维护冻结集中元素的顺序

Question

I have a list of tuples, each tuple of which contains one string and two integers. The list looks like this:

x = [('a',1,2), ('b',3,4), ('x',5,6), ('a',2,1)]

The list contains thousands of such tuples. Now if I want to get unique combinations, I can do the frozenset on my list as follows:

y = set(map(frozenset, x))

This gives me the following result:

{frozenset({'a', 2, 1}), frozenset({'x', 5, 6}), frozenset({3, 'b', 4})}

I know that set is an unordered data structure and this is normal case but I want to preserve the order of the elements here so that I can thereafter insert the elements in a pandas dataframe. The dataframe will look like this:

 Name  Marks1  Marks2
0    a       1       2
1    b       3       4
2    x       5       6

Answer 1

Instead of operating on the set of frozenset s directly you could use that only as a helper data-structure - like in the unique_everseen recipe in the itertools section (copied verbatim):

from itertools import filterfalse

def unique_everseen(iterable, key=None):
    "List unique elements, preserving order. Remember all elements ever seen."
    # unique_everseen('AAAABBBCCDAABBB') --> A B C D
    # unique_everseen('ABBCcAD', str.lower) --> A B C D
    seen = set()
    seen_add = seen.add
    if key is None:
        for element in filterfalse(seen.__contains__, iterable):
            seen_add(element)
            yield element
    else:
        for element in iterable:
            k = key(element)
            if k not in seen:
                seen_add(k)
                yield element

Basically this would solve the issue when you use key=frozenset :

>>> x = [('a',1,2), ('b',3,4), ('x',5,6), ('a',2,1)]

>>> list(unique_everseen(x, key=frozenset))
[('a', 1, 2), ('b', 3, 4), ('x', 5, 6)]

This returns the elements as-is and it also maintains the relative order between the elements.

Answer 2

No ordering with frozensets. You can instead create sorted tuples to check for the existence of an item, adding the original if the tuple does not exist in the set:

y = set()
lst = []
for i in x:
    t = tuple(sorted(i, key=str)
    if t not in y:
         y.add(t)
         lst.append(i)
print(lst)
# [('a', 1, 2), ('b', 3, 4), ('x', 5, 6)]

The first entry gets preserved.

Answer 3

There are some quite useful functions in NumPy which can help you to solve this problem.

import numpy as np
chrs, indices = np.unique(list(map(lambda x:x[0], x)), return_index=True)
chrs, indices
>> (array(['a', 'b', 'x'], 
   dtype='<U1'), array([0, 1, 2]))
[x[indices[i]] for i in range(indices.size)]
>> [('a', 1, 2), ('b', 3, 4), ('x', 5, 6)]