Python：基于项目对将列表分组

Question

My program generates lists like this:

mydata = ["foo", "bar", "baz", "quux", "quid", "quo"]

And I know from other data that these can be grouped in couples (here a list of tuples, but can be changed to whatever):

static_mapping = [("foo", "quo"), ("baz", "quux"), ("quid", "bar")]

There's no ordering in the couples.

Now on to the problem: my program generates mydata and I need to group data by couple but keeping a separate list of non-matched items. The reason is that at any moment mydata may not contain all items that are part of the couples.

Expected results on such a hypothetical function:

 mydata = ["foo", "bar", "quo", "baz"]
 couples, remainder = group_and_split(mydata, static_mapping)
 print(couples)
 [("foo", "quo")]
 print(remainder)
 ["bar", "baz"]

EDIT: Examples of what I've tried (but they stop at finding the coupling):

found_pairs = list()
for coupling in static_mapping:
    pairs = set(mydata).intersect(set(coupling))
    if not pairs or len(pairs) != 2:
        continue
    found_pairs.append(pairs)

I got stuck at finding a reliable way to get the reminder out.

Answer 1

You may try this:

import copy

def group_and_split(mydata, static_mapping):
    remainder = copy.deepcopy(mydata)
    couples = []
    for couple in static_mapping:
        if couple[0] in mydata and couple[1] in mydata:
            remainder.remove(couple[0])
            remainder.remove(couple[1])
            couples.append(couple)
    return [couples, remainder]

Answer 2

Set gives you faster runtime if values are big, but takes memory, and deepcopy keeps the original data intact.

One of the implementations of the hypothetical functions could be :-

from copy import deepcopy

def group_and_split(mydata, static_mapping):
        temp  = set(mydata)
        couples = []
        remainder = deepcopy(mydata)
        for value1,value2 in static_mapping:
                if value1 in temp and value2 in temp:
                        couples.append((value1,value2))
                        remainder.remove(value1)
                        remainder.remove(value2)
        return couples, remainder

---

Answer 3

Here you have how to do it with sets using symmetric_difference function:

>>> full =  ["foo", "quo", "baz", "quux", "quid", "bar", 'newone', 'anotherone']
>>> couples = [("foo", "quo"), ("baz", "quux"), ("quid", "bar")]

## now for the remainder. Note you have to flatten your couples:
>>> set(full).symmetric_difference([item for sublist in couples for item in sublist])
set(['anotherone', 'newone'])
>>>