print unique numbers from a sorted list python

Question

I'm trying to print all elements in a sorted list that only occur once.

My code below works but I'm sure there is a better way:

def print_unique(alist):
    i = 0
    for i in range(len(alist)):
        if i < (len(alist)-1):
            if alist[i] == alist[i+1]:
                i+=1
                if alist[i] == alist[i-1]:
                    i+=1
            elif  alist[i] == alist[i-1]:
                  i+=1    
            else:
              print alist[i]
        else:
            if alist[-1]!= alist[-2]:
                print alist[-1]

randomlist= [1,2,3,3,3,4,4,5,6,7,7,7,7,8,8,8,9,11,12,14,42]
print_unique(randomlist)

This produces

e.g. all values that only appear once in a row.

Martijn's code is very efficient, but (like your code) it does assume that `alist` is already sorted. — PM 2Ring, Jul 16 '16 at 10:10
@MartijnPieters OP's results only includes items that are not duplicates. Too bad the question was not well delivered. So the link to the dupe does not actually solve their problem. A slightly modified extension of the solution you proposed does it: `print [k for k, g in groupby(randomlist) if len(list(g)) == 1]` — Moses Koledoye, Jul 16 '16 at 10:15
@MosesKoledoye: right, yes, and the OP was entirely to blame by not including a clear problem statement. — Martijn Pieters, Jul 16 '16 at 10:16

Martijn Pieters · Answer 1 · 2016-07-16T10:30:41.760

You could use the itertools.groupby() function to group your inputs and filter on groups that are one element long:

from itertools import groupby

def print_unique(alist):
    for elem, group in groupby(alist):
        if sum(1 for _ in group) == 1:  # count without building a new list
            print elem

or if you want to do it 'manually', track the last item seen and if you have seen it more than once:

def print_unique(alist, _sentinel=object()):
    last, once = _sentinel, False
    for elem in alist:
        if elem == last:
            once = False
        else:
            if once:
                print last
            last, once = elem, True
    if last is not _sentinel and once:
        print last

You may want to replace the print statements with yield and leave printing to the caller:

def filter_unique(alist):
    for elem, group in groupby(alist):
        if sum(1 for _ in group) == 1:  # count without building a new list
            yield elem

for unique in filter_unique(randomlist):
    print unique

Amal Rajan · Answer 2 · 2016-07-16T10:57:30.437

-1

This question seems to have duplicates.

If you do not wish to preserve the order of your list, you can do

print list(set(sample_list))

You can also try,

unique_list = []
for i in sample_list:
    if i not in unique_list:
        unique_list.append(i)

EDIT:

In order to print all the elements in the list so that they appear once in a row, you can try this

print '\n'.join([str(i) for i in unique_list])

And, as @martijn-pieters mentioned it in the comments, the first code was found to be very fast compared to the second one, when I did a small benchmark. On a list of 10^5 elements, the second code took 63.66 seconds to complete whereas the first one took a mere 0.2200 seconds. (on a list generated using random.random())

edited Jul 16 '16 at 10:57

answered Jul 16 '16 at 09:57

Amal Rajan

153
1
11

3

So don't answer, flag as a duplicate. And your second solution takes O(NK) time as the `unique_list` is scanned again and again to test membership. – Martijn Pieters Jul 16 '16 at 09:57
I still like this answer, pretty sure the OP can learn alot from it – RoadRunner Jul 16 '16 at 10:05
Thanks for the info, @martijn-pieters. But, I cannot mark it as duplicate with my current reputation :) – Amal Rajan Jul 16 '16 at 10:08
You only need 15 points to [flag posts](http://stackoverflow.com/help/privileges/flag-posts). – PM 2Ring Jul 16 '16 at 10:12
@PM2Ring: but more to flag as duplicate. – Martijn Pieters Jul 16 '16 at 10:24
@MartijnPieters: Ah, ok. I should have read the fine print. :oops: – PM 2Ring Jul 16 '16 at 10:28
1

Sorry for the misleading info in my previous comment, Amal. OTOH, even when you don't have sufficient rep to flag as duplicate, you can still do a search for dupe targets and post a link in a comment. Sure, it's not as good as a dupe flag, but it still lets potential answerers know that the question may be a dupe, and it assists those who _do_ have sufficient rep to flag or close-vote dupes. – PM 2Ring Jul 16 '16 at 10:30
BTW the specs of the problem have changed slightly since you wrote your answer. – PM 2Ring Jul 16 '16 at 10:33

score -3 · Answer 3 · answered Jul 16 '16 at 09:51

-3

you can do by this:

print (set(YOUR_LIST))

or if you need a list use this:

print (list(set(YOUR_LIST)))

answered Jul 16 '16 at 09:51

Thomas Anderson

74
1
1
9

score -4 · Answer 4 · answered Jul 16 '16 at 09:52

-4

Sets are lists containing unique items. If you construct a set from the array, it will contain only the unique items:

def print_unique(alist):
   print set( alist )

The input list does not need to be sorted.

answered Jul 16 '16 at 09:52

Tyler Durden

11,156
9
64
126

print unique numbers from a sorted list python

4 Answers4