Python filter a list to only leave objects that occur once

Question

I would like to filter this list,

l = [0,1,1,2,2]

to only leave,

[0].

I\'m struggling to do it in a \'pythoni

Answer 1

I think the actual timings are kind of interesting:

Alex' answer:

python -m timeit -s "l = range(1,1000,2) + range(1,1000,3); import collections" "d = collections.defaultdict(int)" "for x in l: d[x] += 1" "l[:] = [x for x in l if d[x] == 1]"
1000 loops, best of 3: 370 usec per loop

Mine:

python -m timeit -s "l = range(1,1000,2) + range(1,1000,3)" "once = set()" "more = set()" "for x in l:" " if x not in more:" "  if x in once:" "   more.add(x)" "   once.remove( x )" "  else:" "   once.add( x )"
1000 loops, best of 3: 275 usec per loop

sepp2k's O(n**2) version, to demonstrate why compexity matters ;-)

python -m timeit -s "l = range(1,1000,2) + range(1,1000,3)" "[x for x in l if l.count(x)==1]"
100 loops, best of 3: 16 msec per loop

Roberto's + sorted:

python -m timeit -s "l = range(1,1000,2) + range(1,1000,3); import itertools" "[elem[0] for elem in itertools.groupby(sorted(l)) if elem[1].next()== 0]"
1000 loops, best of 3: 316 usec per loop 

mhawke's:

python -m timeit -s "l = range(1,1000,2) + range(1,1000,3)" "d = {}" "for i in l: d[i] = d.has_key(i)" "[k for k in d.keys() if not d[k]]"
1000 loops, best of 3: 251 usec per loop

I like the last, clever and fast ;-)

Answer 2

Here's another dictionary oriented way:

l = [0, 1, 1, 2, 2]
d = {}
for i in l: d[i] = i in d

[k for k in d if not d[k]]  # unordered, loop over the dictionary
[k for k in l if not d[k]]  # ordered, loop over the original list

Answer 3

In the same spirit as Alex's solution you can use a Counter/multiset (built in 2.7, recipe compatible from 2.5 and above) to do the same thing:

In [1]: from counter import Counter

In [2]: L = [0, 1, 1, 2, 2]

In [3]: multiset = Counter(L)

In [4]: [x for x in L if multiset[x] == 1]
Out[4]: [0]

Answer 4

[x for x in the_list if the_list.count(x)==1]

Though that's still a nested loop behind the scenes.

Answer 5

l = [0,1,2,1,2]
def justonce( l ):
    once = set()
    more = set()
    for x in l:
        if x not in more:
            if x in once:
                more.add(x)
                once.remove( x )
            else:
                once.add( x )
    return once

print justonce( l )

Answer 6

>>> l = [0,1,1,2,2]
>>> [x for x in l if l.count(x) is 1]
[0]