Remove duplicate tuples from a list if they are exactly the same including order of items

Question

I know questions similar to this have been asked many, many times on Stack Overflow, but I need to remove duplicate tuples from a list, but not just if their elements match up, their elements have to be in the same order. In other words, (4,3,5) and (3,4,5) would both be present in the output, while if there were both(3,3,5) and (3,3,5), only one would be in the output.

Specifically, my code is:

import itertools

x = [1,1,1,2,2,2,3,3,3,4,4,5]
y = []

for x in itertools.combinations(x,3):
    y.append(x)
print(y)

of which the output is quite lengthy. For example, in the output, there should be both (1,2,1) and (1,1,2). But there should only be one (1,2,2).

Answer 1

What you need is unique permutations rather than combinations:

y = list(set(itertools.permutations(x,3)))

That is, (1,2,2) and (2,1,2) will be considered as same combination and only one of them will be returned. They are, however, different permutations. Use set() to remove duplicates.

If afterwards you want to sort elements within each tuple and also have the whole list sorted, you can do:

y = [tuple(sorted(q)) for q in y]
y.sort()

Answer 2

You were really close. Just get permutations, not combinations. Order matters in permutations, and it does not in combinations. Thus (1, 2, 2) is a distinct permutation from (2, 2, 1). However (1, 2, 2) is considered a singular combination of one 1 and two 2s. Therefore (2, 2, 1) is not considered a distinct combination from (1, 2, 2).

You can convert your list y to a set so that you remove duplicates...

import itertools

x = [1,1,1,2,2,2,3,3,3,4,4,5]
y = []

for x in itertools.permutations(x,3):
    y.append(x)
print(set(y))

And voila, you are done. :)

Answer 3

This will probably do what you want, but it's vast overkill. It's a low-level prototype for a generator that may be added to itertools some day. It's low level to ease re-implementing it in C. Where N is the length of the iterable input, it requires worst-case space O(N) and does at most N*(N-1)//2 element comparisons, regardless of how many anagrams are generated. Both of those are optimal ;-)

You'd use it like so:

>>> x = [1,1,1,2,2,2,3,3,3,4,4,5]
>>> for t in anagrams(x, 3):
...     print(t)
(1, 1, 1)
(1, 1, 2)
(1, 1, 3)
(1, 1, 4)
(1, 1, 5)
(1, 2, 1)
...

There will be no duplicates in the output. Note: this is Python 3 code. It needs a few changes to run under Python 2.

import operator

class ENode:
    def __init__(self, initial_index=None):
        self.indices = [initial_index]
        self.current = 0
        self.prev = self.next = self

    def index(self):
        "Return current index."
        return self.indices[self.current]

    def unlink(self):
        "Remove self from list."
        self.prev.next = self.next
        self.next.prev = self.prev

    def insert_after(self, x):
        "Insert node x after self."
        x.prev = self
        x.next = self.next
        self.next.prev = x
        self.next = x

    def advance(self):
        """Advance the current index.

        If we're already at the end, remove self from list.

        .restore() undoes everything .advance() did."""

        assert self.current < len(self.indices)
        self.current += 1
        if self.current == len(self.indices):
            self.unlink()

    def restore(self):
        "Undo what .advance() did."
        assert self.current <= len(self.indices)
        if self.current == len(self.indices):
            self.prev.insert_after(self)
        self.current -= 1

def build_equivalence_classes(items, equal):
    ehead = ENode()
    for i, elt in enumerate(items):
        e = ehead.next
        while e is not ehead:
            if equal(elt, items[e.indices[0]]):
                # Add (index of) elt to this equivalence class.
                e.indices.append(i)
                break
            e = e.next
        else:
            # elt not equal to anything seen so far:  append
            # new equivalence class.
            e = ENode(i)
            ehead.prev.insert_after(e)
    return ehead

def anagrams(iterable, count=None, equal=operator.__eq__):
    def perm(i):
        if i:
            e = ehead.next
            assert e is not ehead
            while e is not ehead:
                result[count - i] = e.index()
                e.advance()
                yield from perm(i-1)
                e.restore()
                e = e.next
        else:
            yield tuple(items[j] for j in result)

    items = tuple(iterable)
    if count is None:
        count = len(items)
    if count > len(items):
        return

    ehead = build_equivalence_classes(items, equal)
    result = [None] * count
    yield from perm(count)

Answer 4

Using a set should probably work. A set is basically a container that doesn't contain any duplicated elements.

Python also includes a data type for sets. A set is an unordered collection with no duplicate elements. Basic uses include membership testing and eliminating duplicate entries. Set objects also support mathematical operations like union, intersection, difference, and symmetric difference.

import itertools

x = [1,1,1,2,2,2,3,3,3,4,4,5]
y = set()

for x in itertools.combinations(x,3):
    y.add(x)
print(y)

Answer 5

No need to do for loop, combinations gives a generator.

x = [1,1,1,2,2,2,3,3,3,4,4,5]
y = list(set(itertools.combinations(x,3)))

Answer 6

set will take care of that:

>>> a = [(1,2,2), (2,2,1), (1,2,2), (4,3,5), (3,3,5), (3,3,5), (3,4,5)]
>>> set(a)
set([(1, 2, 2), (2, 2, 1), (3, 4, 5), (3, 3, 5), (4, 3, 5)])
>>> list(set(a))
[(1, 2, 2), (2, 2, 1), (3, 4, 5), (3, 3, 5), (4, 3, 5)]
>>>

set will remove only exact duplicates.