Python-将重复对象分离到不同的列表中
假设我有一门课:Python-将重复对象分离到不同的列表中,python,duplicates,unique,Python,Duplicates,Unique,假设我有一门课: class Spam(object): def __init__(self, a): self.a = a 现在我有了这些物体: 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
class Spam(object):
def __init__(self, a):
self.a = a
现在我有了这些物体:
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]
因此,在运行某种方法之后,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
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)
像这样:
dups = [s2, s3, s5, s6]
normal = [s1, s4]
所以这有点像获取重复项,但除此之外,它还应该添加共享相同属性值的对象的第一次出现
我已经写了这个方法,它似乎是有效的,但在我看来,它是相当丑陋的(可能不是很理想)
有人能为这样的问题编写更合适的pythonic方法吗?我会将字典中的对象分组,使用a
属性作为键。然后,我会根据组的大小将它们分开
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))]
我会使用a
属性作为键,将字典中的对象分组。然后,我会根据组的大小将它们分开
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))]
如果将方法添加到垃圾邮件
,定义为
def __eq__(self, other):
return self.a == other.a
然后你可以很简单地用这样的方法来做
# 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)]
如果将方法添加到垃圾邮件
,定义为
def __eq__(self, other):
return self.a == other.a
然后你可以很简单地用这样的方法来做
# 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)]
使用,以下是多个公用键:
import collections
common = [k for (k, v) in collections.Counter([o.a for o in objects]).items() if v > 1]
你的两份名单,现在是
[o for o in objects if o.a in common], [o for o in objects if o.a not in common]
使用,以下是多个公用键:
import collections
common = [k for (k, v) in collections.Counter([o.a for o in objects]).items() if v > 1]
你的两份名单,现在是
[o for o in objects if o.a in common], [o for o in objects if o.a not in common]
如果对象列表不太大,一种方法是对对象列表进行排序,然后对其应用groupby
,以获取重复项。为了对列表进行排序,我们提供了一个键函数,用于提取对象的.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)
输出
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))]
如果对象列表不太大,一种方法是对对象列表进行排序,然后对其应用groupby
,以获取重复项。为了对列表进行排序,我们提供了一个键函数,用于提取对象的.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)
输出
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))]
这似乎给我带来了TypeError:“垃圾邮件”对象不可订阅
我不知道它是否在我这边,当我有时间使用你的功能时,我会调查它,除了
功能。我很抱歉!我将错误的变量传递到中,但在处除外。修复了它。这似乎让我陷入了类型错误:“垃圾邮件”对象不可订阅
我不知道它是否在我这边,当我有时间使用你的功能时,我会调查它,除了
功能。我道歉!我将错误的变量传递到中,但在处除外。修好了。