Removing duplicates from a list of lists

>>> k = [[1, 2], [4], [5, 6, 2], [1, 2], [3], [4]]
>>> k = sorted(k)
>>> k
[[1, 2], [1, 2], [3], [4], [4], [5, 6, 2]]
>>> dedup = [k[i] for i in range(len(k)) if i == 0 or k[i] != k[i-1]]
>>> dedup
[[1, 2], [3], [4], [5, 6, 2]]

I don’t know if it’s necessarily faster, but you don’t have to use to tuples and sets.

Doing it manually, creating a new k list and adding entries not found so far:

k = [[1, 2], [4], [5, 6, 2], [1, 2], [3], [4]]
new_k = []
for elem in k:
    if elem not in new_k:
        new_k.append(elem)
k = new_k
print k
# prints [[1, 2], [4], [5, 6, 2], [3]]

Simple to comprehend, and you preserve the order of the first occurrence of each element should that be useful, but I guess it’s quadratic in complexity as you’re searching the whole of new_k for each element.

Even your “long” list is pretty short. Also, did you choose them to match the actual data? Performance will vary with what these data actually look like. For example, you have a short list repeated over and over to make a longer list. This means that the quadratic solution is linear in your benchmarks, but not in reality.

For actually-large lists, the set code is your best bet—it’s linear (although space-hungry). The sort and groupby methods are O(n log n) and the loop in method is obviously quadratic, so you know how these will scale as n gets really big. If this is the real size of the data you are analyzing, then who cares? It’s tiny.

Incidentally, I’m seeing a noticeable speedup if I don’t form an intermediate list to make the set, that is to say if I replace

kt = [tuple(i) for i in k]
skt = set(kt)

with

skt = set(tuple(i) for i in k)

The real solution may depend on more information: Are you sure that a list of lists is really the representation you need?

Another probably more generic and simpler solution is to create a dictionary keyed by the string version of the objects and getting the values() at the end:

>>> dict([(unicode(a),a) for a in [["A", "A"], ["A", "A"], ["A", "B"]]]).values()
[['A', 'B'], ['A', 'A']]

The catch is that this only works for objects whose string representation is a good-enough unique key (which is true for most native objects).

List of tuple and {} can be used to remove duplicates

>>> [list(tupl) for tupl in {tuple(item) for item in k }]
[[1, 2], [5, 6, 2], [3], [4]]
>>> 

Create a dictionary with tuple as the key, and print the keys.

• create dictionary with tuple as key and index as value
• print list of keys of dictionary

k = [[1, 2], [4], [5, 6, 2], [1, 2], [3], [4]]

dict_tuple = {tuple(item): index for index, item in enumerate(k)}

print [list(itm) for itm in dict_tuple.keys()]

# prints [[1, 2], [5, 6, 2], [3], [4]]

All the set-related solutions to this problem thus far require creating an entire set before iteration.

It is possible to make this lazy, and at the same time preserve order, by iterating the list of lists and adding to a “seen” set. Then only yield a list if it is not found in this tracker set.

This unique_everseen recipe is available in the itertools docs. It’s also available in the 3rd party toolz library:

from toolz import unique

k = [[1, 2], [4], [5, 6, 2], [1, 2], [3], [4]]

# lazy iterator
res = map(list, unique(map(tuple, k)))

print(list(res))

[[1, 2], [4], [5, 6, 2], [3]]

Note that tuple conversion is necessary because lists are not hashable.

This should work.

k = [[1, 2], [4], [5, 6, 2], [1, 2], [3], [4]]

k_cleaned = []
for ele in k:
    if set(ele) not in [set(x) for x in k_cleaned]:
        k_cleaned.append(ele)
print(k_cleaned)

# output: [[1, 2], [4], [5, 6, 2], [3]]

Strangely, the answers above removes the ‘duplicates’ but what if I want to remove the duplicated value also??
The following should be useful and does not create a new object in memory!

def dictRemoveDuplicates(self):
    a=[[1,'somevalue1'],[1,'somevalue2'],[2,'somevalue1'],[3,'somevalue4'],[5,'somevalue5'],[5,'somevalue1'],[5,'somevalue1'],[5,'somevalue8'],[6,'somevalue9'],[6,'somevalue0'],[6,'somevalue1'],[7,'somevalue7']]


print(a)
temp = 0
position = -1
for pageNo, item in a:
    position+=1
    if pageNo != temp:
        temp = pageNo
        continue
    else:
        a[position] = 0
        a[position - 1] = 0
a = [x for x in a if x != 0]         
print(a)

and the o/p is:

[[1, 'somevalue1'], [1, 'somevalue2'], [2, 'somevalue1'], [3, 'somevalue4'], [5, 'somevalue5'], [5, 'somevalue1'], [5, 'somevalue1'], [5, 'somevalue8'], [6, 'somevalue9'], [6, 'somevalue0'], [6, 'somevalue1'], [7, 'somevalue7']]
[[2, 'somevalue1'], [3, 'somevalue4'], [7, 'somevalue7']]

k=[[1, 2], [4], [5, 6, 2], [1, 2], [3], [5, 2], [3], [8], [9]]
kl=[]
kl.extend(x for x in k if x not in kl)
k=list(kl)
print(k)

which prints,

[[1, 2], [4], [5, 6, 2], [3], [5, 2], [8], [9]]

A bit of a background, I just started with python and learnt comprehensions.

k = [[1, 2], [4], [5, 6, 2], [1, 2], [3], [4]]
dedup = [elem.split('.') for elem in set(['.'.join(str(int_elem) for int_elem in _list) for _list in k])]

a_list = [
          [1,2],
          [1,2],
          [2,3],
          [3,4]
]

print (list(map(list,set(map(tuple,a_list)))))

outputs: [[1, 2], [3, 4], [2, 3]]

The simplest solution is to convert a list of lists into a list of tuples and then apply dict.fromkeys() method then convert it back to the list.

for example:

you have k = [[1, 2], [4], [5, 6, 2], [1, 2], [3], [4]]

Convert to list of tuples k = list(map(tuple, k))

This will give you [(1, 2), (4,), (5, 6, 2), (1, 2), (3,), (4,)]

Then do the following: unique = list(dict.fromkeys(k))

You will have [(1, 2), (4,), (5, 6, 2), (3,)]

That’s all.

If the complaint is not with ‘not fast’ per se but with ‘not concise enough’ part of your proposed solution, then in Python 3.5+ with help of unpacking operator and concise tuple notation you can make chained data structure conversions extremely brief (granted, this is still O(n^2), but still unpacking is slightly faster than direct conversion):

Input:

k = [[1, 2], [4], [5, 6, 2], [1, 2], [3], [4]]
k = [*map(list, {*map(tuple, k)})]

# If you prefer comprehensions to map()
# k = [[*t] for t in {(*l,) for l in k}]

# Order-preserving alternative:
# k = [*map(list, dict.fromkeys(map(tuple, k)))]

print(k)

Output:

[[1, 2], [4], [5, 6, 2], [3]]