Python -将重复的对象分离到不同的列表中

Question

假设我有这样一个类:

class Spam(object):
    def __init__(self, a):
        self.a = a

现在我有了这些对象:

s1 = Spam((1, 1, 1, 4))
s2 = Spam((1, 2, 1, 4))
s3 = Spam((1, 2, 1, 4))
s4 = Spam((2, 2, 1, 4))
s5 = Spam((2, 1, 1, 8))
s6 = Spam((2, 1, 1, 8))
objects = [s1, s2, s3, s4, s5, s6]

所以在运行某种方法后，我需要有两个列表，其中一个列表中的对象具有相同的a属性值，另一个列表中的对象具有唯一的a属性值。

:

dups = [s2, s3, s5, s6]
normal = [s1, s4]

所以这有点像获得重复，但除此之外，它还应该添加甚至第一次出现的对象共享相同的a属性值。

我已经写了这个方法，它似乎是工作的，但在我看来它是相当丑陋的(可能不是很理想)。

def eggs(objects):
    vals = []
    dups = []
    normal = []
    for obj in objects:
        if obj.a in vals:
            dups.append(obj)
        else:
            normal.append(obj)
            vals.append(obj.a)
    dups_vals = [o.a for o in dups]
    # separate again
    new_normal = []
    for n in normal:
        if n.a in dups_vals:
            dups.append(n)
        else:
            new_normal.append(n)
    return dups, new_normal

有谁能写出更合适的python方法来解决这个问题吗?

Answer 1

我将把字典中的对象组合在一起，使用a属性作为键。然后我将它们按组的大小分开。

import collections
def separate_dupes(seq, key_func):
    d = collections.defaultdict(list)
    for item in seq:
        d[key_func(item)].append(item)
    dupes   = [item for v in d.values() for item in v if len(v) > 1]
    uniques = [item for v in d.values() for item in v if len(v) == 1]
    return dupes, uniques
class Spam(object):
    def __init__(self, a):
        self.a = a
    #this method is not necessary for the solution, just for displaying the results nicely
    def __repr__(self):
        return "Spam({})".format(self.a)
s1 = Spam((1, 1, 1, 4))
s2 = Spam((1, 2, 1, 4))
s3 = Spam((1, 2, 1, 4))
s4 = Spam((2, 2, 1, 4))
s5 = Spam((2, 1, 1, 8))
s6 = Spam((2, 1, 1, 8))
objects = [s1, s2, s3, s4, s5, s6]
dupes, uniques = separate_dupes(objects, lambda item: item.a)
print(dupes)
print(uniques)

结果:

[Spam((2, 1, 1, 8)), Spam((2, 1, 1, 8)), Spam((1, 2, 1, 4)), Spam((1, 2, 1, 4))]
[Spam((1, 1, 1, 4)), Spam((2, 2, 1, 4))]

Answer 2

如果将__eq__方法添加到Spam中，则定义为

def __eq__(self, other):
    return self.a == other.a

那么你可以很简单地使用

# you can inline this if you want, just wanted to give it a name
def except_at(elems, ind):
    return elems[:ind] + elems[ind+1:]
dups = [obj for (i, obj) in enumerate(objects) if obj in except_at(objects, i)]
normal = [obj for (i, obj) in enumerate(objects) if obj not in except_at(objects, i)]

Answer 3

使用collections.Counter，这些键是多个键共用的:

import collections
common = [k for (k, v) in collections.Counter([o.a for o in objects]).items() if v > 1]

你的两个列表，现在是

[o for o in objects if o.a in common], [o for o in objects if o.a not in common]

Answer 4

如果对象列表不是太大，一种方法是对对象列表进行排序，然后对其应用groupby以获得重复项。为了对列表进行排序，我们提供了一个key函数，该函数提取对象的.a属性的值。

from operator import attrgetter
from itertools import groupby
class Spam(object):
    def __init__(self, a):
        self.a = a
    def __repr__(self):
        return 'Spam({})'.format(self.a)
s1 = Spam((1, 1, 1, 4))
s2 = Spam((1, 2, 1, 4))
s3 = Spam((1, 2, 1, 4))
s4 = Spam((2, 2, 1, 4))
s5 = Spam((2, 1, 1, 8))
s6 = Spam((2, 1, 1, 8))
objects = [s1, s2, s3, s4, s5, s6]
keyfunc = attrgetter('a')
dupe, unique = [], []
for k, g in groupby(sorted(objects, key=keyfunc), key=keyfunc):
    g = list(g)
    target = unique if len(g) == 1 else dupe
    target.extend(g)
print('dupe', dupe)
print('unique', unique)

dupe [Spam((1, 2, 1, 4)), Spam((1, 2, 1, 4)), Spam((2, 1, 1, 8)), Spam((2, 1, 1, 8))]
unique [Spam((1, 1, 1, 4)), Spam((2, 2, 1, 4))]