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pythonobject
Arturas Lapinskas
Arturas Lapinskas

Reputation: 336

Python, how to check object is in list?

I have an object to store data: 

Vertex(key, children)

and a list to store this objects 

vertices = []

im using another list 

key_vertices = []

to store keys of my vertices, so i can easy access(withot looping every object), each time i need to check vertex with such key exist, like that:

if key not in self.key_vertices:
    # add a new key to the array
    self.key_vertices.append(key)
    # create a vertex object to store dependency information
    self.verteces.append(Vertex(key, children))

i think it a bit complicated, maybe someone now better way, to store multiple Vertices objects with ability easy to check and access them

Thanks

Upvotes: 0

Views: 475

Answers (3)

Jean-François Fabre
Jean-François Fabre

Reputation: 140148

your example works fine, the only problem you could have is a performance issue with the in operator for list which is O(n).

If you don't care about the order of the keys (which is likely), just do this:

self.key_vertices = set()

then:

if key not in self.key_vertices:
    # add a new key to the array
    self.key_vertices.add(key)
    # create a vertex object to store dependency information
    self.verteces.append(Vertex(key, children))

you'll save a lot of time in the in operator because set in is way faster due to key hashing.

And if you don't care about order in self.verteces, just do a dictionary, and in that case, you probably don't need the first key parameter to your Vertex structure.

self.verteces = dict()

if key not in self.verteces:
    # create a vertex object to store dependency information
    self.verteces[key] = Vertex(children)

Upvotes: 4

Matt3o12
Matt3o12

Reputation: 4332

If I understand correctly, you want to check if an element has already been added for O(1) (i.e. that you do not have to check every element in the list).

The easiest way to do that is use a set. A set is an unordered list that allows you to check if an element exists with a constant time O(1). You can think of a set like a dict with keys only but it works just like a list:

for value in mySet:
     print(value)

print("hello" in mySet)

If you need an ordered list (most of the time, you don't), your approach is pretty good but I would use a set instead:

self.vertece_set = set() # Init somewhere else ;)

if key not in self.vertece_set:
    # add a new key to the array
    self.vertece_set.add(key)
    # create a vertex object to store dependency information
    self.verteces.append(Vertex(key, children))

Upvotes: 1

Javier
Javier

Reputation: 2776

When you need to check for membership, a list is not the best choice as every object in the list will be checked.

If key is hashable, use a set. If it's not hashable but is comparable, use a tree (unavailable in the standard library). Try to make it hashable.

Upvotes: 1

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