Grab unique tuples in python list, irrespective of order

Using a set will remove duplicates, and you create a list from it afterwards:

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

you could simply do

y = np.unique(x, axis=0)
z = [] 
for i in y:
   z.append(tuple(i))

The reason is that a list of tuples is interpreted by numpy as a 2D array. By setting axis=0, you'd be asking numpy not to flatten the array and return unique rows.

If order does not matter

If the order of the result is not critical, you can convert your list to a set (because tuples are hashable) and convert the set back to a list:

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

If order matters

(UPDATE)

As of CPython 3.6 (or any Python 3.7 version) regular dictionaries remember their insertion order, so you can simply issue.

>>> l = [(2,2),(2,3),(1,4),(2,2)]
>>> list(dict.fromkeys(l))
[(2, 2), (2, 3), (1, 4)]

---

(OLD ANSWER)

If the order is important, the canonical way to filter the duplicates is this:

>>> seen = set()
>>> result = []
>>> for item in l:
...     if item not in seen:
...         seen.add(item)
...         result.append(item)
... 
>>> result
[(2, 2), (2, 3), (1, 4)]

Finally, a little slower and a bit more hackish, you can abuse an OrderedDict as an ordered set:

>>> from collections import OrderedDict
>>> OrderedDict.fromkeys(l).keys() # or list(OrderedDict.fromkeys(l)) if using a version where keys() does not return a list
[(2, 2), (2, 3), (1, 4)]