在Python中，如何减去两个非唯一的无序列表？假设我们有a = [0,1,2,1,0]和b = [0, 1, 1]我想做c = a - b之类的事情并c [2, 0]或[0, 2]订单不对我来说很重要如果a不包含b中的所有元素，则应抛出异常。
注意这与集合不同！我对找到a和b中元素集的差异不感兴趣，我对实际元素集合之间的差异感兴趣a和b。
我可以用for循环来做这个，在a中查找b的第一个元素，然后从a中删除元素，然后从a中删除等等。但是这对我没有吸引力，它会非常低效（顺序） O(n^2)时间的O(n log n)时间，但在{{1}}时间内执行此操作应该没问题。
答案 0 :(得分：55)
我知道“for”不是你想要的，但它简单明了：
for x in b:
a.remove(x)
如果b的成员可能不在a，请使用：
for x in b:
if x in a:
a.remove(x)
答案 1 :(得分：29)
Python 2.7和3.2将添加collections.Counter类，这是一个将元素映射到元素出现次数的字典。这可以用作多重集。
根据文档，你应该可以做这样的事情（未经测试，因为我没有安装任何版本）。
from collections import Counter
a = Counter(0,1,2,1)
b = Counter(0,1,1)
print a - b # ignores items in b missing in a
# check every element in a is in b
# a[key] returns 0 if key not in a, instead of raising an exception
assert all(a[key] > b[key] for key in b)
修改：
由于你坚持使用2.5，你可以尝试导入它并定义你自己的版本，如果失败了。这样一来，如果没有，你肯定会得到最新版本，如果没有，你会回到工作版本。如果将来转换为C实现，您也将受益于速度改进。
即
try:
from collections import Counter
except ImportError:
class Counter(dict):
...
您可以找到当前的Python源here。
答案 2 :(得分：28)
我会以一种更简单的方式做到这一点：
a_b = [e for e in a if not e in b ]
<击>
..正如Wich写的那样，这是错误的 - 只有当项目在列表中是唯一的时才有效。如果是，最好使用
a_b = list(set(a) - set(b))
答案 3 :(得分：6)
我不确定for循环的异议是什么：Python中没有多重集，所以你不能使用内置容器来帮助你。
在我看来，任何一行（如果可能的话）都可能很难理解。寻求可读性和KISS。 Python不是C：）
答案 4 :(得分：5)
Python 2.7+和3.0有collections.Counter（a.k.a。multiset）。文档链接到Python {的Recipe 576611: Counter class：
from operator import itemgetter
from heapq import nlargest
from itertools import repeat, ifilter
class Counter(dict):
'''Dict subclass for counting hashable objects. Sometimes called a bag
or multiset. Elements are stored as dictionary keys and their counts
are stored as dictionary values.
>>> Counter('zyzygy')
Counter({'y': 3, 'z': 2, 'g': 1})
'''
def __init__(self, iterable=None, **kwds):
'''Create a new, empty Counter object. And if given, count elements
from an input iterable. Or, initialize the count from another mapping
of elements to their counts.
>>> c = Counter() # a new, empty counter
>>> c = Counter('gallahad') # a new counter from an iterable
>>> c = Counter({'a': 4, 'b': 2}) # a new counter from a mapping
>>> c = Counter(a=4, b=2) # a new counter from keyword args
'''
self.update(iterable, **kwds)
def __missing__(self, key):
return 0
def most_common(self, n=None):
'''List the n most common elements and their counts from the most
common to the least. If n is None, then list all element counts.
>>> Counter('abracadabra').most_common(3)
[('a', 5), ('r', 2), ('b', 2)]
'''
if n is None:
return sorted(self.iteritems(), key=itemgetter(1), reverse=True)
return nlargest(n, self.iteritems(), key=itemgetter(1))
def elements(self):
'''Iterator over elements repeating each as many times as its count.
>>> c = Counter('ABCABC')
>>> sorted(c.elements())
['A', 'A', 'B', 'B', 'C', 'C']
If an element's count has been set to zero or is a negative number,
elements() will ignore it.
'''
for elem, count in self.iteritems():
for _ in repeat(None, count):
yield elem
# Override dict methods where the meaning changes for Counter objects.
@classmethod
def fromkeys(cls, iterable, v=None):
raise NotImplementedError(
'Counter.fromkeys() is undefined. Use Counter(iterable) instead.')
def update(self, iterable=None, **kwds):
'''Like dict.update() but add counts instead of replacing them.
Source can be an iterable, a dictionary, or another Counter instance.
>>> c = Counter('which')
>>> c.update('witch') # add elements from another iterable
>>> d = Counter('watch')
>>> c.update(d) # add elements from another counter
>>> c['h'] # four 'h' in which, witch, and watch
4
'''
if iterable is not None:
if hasattr(iterable, 'iteritems'):
if self:
self_get = self.get
for elem, count in iterable.iteritems():
self[elem] = self_get(elem, 0) + count
else:
dict.update(self, iterable) # fast path when counter is empty
else:
self_get = self.get
for elem in iterable:
self[elem] = self_get(elem, 0) + 1
if kwds:
self.update(kwds)
def copy(self):
'Like dict.copy() but returns a Counter instance instead of a dict.'
return Counter(self)
def __delitem__(self, elem):
'Like dict.__delitem__() but does not raise KeyError for missing values.'
if elem in self:
dict.__delitem__(self, elem)
def __repr__(self):
if not self:
return '%s()' % self.__class__.__name__
items = ', '.join(map('%r: %r'.__mod__, self.most_common()))
return '%s({%s})' % (self.__class__.__name__, items)
# Multiset-style mathematical operations discussed in:
# Knuth TAOCP Volume II section 4.6.3 exercise 19
# and at http://en.wikipedia.org/wiki/Multiset
#
# Outputs guaranteed to only include positive counts.
#
# To strip negative and zero counts, add-in an empty counter:
# c += Counter()
def __add__(self, other):
'''Add counts from two counters.
