In Java we have HashSet<Integer>
, I need similar structure in Python to use contains like below:
A = [1, 2, 3] S = set() S.add(2) for x in A: if S.contains(x): print "Example"
Could you please help?
HashSet is used to store the values using a hash table. In this tutorial, we will cover HashSet in Python. We will also learn about how we can design HashSet in Python. HashSet data structure can be created without using any built-in hash table libraries.
On average, the contains() of HashSet runs in O(1) time. Getting the object's bucket location is a constant time operation. Taking into account possible collisions, the lookup time may rise to log(n) because the internal bucket structure is a TreeMap.
To check if the Set contains an element in Python, use the in keyword, which returns True if the specified Set contains an element and False otherwise. The in keyword checks if the item is present in a sequence like a list, range, string, set, etc.
Just use a set:
>>> l = set() >>> l.add(1) >>> l.add(2) >>> 1 in l True >>> 34 in l False
The same works for lists:
>>> ll = [1,2,3] >>> 2 in ll True >>> 23 in ll False
Edit: Note @bholagabbar's comment below that the time complexity for in
checks in lists and tuples is O(n) on average (see the python docs here), whereas for sets it is on average O(1) (worst case also O(n), but is very uncommon and might only happen if __hash__
is implemented poorly).
In Python, there is a built-in type, set
. The major difference from the hashmap
in Java is that the Python set
is not typed, i.e., it is legal to have a set {'2', 2}
in Python.
Out of the box, the class set
does not have a contains()
method implemented. We typically use the Python keyword in
to do what you want, i.e.,
A = [1, 2, 3] S = set() S.add(2) for x in A: if x in S: print("Example")
If that does not work for you, you can invoke the special method __contains__()
, which is NOT encouraged.
A = [1, 2, 3] S = set() S.add(2) for x in A: if S.__contains__(x): print("Example")
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