比较列表 [英] Comparing lists
问题描述
我必须列出A和B，可能或可能不相同。如果它们不是相同的
，我希望输出为三个新列表，X，Y和Z，其中X具有
所有A中的元素，但不是在B中，Y包含B中但不在A中的所有
元素.Z将在A和B中具有
的元素。
这样做的一种方法当然是通过列表迭代并比较每个元素的每个元素，但是有更有效的方法吗？
提前致谢！
-
Har du et kj？leskap，har du en TV
S？ har du alt du trenger for？ leve
-Jokke& Valentinerne
Odd-R。写道：
我必须列出A和B，它们可能相同，也可能不相同。如果它们不相同，我希望输出为三个新列表，X，Y和Z，其中X具有A中的所有元素，但不包含在B中，并且Y包含所有
B中但不在A.中的元素Z将在A和B中具有
元素。
这些是设置操作。 />
这样做的一种方法当然是通过列表迭代并比较每个元素，但是有更有效的方法吗？
也许，使用套装？
L1 = [1,2,3,4]
L2 = [3,4,5,6 ]
diff1 = list（set（L1）-set（L2））＃[1,2]
diff2 = list（set（L2）-set（L1） ）＃[5,6]
symdiff = diff1 + diff2＃对称差异[1,2,5,6]
intersect = set（L1 + L2） - set （symdiff）#Interect [3,4]
Best，
Les
尝试使用set。
L1 = [1,1,2,3,4]
L2 = [1,3,99]
A =设定（L1）
B =设定（L2）
X = AB
打印X
Y = BA
打印Y
Z = A | B
打印Z
干杯，
pujo
< AJ **** @ gmail.com>在消息中写道
news：11 ********************** @ g47g2000cwa.googlegr oups.com ...尝试使用set。
L1 = [1,1,2,3,4]
L2 = [1,3,99]
A = set（L1）< br => B =设定（L2）
X = AB
打印X
Y = BA
打印Y
Z = A | B
打印Z
但是如何高效这是？在这一点上，你能否更明确一点
？什么是订单的复杂性（[...]）或A-B，B-A，
A | B，A ^ B和A& B' - Python文档
在这方面让我完全处于黑暗中。
对两个列表进行排序然后解压缩
AB ，BA，A | B，A& B和A ^ B在单一
通行证中对我来说很可能更快更好
对于大型列表。
此致，
Christian
I have to lists, A and B, that may, or may not be equal. If they are not
identical, I want the output to be three new lists, X,Y and Z where X has
all the elements that are in A, but not in B, and Y contains all the
elements that are B but not in A. Z will then have the elements that are
in both A and B.
One way of doing this is of course to iterate throug the lists and compare
each of the element, but is there a more efficient way?
Thanks in advance!
--
Har du et kj?leskap, har du en TV
s? har du alt du trenger for ? leve
-Jokke & Valentinerne
Odd-R. wrote:
I have to lists, A and B, that may, or may not be equal. If they are not
identical, I want the output to be three new lists, X,Y and Z where X has
all the elements that are in A, but not in B, and Y contains all the
elements that are B but not in A. Z will then have the elements that are
in both A and B.
These are set operations.
One way of doing this is of course to iterate throug the lists and compare
each of the element, but is there a more efficient way?
Maybe, using sets?
L1 = [1,2,3,4]
L2=[3,4,5,6]
diff1 = list(set(L1)-set(L2)) # [1,2]
diff2 = list(set(L2)-set(L1)) # [5,6]
symdiff = diff1+diff2 # Symmetric difference [1,2,5,6]
intersect = set(L1+L2) - set(symdiff) # Intersect [3,4]
Best,
Les
try to use set.
L1 = [1,1,2,3,4]
L2 = [1,3, 99]
A = set(L1)
B = set(L2)
X = A-B
print X
Y = B-A
print Y
Z = A | B
print Z
Cheers,
pujo
<aj****@gmail.com> wrote in message
news:11**********************@g47g2000cwa.googlegr oups.com...try to use set.
L1 = [1,1,2,3,4]
L2 = [1,3, 99]
A = set(L1)
B = set(L2)
X = A-B
print X
Y = B-A
print Y
Z = A | B
print Z
But how "efficient" is this? Could you be a bit
more explicit on that point? What is the order
of complexity of set([...]) or of A-B, B-A,
A | B, A ^ B and A & B? - The Python Documentation
leaves me completely in the dark in this regard.
Sorting the two lists and then extracting
A-B, B-A, A|B, A & B and A ^ B in one single
pass seems to me very likely to be much faster
for large lists.
Regards,
Christian
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