How can i get unique list?

Question

https://code.sololearn.com/cnc5HJk9KngK/?ref=app

Answer 1

Like this convert to set and back to list:
no_list = [22,22,2,1,11,11,2,2,3,3,3,4,5,5,5,55,55,66]
def unique_list(l):
#complete the function's body to return the unique list of numbers
return list(set(l))
print(unique_list(no_list))

Answer 2

You can transform the list into dictionary keys and back into a list.
https://code.sololearn.com/c6otnBb5ELXS/?ref=app

Answer 3

Ipang set is defined as unordered and unique. It doesn't care about the order of the input. But you can always sort the result.

Answer 4

HonFu here is a generator solution for unique elements
https://code.sololearn.com/cZ1BL2zAPXj2/?ref=app

Answer 5

https://code.sololearn.com/c4cIWEspujsd/?ref=app

Answer 6

I am using Paul's list to explain the difference
list1 = [22,22,2,1,11,11,2,2,3,3,3,4,5,5,5,55,55,66]
#We can get unique lists by two ways
1.By using Sets
def func1(l):
return list(set(l))
print(func1(list1))
But here our result is going to be unordered, because sets are unordered.
So to preserve the order of the list we can use 2nd way.
2.With Python 3.7 Insertion and deletion in dictionaries is ordered.
So we can use 'fromkeys' method to create a unique list as following -
def func2(l):
return list({}.fromkeys(l))
print(func2(list1))
With this way the order of the list is preserved.
If anybody have still doubt you can message me anytime.

Answer 7

Paul Jacobs
How can we preserve the order for the resulting list? as I tried this code it appears the order for the numbers changed (sequentially)

Answer 8

If you want to maintain the order, loop over list a and append an element to list b if it's not 'in' there yet.

Answer 9

HonFu
Thank you.
I tried that, and it works, I was wondering if there's a way without an additional list.
Paul Jacobs
Thank you.
Yes I understand that much, I just thought that the set will add the numbers sequentialy, as they appear in the source list, ignoring duplicates of course.

Answer 10

Ipang, you could also go by generator function, yielding every value separately if it's not 'in' the original list up to index.
Or you loop and print out each item if it's not in the part of the list up to that index, but if you use slices for that, you'll end up copying even more.
You can manually inner-loop until index and check by == if it was there...
It seems all cumbersome. I think it's natural to create a subsum list, or tuple, or if you strictly want no copy, generator.
In short: If you don't create a 'sequence', you can't rely well on the order.
EDIT: Not even sure anymore that I'd know a way to create a generator for that.

Answer 11

Ipang problem solved !
Thank you

Answer 12

Tibor Santa, ha, nice! 😀

Answer 13

Step 1 - Convert List into Set (It remove the duplicate value ).
Step 2 - Convert Set into list .

Answer 14

What's wrong in this guys?
text = str(input())
def reverse(text):
return text[::-1]
print("the reversed text is: "+reverse(text))

Answer 15

Yessine Agrebi
Just replace the + with a comma when printing the result
print("the reverse text is:", reverse(text))

Answer 16

Ipang
This is the error :
StdinNotImplementedError: raw_input was called, but this frontend does not support input requests.

Answer 17

Yessine Agrebi
`raw_input` is for Python 2, `input` is for Python 3. I don't know what the "frontend" word there even mean.

Answer 18

use set() method

Answer 19

2²we or