从排序列表python打印唯一数字

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

1
2
5
6
9
11
12
14
42

eg all values that only appear once in a row.

Answer 1

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

Answer 2

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() )

Answer 3

you can do by this:

print (set(YOUR_LIST))

or if you need a list use this:

print (list(set(YOUR_LIST)))

Answer 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.