Sets cannot contain lists, how to circumvent?

[meijeru]

The easiest way to find out if a list is unique (does not contain duplicates) is to convert it into a set and check the size. However this does not work for a list of lists! Apparently set elements must be hashable... Seems a pity. Are there any plans to remove this constraint? A set of lists must surely have its uses. In the meantime, how could one circumvent this?

[noisefloor]

Please describe in more detail what your are in the end trying to achieve. Very preferable with an example demonstrating your vision of comparing lists of lists.

Regards, noisefloor

[bowlofred]

Do you understand why sets must contain only hashable elements?

While you mention wanting it only for a temporary purpose, sets could be longer lasting and having mutable, non-hashable elements could cause problems.

What would you expect this to output if sets could contain normal lists?

l1 = ["A"]
l2 = ["B"]
s = {l1, l2}
l2[0] = "B"
print(s)

As a workaround, if your list-of-lists is consistent (every element is a list, and every inner list consists of immutable objects (like numbers or strings), then you could do a set comprehension that converts the lists to tuples first.

>>> l = [[1, 2, 3], ["A", "B", "C"], [1, 2, 3]]
>>> {l}
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: unhashable type: 'list'
# Can't add the inner lists as set elements.   So instead we build a set where they are tuples...

>>> s = {tuple(x) for x in l}
>>> s
{(1, 2, 3), ('A', 'B', 'C')}

[Pedroski55]

It is easy to make a list of lists free of duplicate lists:

from random import choice

# simple list
#listig = [i for i in range(9)] * 2
# convoluted list
listig = [[choice([1,2,3]) for j in range(3)] for i in range(20)]

# remove dups
def unique_me(alist):
    uni = []
    for e in alist :
        if not e in uni:
            uni.append(e)
    uni.sort()
    return uni

unique = unique_me(listig)