Python实现求多个集合之间并集的方法

目的:求多个集合之前的并集,例如:现有四个集合C1 = {11, 22, 13, 14}、C2 = {11, 32, 23, 14, 35}、C3 = {11, 22, 38}、C4 = {11, 22, 33, 14, 55, 66},则它们之间的并集应该为:

C1 & C2 & C3 = {11}、C1 & C2 & C4 = {14}、C1 & C3 & C4 = {22}。

如下图所示:


实现方法:Python自带了set数据类型,并且可以实现求集合的并集、交集、差集等,十分好用。按照一般的数学方法实现,实现的步骤如下:

(1)先求4个集合共有的成员;

(2)每个集合减去所有集合的共有成员,在求其中任意3个集合共有的成员;

(3)每个集合减去包含自己的任意三个集合的共有成员,最后求其中任意两个集合共有的成员。

具体的代码如下:

# encoding: utf-8

def func(content):
    # 使用集合实现, 使用集合真是太方便了

    c1 = set(content[0])  # [11, 22, 13, 14]
    c2 = set(content[1])  # [11, 32, 23, 14, 35]
    c3 = set(content[2])  # [11, 22, 38]
    c4 = set(content[3])  # [11, 22, 33, 14, 55, 66]

    # all collections have element
    all_union_elems = c1 & c2 & c3 & c4
    if all_union_elems:
        print ('all collections have elems: ', all_union_elems)

    # three collections have
    c1 = c1 - all_union_elems
    c2 = c2 - all_union_elems
    c3 = c3 - all_union_elems
    c4 = c4 - all_union_elems
    c123_union_elems = c1 & c2 & c3
    c124_union_elems = c1 & c2 & c4
    c134_union_elems = c1 & c3 & c4
    c234_union_elems = c2 & c3 & c4
    if c123_union_elems:
        print ("c123_union_elems ", c123_union_elems)
    if c124_union_elems:
        print ("c124_union_elems ", c124_union_elems)
    if c134_union_elems:
        print ("c134_union_elems ", c134_union_elems)
    if c234_union_elems:
        print ("c234_union_elems ", c234_union_elems)

    # two collections have
    c1 = c1 - c123_union_elems - c124_union_elems - c134_union_elems
    c2 = c2 - c123_union_elems - c124_union_elems - c234_union_elems
    c3 = c3 - c123_union_elems - c134_union_elems - c234_union_elems
    c4 = c4 - c124_union_elems - c134_union_elems - c234_union_elems
    c12_union_have = c1 & c2
    c13_union_have = c1 & c3
    c14_union_have = c1 & c4
    c23_union_have = c2 & c3
    c24_union_have = c2 & c4
    c34_union_have = c3 & c4
    if c12_union_have:
        print ("c12_union_have ", c12_union_have)
    if c13_union_have:
        print ("c13_union_have ", c13_union_have)
    if c14_union_have:
        print ("c14_union_have ", c14_union_have)
    if c23_union_have:
        print ("c23_union_have ", c23_union_have)
    if c24_union_have:
        print ("c24_union_have ", c24_union_have)
    if c34_union_have:
        print ("c34_union_have ", c34_union_have)

    c1 = c1 - c12_union_have - c13_union_have - c14_union_have
    c2 = c2 - c12_union_have - c23_union_have - c24_union_have
    c3 = c3 - c13_union_have - c23_union_have - c34_union_have
    c4 = c4 - c14_union_have - c24_union_have - c34_union_have
    if c1:
        print ('only c1 have ', c1)
    if c2:
        print ('only c2 have ', c2)
    if c3:
        print ('only c3 have ', c3)
    if c4:
        print ('only c4 have ', c4)

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if __name__ == "__main__":

    content = [[11, 22, 13, 14], [11, 32, 23, 14, 35], [11, 22, 38], [11, 22, 33, 14, 55, 66]]

    func(content)

输出结果如下:

all collections have elems:  {11}
c124_union_elems  {14}
c134_union_elems  {22}
only c1 have  {13}
only c2 have  {32, 35, 23}
only c3 have  {38}
only c4 have  {33, 66, 55}

这种实现方法其实效率不高,需要比较集合的次数为:1 + 4 + 6 = 11次,另外代码也很冗余,并不是一种好的实现方式。

还有另外一种效率高的实现方式:

(1)首先,先找出成员数最多的那个集合,这里就是集合C4;

(2)将集合C4中的每个成员依次和其它集合进行比较,看其它集合中是否包含此成员;

(3)若其它集合中包括这个成员,就将这个成员从集合中去除,依次这样比较每个集合;

(4)比较一轮之后,集合C4中剩余的成员就是只有自己的成员。

(5)再在除C4以外剩下的集合中,找出成员数最多的集合,重复上诉操作。依次类推,就可以求出各集合之间的并集了。

上述算法中需要比较的次数只有3 + 2 + 1 = 6次。

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