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Removing duplicates in lists

Essentially I need to compose a program to check if a rundown has any copies and in the event that it does it eliminates them and returns another rundown with the things that weren't copied/taken out. This is the thing that I have however to be straightforward I don't have a clue what to do.
def remove_duplicates():
t = ['a', 'b', 'c', 'd']
t2 = ['a', 'c', 'd']
for t in t2:
t.append(t.remove())
return t
by

2 Answers

Shahlar1vxp
One of the best ways of removing duplicates is by using the set() function in python, and then converting it back into list. The code is as follows-
In [2]: some_list = ['a', 'a', 'v', 'v', 'v', 'c', 'c', 'd']
In [3]: list(set(some_list))
Out [3]: ['a', 'c', 'd', 'v']

Simplifying it,
mylist = list(set(mylist))
You may also use the iterator to remove the duplicates-
def uniqify(iterable):
seen = set()
for item in iterable:
if item not in seen:
seen.add(item)
yield item

This will return an iterator or generator so that you may use it wherever you can use an iterator.
For list:
unique_list = list(uniqify([1, 2, 3, 4, 3, 2, 4, 5, 6, 7, 6, 8, 8]))
print(unique_list)
Sonali7
The best approach to get a unique collection of items is to use a set. Sets are basically the unordered collections of distinct objects. If you want to create a set from any iterable, then simply pass it to the built-in set() function.
For example,
>>> t = [1, 2, 3, 1, 2, 5, 6, 7, 8]
>>> t
[1, 2, 3, 1, 2, 5, 6, 7, 8]
>>> list(set(t))
[1, 2, 3, 5, 6, 7, 8]
>>> s = [1, 2, 3]
>>> list(set(t) - set(s))
[8, 5, 6, 7]

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