比较两个列表并找到不同的对

Question

我正在尝试比较两个列表（包含不同大小的子列表），并找到另一个列表中不存在的对（用圆括号括起来）。

这是代码：

s1 = [
    [('RESOLVED - DUPLICATE', 'VERIFIED')], [('NEW', 'RESOLVED - FIXED'), ('RESOLVED - FIXED', 'VERIFIED')], [('NEW', 'RESOLVED - DUPLICATE'), ('RESOLVED - DUPLICATE', 'VERIFIED')], [('ASSIGNED', 'RESOLVED - FIXED'), ('RESOLVED - FIXED', 'VERIFIED')], [('NEW', 'RESOLVED - WONTFIX'), ('RESOLVED - WONTFIX', 'VERIFIED')], [('NEW', 'RESOLVED - INVALID'), ('RESOLVED - INVALID', 'VERIFIED')]
]

s2 = [
    [('RESOLVED - DUPLICATE', 'VERIFIED')], [('NEW', 'RESOLVED - DUPLICATE'), ('RESOLVED - DUPLICATE', 'VERIFIED')], [('NEW', 'ASSIGNED'), ('ASSIGNED', 'RESOLVED - FIXED')], [('ASSIGNED', 'RESOLVED - FIXED'), ('RESOLVED - FIXED', 'VERIFIED')], [('NEW', 'RESOLVED - WONTFIX'), ('RESOLVED - WONTFIX', 'VERIFIED')]
]

a = []

for item in s2:
    i = 0
    print item
    while (i < len(item)):
        for item1 in s1:
        print item[i]
        if item[i] not in s1:
            a.append(item[i])
        i = i + 1

print a

上面提到的代码没有给出s2中存在的唯一对，而s1中没有。
任何帮助将非常感激。 谢谢！

Answer 1

将您的列表平整化为set对象及其实现的集合操作，在这种情况下，您需要s2和s1之间的区别 ，

>>> from itertools import chain
>>> set(chain.from_iterable(s2)).difference(chain.from_iterable(s1))
{('NEW', 'ASSIGNED')}

请注意，您实际上可能希望两者之间存在对称差异 ，即两者中不存在的项目：

>>> set(chain.from_iterable(s1)).symmetric_difference(chain.from_iterable(s2))
{('NEW', 'RESOLVED - FIXED'), ('NEW', 'RESOLVED - INVALID'), ('NEW', 'ASSIGNED'), ('RESOLVED - INVALID', 'VERIFIED')}

Answer 2

假设嵌套结构不相关，并且列表可以展平：

s1flat = [x for sublist in s1 for x in sublist]
s2flat = [x for sublist in s2 for x in sublist]

new_pair = list(set(s2flat) - set(s1flat))

Answer 3

如果您发现s1中不存在的s2项目，则只需使用：

一条线

print([item for item in s2 if item not in s1])

详细：

该列表的理解与：

new_list=[]
for i in s2:
    if i not in s1:
        new_list.append(i)
print(new_list)

输出：

[[('NEW', 'ASSIGNED'), ('ASSIGNED', 'RESOLVED - FIXED')]]