>>> Counter('abbb') + Counter('bcc')
Counter({'b': 4, 'c': 2, 'a': 1})
'''
if not isinstance(other, Counter):
return NotImplemented
result = Counter()
for elem in set(self) | set(other):
newcount = self[elem] + other[elem]
if newcount > 0:
result[elem] = newcount
return result
def __sub__(self, other):
''' Subtract count, but keep only results with positive counts.
>>> Counter('abbbc') - Counter('bccd')
Counter({'b': 2, 'a': 1})
'''
if not isinstance(other, Counter):
return NotImplemented
result = Counter()
for elem in set(self) | set(other):
newcount = self[elem] - other[elem]
if newcount > 0:
result[elem] = newcount
return result
def __or__(self, other):
'''Union is the maximum of value in either of the input counters.
>>> Counter('abbb') | Counter('bcc')
Counter({'b': 3, 'c': 2, 'a': 1})
'''
if not isinstance(other, Counter):
return NotImplemented
_max = max
result = Counter()
for elem in set(self) | set(other):
newcount = _max(self[elem], other[elem])
if newcount > 0:
result[elem] = newcount
return result
def __and__(self, other):
''' Intersection is the minimum of corresponding counts.
>>> Counter('abbb') & Counter('bcc')
Counter({'b': 1})
'''
if not isinstance(other, Counter):
return NotImplemented
_min = min
result = Counter()
if len(self) < len(other):
self, other = other, self
for elem in ifilter(self.__contains__, other):
newcount = _min(self[elem], other[elem])
if newcount > 0:
result[elem] = newcount
return result
if __name__ == '__main__':
import doctest
print doctest.testmod()
然后你可以写
a = Counter([0,1,2,1,0])
b = Counter([0, 1, 1])
c = a - b
print list(c.elements()) # [0, 2]
答案 5 :(得分：4)
使用列表理解：
[i for i in a if not i in b or b.remove(i)]
会做到这一点。但它会在这个过程中改变b。但我同意jkp和Dyno Fu的观点，即使用for循环会更好。
也许有人可以创建一个更好的例子，使用列表理解但仍然是KISS？
答案 6 :(得分：2)
为了证明jkp的观点，'在一条线上的任何东西都可能是非常复杂的理解'，我创造了一个单行。请不要模仿我，因为我知道这不是你应该实际使用的解决方案。这只是为了示范目的。
我们的想法是逐个添加值，只要您添加该值的总次数小于此值的总次数减去它在b中的次数：
[ value for counter,value in enumerate(a) if a.count(value) >= b.count(value) + a[counter:].count(value) ]
恐怖！但也许有人可以改进它？它甚至没有bug吗？
编辑：看到Devin Jeanpierre关于使用字典数据结构的评论，我想出了这个oneliner：
sum([ [value]*count for value,count in {value:a.count(value)-b.count(value) for value in set(a)}.items() ], [])
更好，但仍然不可读。
答案 7 :(得分：0)
您可以尝试这样的事情：
class mylist(list):
def __sub__(self, b):
result = self[:]
b = b[:]
while b:
try:
result.remove(b.pop())
except ValueError:
raise Exception("Not all elements found during subtraction")
return result
a = mylist([0, 1, 2, 1, 0] )
b = mylist([0, 1, 1])
>>> a - b
[2, 0]
你必须定义[1,2,3] - [5,6]应输出的内容，我想你想要[1,2,3]这就是为什么我忽略了ValueError。
编辑：
现在我看到如果a不包含所有元素，则需要一个例外，添加它而不是传递ValueError。
答案 8 :(得分：0)
我试图找到一个更优雅的解决方案，但我能做的最好的事情与Dyno Fu所说的基本相同：
from copy import copy
def subtract_lists(a, b):
"""
>>> a = [0, 1, 2, 1, 0]
>>> b = [0, 1, 1]
>>> subtract_lists(a, b)
[2, 0]
>>> import random
>>> size = 10000
>>> a = [random.randrange(100) for _ in range(size)]
>>> b = [random.randrange(100) for _ in range(size)]
>>> c = subtract_lists(a, b)
>>> assert all((x in a) for x in c)
"""
a = copy(a)
for x in b:
if x in a:
a.remove(x)
return a
答案 9 :(得分：0)
这是一个相对较长但有效且易读的解决方案。它是O（n）。
def list_diff(list1, list2):
counts = {}
for x in list1:
try:
counts[x] += 1
except:
counts[x] = 1
for x in list2:
try:
counts[x] -= 1
if counts[x] < 0:
raise ValueError('All elements of list2 not in list2')
except:
raise ValueError('All elements of list2 not in list1')
result = []
for k, v in counts.iteritems():
result += v*[k]
return result
a = [0, 1, 1, 2, 0]
b = [0, 1, 1]
%timeit list_diff(a, b)
%timeit list_diff(1000*a, 1000*b)
%timeit list_diff(1000000*a, 1000000*b)
100000 loops, best of 3: 4.8 µs per loop
1000 loops, best of 3: 1.18 ms per loop
1 loops, best of 3: 1.21 s per loop
答案 10 :(得分：0)
您可以使用map构造来执行此操作。它看起来很不错，但请注意map行本身将返回None的列表。
a = [1, 2, 3]
b = [2, 3]
map(lambda x:a.remove(x), b)
a
答案 11 :(得分：-3)
c = [i for i in b if i not in a]
答案 12 :(得分：-4)
list(set([x for x in a if x not in b]))
a和b。