list并集:1
print list(set(a).union(set(b)))
或者:1
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8k = [x for x in s1 if x in s2]
>>> x1 = [1,2,3,4,5,6,7]
>>> s1 = x1
>>> s2 = (3,4,6,8)
>>> k = [x for x in s1 if x in s2]
>>> k
[3, 4, 6]
list差集:1
2print list(set(b).difference(set(a)))
# b中有而a中没有的