带有条件的列表中的唯一项目

Question

If i have a list in python say

thing = [[20,0,1],[20,0,2],[20,1,1],[20,0],[30,1,1]]

I would want to have a resulting list

thing = [[20,1,1],[20,0,2],[30,1,1]]

That is if the first element is the same, remove duplicates and give priority to the number 1 in the second element. Lastly the 3rd element must also be unique to the first element.

In this previous question we solved a complicated method where for a transaction it details a purchased unit. I want to output other units in that course. If two transactions exist that relate to two units in one course it will display them a duplicate (or times each subsequent unit).

The aim of this question it to ensure that this duplication is stopped. Because of the complication of this solution it has resulted in a series of question. Thanks for everyone that has helped so far.

Answer 1

I am not sure you would like this, but it works with your example:

[list(i) + j for i, j in dict([(tuple(x[:2]), x[2:]) for x in sorted(thing, key=lambda x:len(x))]).items()]

EDIT:

Here a bit more detailed (note that it fits better to your description of the problem, sorting ONLY by the length of each sublist, may not be the best solution):

thing = [[20,0,1],[20,0,2],[20,1,1],[20,0],[30,1,1]]
dico = {}
for x in thing:
    if not tuple(x[:2]) in dico:
        dico[tuple(x[:2])] = x[2:]
        continue
    if tuple(x[:2])[1] < x[1]:
        dico[tuple(x[:2])] = x[2:]

new_thing = []
for i, j in dico.items():
    new_thing.append(list(i) + j)

Answer 2

You might want to try using the unique_everseen function from the itertools recipes .

As a first step, here is a solution excluding [20, 0] :

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

thing = [[20,0,1],[20,0,2],[20,1,1],[30,1,1]]

thing.sort(key=lambda x: 0 if x[1] == 1 else 1)

print(list(unique_everseen(thing, key=lambda x: (x[0], x[2]))))

Output:

[[20, 1, 1], [30, 1, 1], [20, 0, 2]]

Answer 3

thing = [[20,0,1],[20,0,2],[20,1,1],[20,0,1],[30,1,1]]

d = {}
for e in thing:
    k = (e[0], e[2])
    if k not in d or (d[k][1] != 1 and e[1] == 1):
        d[k] = list(e)

print d.values()

[[20, 0, 2], [30, 1, 1], [20, 1, 1]]

if you don't need initial list:

thing = [[20,0,1],[20,0,2],[20,1,1],[20,0,1],[30,1,1]]

d = {}
for e in thing:
    k = (e[0], e[2])
    if k not in d or (d[k][1] != 1 and e[1] == 1):
        d[k] = e

thing = d.values()

[[20, 0, 2], [30, 1, 1], [20, 1, 1]]

if you want to keep order of your lists, use OrderedDict

from collections import OrderedDict
d = OrderedDict()