s.remove(v)

Answer 2

First, using a set, as Zero Piraeus told us, you can

myset = set([3,4,5,6,2])
while myset:
    myset.pop()
    print(myset)

I added a print method giving these outputs

>>> 
set([3, 4, 5, 6])
set([4, 5, 6])
set([5, 6])
set([6])
set([])

If you want to stick to your choice for a list, I suggest you deep copy the list using a list comprehension, and loop over the copy, while removing items from original list. In my example, I make length of original list decrease at each loop.

l = list(myset)
l_copy = [x for x in l]
for k in l_copy:
    l = l[1:]
    print(l)

gives

>>> 
[3, 4, 5, 6]
[4, 5, 6]
[5, 6]
[6]
[]

Answer 3

Use the copy library to make a copy of the set, iterate over the copy and remove from the original one. In case you want to stick to the for loop and you want to iterate over one element only once - comes in handy if you don't necessarily want to remove all the elements.

import copy
for item in copy.copy(myset):
    myset.remove(item)

Answer 4

Let's return all even numbers while modifying current set.

myset = set(range(1,5))
myset = filter(lambda x:x%2==0, myset)
print myset

Will return

>>> [2, 4]

If there is opportunity use always use lambda it will make your life easier.

Answer 5

This ought to work:

while myset:
    item = myset.pop()
    # do something

Or, if you need to remove items conditionally:

def test(item):
    return item != "foo"  # or whatever

myset = set(filter(test, myset))