[Tutor] Duplicate items in list

[Manprit Singh]

Dear Mats Wichmann,

Here in this question my purpose is not to remove duplicates from the
list.This question is about removing duplicates from the groups.
see  the list is lst = [3, 3, 5, 5, 5, 6, 6, 6, 5, 9, 3, 3, 3, 5, 5], the
output i need is :
3
5
6
5
9
3
5
As you can see 3 and 5 are still repeating in the output ,  there are 2
groups of 3 in the list, i have to write a program that gives a single
three for each group of 3, in that way 3 will occur in the output  for 2
times, for the same reason 5 will occur in the output for 3 times .

Regards
Manprit Singh


On Fri, Oct 9, 2020 at 9:47 PM Mats Wichmann <mats at wichmann.us> wrote:

> 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)]
> _______________________________________________
> Tutor maillist  -  Tutor at python.org
> To unsubscribe or change subscription options:
> https://mail.python.org/mailman/listinfo/tutor
>