[Tutor] Duplicate items in list

[Mats Wichmann]

On 10/9/20 9:37 AM, Manprit Singh wrote:
> Dear sir,
> 
> Consider a list as given below :
> 
> lst = [3, 3, 5, 5, 5, 6, 6, 6, 5, 9, 3, 3, 3, 5, 5]
> I need to print the values  as given below  and in same order:
> 
> 3       # print 3 only once as there are two occurrence of 3 in the
> beginning  in list
> 5       # print 5 only once as there are 3 occurrence of 5 after 3 in the
> list
> 6       # print 6 only once as there are 3 occurrence of 6 after 5 in the
> list
> 5       # print 5 only once as there is single occurrence of 5 after 6 in
> the list
> 9       # print 9 only once as there is single occurrence of 9 after 5 in
> the list
> 3       # print 3 only once as there  are 3 occurrence of 3 after 9 in the
> list
> 5       # print 5 only once as there are 2 occurrence of 5 in the last
> 
> I have written the code as given below:
> lst = [3, 3, 5, 5, 5, 6, 6, 6, 5, 9, 3, 3, 3, 5, 5]
> for i, j in enumerate(lst[:-1]):
>     if lst[i+1] != j:
>         print(j)
> print(lst[-1])
> 
> which gives the answer as given above
> I feel that a better code can be written for this problem , Need your
> guidance.

Turns out there are a ton of ways to get the unique values from a list.
 The problem comes up often, usually in the context of "what is the
fastest way...".  The answer depends on whether you need to preserve the
order or not, and what degree of cleverness feels comfortable in your code.

 If order doesn't matter, using a set will give you unique

uniques = set(lst)
# and to print:
print(*uniques)

If order does matter - as you have indicated it does, it is now possible
to do with with a dictionary operation. This wins for speed, but isn't
necessarily the "most obvious":

uniques = list(dict.fromkeys(lst))

dict.fromkeys constructs (using dict's alternate constructor) a
dictionary given a list of keys (the values will all be None).  When you
then convert the dictionary to a list, that is implicitly like doing
list(d.keys()), but you don't have to specify the keys(). Since
dictionary keys by definition must be unique, this uniquifies.

Other options exist, as noted...

One is to use a set, but only as a tracking devices:

seen = set()
uniques = [x for x in lst if x not in seen and not seen.add(x)]