python技巧——列出列表（集合）中的所有子集的方法总结_py 生成所有子集

问题定义：给定一个列表a=[1,2,3,4]，请获取它的所有子集，即获取[[], [1], [1, 2], [1, 2, 3], [1, 2, 3, 4], [1, 2, 4], [1, 3], [1, 3, 4], [1, 4], [2], [2, 3], [2, 3, 4], [2, 4], [3], [3, 4], [4]]

说明：给定一个长度为n的列表，那么它的子集数量就有2**n个

解题方法：

#方法一：使用itertools库函数
def get_subset(mylist):
    n =len(mylist)
    for num in range(n):
        for i in itertools.combinations(mylist,num+1):
            print(i)
# get_subset(mylist=a)
#方法二：二进制位运算
def get_subset2(mylist):
    n = len(mylist)
    for i in range(2**n):
        combi = []
        for j in range(n):
            if (i>>j)%2:
                combi.append(mylist[j])
        print(combi)
# get_subset2(mylist=a)
#方法三：反向位运算
def get_subset3(mylist):
    n = len(mylist)
    subset_nums = 1<<n
    for i in range(subset_nums):
        #&1的作用是判断当前是否能被2整除
        bits = [(i>>offset)%2 for offset in range(n-1,-1,-1)]
        # bits = [(i>>offset)&1 for offset in range(n-1,-1,-1)]
        current = [mylist[index] for (index,bit) in enumerate(bits) if bit==1]
        print(current)
#get_subset3(mylist=a)
 
def get_subset4(mylist):
    output = [[]]
    for num in mylist:
        output += [curr + [num] for curr in output]
    print(output)
get_subset4(mylist=a)
 
def get_subset5(mylist):
    def backtrack(first=0,curr=[]):
        #当curr的长度为k时
        if len(curr)==k:
            output.append(curr[:])
        #一般时候，把mylist[i]加入，然后判断mylist[i+1]
        for i in range(first,n):
            curr.append(mylist[i])
            backtrack(i+1,curr)
            curr.pop()
    output=[]
    n = len(mylist)
    for k in range(n+1):
        backtrack()
    print(output)
get_subset5(mylist=a)
 
def get_subset6(mylist):
    n = len(mylist)
    output=[]
    for i in range(2**n,2**(n+1)):
        bitmask = bin(i)[3:]
        curr = [mylist[i] for i in range(n) if bitmask[i]=='1']
        output.append(curr)
    print(output)
get_subset6(mylist=a)
 
def get_subset7(mylist):
    res = [[]]
    for i in mylist:
        res = res+ [[i] + num for num in res]
    print(res)
get_subset7(mylist=a)
 
#递归
def get_subset8(mylist):
    res = []
    n = len(mylist)
 
    def helper(i,tmp):
        res.append(tmp)
        for j in range(i,n):
            helper(j+1,tmp+[mylist[j]])
    helper(0,[])
    print(res)
get_subset8(mylist=a)