我在Python中有两个列表:
temp1 = ['One', 'Two', 'Three', 'Four']
temp2 = ['One', 'Two']
假设每个列表中的元素都是唯一的，我想用第一个列表中的项创建第三个列表，这些项不在第二个列表中:
temp3 = ['Three', 'Four']
有没有没有周期和检查的快速方法?
当前回答
如果您想要更类似于变更集的东西……可以使用Counter
from collections import Counter
def diff(a, b):
""" more verbose than needs to be, for clarity """
ca, cb = Counter(a), Counter(b)
to_add = cb - ca
to_remove = ca - cb
changes = Counter(to_add)
changes.subtract(to_remove)
return changes
lista = ['one', 'three', 'four', 'four', 'one']
listb = ['one', 'two', 'three']
In [127]: diff(lista, listb)
Out[127]: Counter({'two': 1, 'one': -1, 'four': -2})
# in order to go from lista to list b, you need to add a "two", remove a "one", and remove two "four"s
In [128]: diff(listb, lista)
Out[128]: Counter({'four': 2, 'one': 1, 'two': -1})
# in order to go from listb to lista, you must add two "four"s, add a "one", and remove a "two"
其他回答
我们可以计算交集减去列表并集:
temp1 = ['One', 'Two', 'Three', 'Four']
temp2 = ['One', 'Two', 'Five']
set(temp1+temp2)-(set(temp1)&set(temp2))
Out: set(['Four', 'Five', 'Three'])
如果要删除列表a中存在于列表b中的所有值。
def list_diff(a, b):
r = []
for i in a:
if i not in b:
r.append(i)
return r
list_diff([1、2、2],[1])
结果(2,2):
or
def list_diff(a, b):
return [x for x in a if x not in b]
由于目前的解决方案都不产生一个元组，我将抛出:
temp3 = tuple(set(temp1) - set(temp2))
另外:
#edited using @Mark Byers idea. If you accept this one as answer, just accept his instead.
temp3 = tuple(x for x in temp1 if x not in set(temp2))
像其他在这个方向上得到答案的非元组一样，它保持了顺序
def diffList(list1, list2): # returns the difference between two lists.
if len(list1) > len(list2):
return (list(set(list1) - set(list2)))
else:
return (list(set(list2) - set(list1)))
例如，如果list1 =[10, 15, 20, 25, 30, 35, 40]和list2 =[25, 40, 35]，那么返回的列表将输出= [10,20,30,15]
如果您想要更类似于变更集的东西……可以使用Counter
from collections import Counter
def diff(a, b):
""" more verbose than needs to be, for clarity """
ca, cb = Counter(a), Counter(b)
to_add = cb - ca
to_remove = ca - cb
changes = Counter(to_add)
changes.subtract(to_remove)
return changes
lista = ['one', 'three', 'four', 'four', 'one']
listb = ['one', 'two', 'three']
In [127]: diff(lista, listb)
Out[127]: Counter({'two': 1, 'one': -1, 'four': -2})
# in order to go from lista to list b, you need to add a "two", remove a "one", and remove two "four"s
In [128]: diff(listb, lista)
Out[128]: Counter({'four': 2, 'one': 1, 'two': -1})
# in order to go from listb to lista, you must add two "four"s, add a "one", and remove a "two"
