CODEWITHSUNDEEP
pythonpython-3.xlistlist-comprehension
the.light_bringer
the.light_bringer

Reputation: 31

Making an array from another defined array with no duplicates using list comprehension

I had tried to make a list free of duplicates from another defined list using list comprehension as follows,

num = [1, 2, 2, 2, 3, 3, 4, 4, 4, 5]
my_list = []
my_list = [x for x in num if x not in my_list]

Upon calling my_list, I get back the same array

[1, 2, 2, 2, 3, 3, 4, 4, 4, 5]

Could anyone kindly explain why this is happening?

Upvotes: 3

Views: 67

Answers (4)

Ch3steR
Ch3steR

Reputation: 20689

You can try this list comprehension.

my_list=[]
num = [1, 2, 2, 2, 3, 3, 4, 4, 4, 5]
my_list=[my_list.append(x) or x for x in num if x not in my_list]
# [1,2,3,4,5]

If you don't care about the order.

my_lst=list(set(num))

From Python3.6 and above dictionaries store the insertion order or use collections.OrderedDict. So, you can try this.

num = [1, 2, 2, 2, 3, 3, 4, 4, 4, 5]
my_list=list(dict.fromkeys(num).keys())

You can use seen set to keep track of the elements that are seen. If an element in in seen don't add it to my_list.

num = [1, 2, 2, 2, 3, 3, 4, 4, 4, 5]
seen=set()
my_list=[]
for x in num:
    if x not in seen:
        seen.add(x)
        my_list.append(x)
my_list
# [1,2,3,4,5]

Upvotes: 2

Bash
Bash

Reputation: 656

my_list isn't updated until after the comprehension is complete. During the comprehension my_list is an empty list.

If order matters the approach you want to take is:

num = [1, 2, 2, 2, 3, 3, 4, 4, 4, 5]
my_list = []
check_set = set()
for x in num:
    if x not in check_set:
        my_list.append(x)
        check_set.add(x)

If order does not matter:

my_list = list(set(num))

Upvotes: 3

Sheri
Sheri

Reputation: 1415

Using set is more feasible than an if condition along with for loop:

num = [1, 2, 2, 2, 3, 3, 4, 4, 4, 5]
my_list = list(set(num))
print(my_list)

Ouput:

[1, 2, 3, 4, 5]

Upvotes: 3

Rajat Mishra
Rajat Mishra

Reputation: 3780

you should try something like this :

In [99]: num = [1, 2, 2, 2, 3, 3, 4, 4, 4, 5]
    ...: my_list = []

In [100]: [my_list.append(item) for item in num if item not in my_list]
Out[100]: [None, None, None, None, None]

In [101]: my_list
Out[101]: [1, 2, 3, 4, 5]

The list is updated after the list comprehension process is complete because which the duplicates are still present.

Upvotes: 2

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