python中求两个List的交集、并集和差集

直接上代码，有三种方法，第三种调用库函数效率最高

# ! /usr/bin/env python
# encoding:utf-8

if __name__ == '__main__':
	a = [1,2,3,4,5]
	b = [2,3,6,7]
	u =[]
	dif =[]
	intersec = []

	'''方法一,最简单的方法,容易想到的'''
	for item in a:
		u.append(item)
		if item in b:
			intersec.append(item)
		if item not in b:
			dif.append(item)
	for key in b:
		u.append(key)
	print('u is ',list(set(u))) #u is  [1, 2, 3, 4, 5, 6, 7]
	print('dif is ', dif) #dif is  [1, 4, 5]
	print('intersec is ', intersec) #intersec is  [2, 3]

	'''方法二'''
	a = [1,2,3,4,5]
	b = [2,3,6,7]
	intersec = [item for item in a if item in b]
	dif = [item for item in a if item not in b] #不能放在最后，因为b.extend时b已经被改变了
	b.extend([i for i in a])
	u = b

	print('u is ', list(set(u)))  # u is  [1, 2, 3, 4, 5, 6, 7]
	print('dif is ', dif)  # dif is  [1, 4, 5]
	print('intersec is ', intersec)  # intersec is  [2, 3]

	'''方法三,调用库函数,集合操作'''
	a = [1,2,3,4,5]
	b = [2,3,6,7]
	intersec = list(set(a).intersection(set(b)))
	u = list(set(a).union(set(b)))
	dif = list(set(a).difference(set(b)))
	print('u is ', u)  # u is  [1, 2, 3, 4, 5, 6, 7]
	print('dif is ', dif)  # dif is  [1, 4, 5]
	print('intersec is ', intersec)  # intersec is  [2, 3]