I lost a little bit of time in this Python for statement:
class MyListContainer:
def __init__(self):
self.list = []
def purge(self):
for object in self.list:
if (object.my_cond()):
self.list.remove(object)
return self.list
container = MyListContainer()
# now suppose both obj.my_cond() return True
obj1 = MyCustomObject(par)
obj2 = MyCustomObject(other_par)
container.list = [obj1, obj2]
# returning not an empty list but [obj2]
container.purge()
It doesn't work as I expected because when the cycle in "purge" delete the first object in list the second one is shifted to the beginning of the list and the cycle is ended.
I solved duplicating self.list before the for cycle:
...
local_list = self.list[:]
for object in local_list:
...
I suppose that the for statement stop working because I'm changing the length of the original list. Can someone clarify this point ?
And is there a more "elegant" way to solve this problem ? If I have more than few elements inside the list, duplicating it every time does not seem a good idea.
Maybe the filter() function is the right one but i whish to have some other approach if any.
I'm a newbie.
To summarize your useful answers:
- Never edit a list you are looping
- Duplicate the list or use list comprehensions
- Duplicating a list could not waste your memory or in this case who's mind about it
Filter (or list comprehension) IS the way to go. If you want to do it inplace, something like this would work:
purge = []
for i,object in enumerate(self.list):
if object.mycond()
purge.append(i)
for i in reversed(purge):
del self.list[i]
Or alternatively, the purge list can be made with a comprehension, a shortcut version looks like:
for i in reversed([ i for (i,o) in enumerate(self.list) if o.mycond() ]):
del self.list[i]
Don't try. Just don't. Make a copy or generate a new list.
Just make yourself a new list:
def purge(self):
self.list = [object for object in self.list if not object.my_cond()]
return self.list
Reserve any optimization until you've profiled and found that this method really is the bottleneck of your application. (I bet it won't be.)
In python variables are actually labels to data. Duplicating a list is, for the most part, making a new set of pointers to the data from the first list. Don't feel too bad about it.
List comprehensions are your friend.
e.g.
>>> a = range(20)
>>> a
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19]
>>> [ x for x in a if x % 2 == 0 ]
[0, 2, 4, 6, 8, 10, 12, 14, 16, 18]
Your second solution in which you duplicate the list is the right way to go. Afterward you can just replace the old list with the duplicate if need be.
It's perfectly safe to shorten the list in place if you do it in reverse!
>>> a=range(20)
>>> for i in reversed(range(len(a))):
... if a[i]%2: del a[i]
...
>>> a
[0, 2, 4, 6, 8, 10, 12, 14, 16, 18]
Another way is to reassign the whole slice
>>> a=range(20)
>>> a[:]=(x for x in a if not x%2)
>>> a
[0, 2, 4, 6, 8, 10, 12, 14, 16, 18]
If the items in the list are unique, this works too
>>> a=range(20)
>>> for item in reversed(a):
... if item%2: a.remove(item)
...
>>> a
[0, 2, 4, 6, 8, 10, 12, 14, 16, 18]
Here is some more explanation in response to yuri's comment
Suppose we have
>>> a=[0,1,2,3,4,5]
Now trying naively to delete the 3rd and 4th items
>>> del a[3]
>>> del a[4]
>>> a
[0, 1, 2, 4] # didn't work because the position of all the item with index >=3 was changed
However if we do the del's in the opposite order
>>> a=[0,1,2,3,4,5]
>>> del a[4]
>>> del a[3]
>>> a
[0, 1, 2, 5] # this is the desired result
Now extend that idea over a for loop with a removal condition, and you see that removal from the live list is possible
indeces = []
minus = 0
for i in range(self.list):
if cond(self.list[i]):
indeces.append(i)
for i in indeces:
self.list = self.list[:(i-minus)].extend(self.list[i-minus+1:])
来源：https://stackoverflow.com/questions/2233388/how-to-delete-list-elements-while-cycling-the-list-itself-without-duplicate-it