Question

我需要编写一个代码，该代码在列表中返回给定大小的集合的子集。因此，首先让我们说我要从集合（0,1,2,3,4,5,6,7,8）中获得大小为3的子集

我想写出列表中的子集：

[[0,1,2]，[0,2,3]，[0,3,4] ....]

然后我想在其中进行递归，将除第一个元素以外的所有元素与我的dictionary（graph）进行比较，以检查我的值中是否包含列表。字典的键是我子集中的第一个元素。

例如：

在[0,1,2]中：

图[0]中的1和2是吗？

字典图就像：{0：[1,2,3,6,7]，1：[0,2,4,6,7] ....}

如果我完成了所有的准备工作，就想检查下一个子集。

所以我的问题如何将其放在列表中？我知道我也有k问题，但不确定如何更改。

def indep(graph,a,b):
    l=list( itertools.combinations(range(a), b))
    for k in l:
        k=list(k)
        while j<=len(k):
           for j in range(len(k)): 
              if k[j]  in graph[k[j]]:
                 j+=1
              else:
                return "no"

Answer 1

这将为您带来预期的结果

从itertools导入组合

original_set = (0,1,2,3,4,5,6,7,8)
final_set = [list(pair) for pair in combinations(l, 3)]

Out[6]: 
[[0, 1, 2],
 [0, 1, 3],
 [0, 1, 4],
 [0, 1, 5],
 [0, 1, 6],
 [0, 1, 7],
 [0, 1, 8],
 [0, 2, 3],
 [0, 2, 4],
 [0, 2, 5],
 [0, 2, 6],
 [0, 2, 7],
 [0, 2, 8],
 [0, 3, 4],
 [0, 3, 5],
 [0, 3, 6],
 [0, 3, 7],
 [0, 3, 8],
 [0, 4, 5],
 [0, 4, 6],
 [0, 4, 7],
 [0, 4, 8],
 [0, 5, 6],
 [0, 5, 7],
 [0, 5, 8],
 [0, 6, 7],
 [0, 6, 8],
 [0, 7, 8],
 [1, 2, 3],
 [1, 2, 4],
 [1, 2, 5],
 [1, 2, 6],
 [1, 2, 7],
 [1, 2, 8],
 [1, 3, 4],
 [1, 3, 5],
 [1, 3, 6],
 [1, 3, 7],
 [1, 3, 8],
 [1, 4, 5],
 [1, 4, 6],
 [1, 4, 7],
 [1, 4, 8],
 [1, 5, 6],
 [1, 5, 7],
 [1, 5, 8],
 [1, 6, 7],
 [1, 6, 8],
 [1, 7, 8],
 [2, 3, 4],
 [2, 3, 5],
 [2, 3, 6],
 [2, 3, 7],
 [2, 3, 8],
 [2, 4, 5],
 [2, 4, 6],
 [2, 4, 7],
 [2, 4, 8],
 [2, 5, 6],
 [2, 5, 7],
 [2, 5, 8],
 [2, 6, 7],
 [2, 6, 8],
 [2, 7, 8],
 [3, 4, 5],
 [3, 4, 6],
 [3, 4, 7],
 [3, 4, 8],
 [3, 5, 6],
 [3, 5, 7],
 [3, 5, 8],
 [3, 6, 7],
 [3, 6, 8],
 [3, 7, 8],
 [4, 5, 6],
 [4, 5, 7],
 [4, 5, 8],
 [4, 6, 7],
 [4, 6, 8],
 [4, 7, 8],
 [5, 6, 7],
 [5, 6, 8],
 [5, 7, 8],
 [6, 7, 8]]

Answer 2

import itertools
a = [0,1,2,3,4,5,6]

# all sets here.
sets = [list(x) for x in itertools.permutations(a, 3) if x[1]==x[2]-1]

#here are all the sets
#[[0, 1, 2], [0, 2, 3], [0, 3, 4], [0, 4, 5], [0, 5, 6], [1, 2, 3], [1, 3, 4], [1, 4, 5], [1, 5, 6], [2, 0, 1], [2, 3, 4],
#[2, 4, 5], [2, 5, 6], [3, 0, 1], [3, 1, 2], [3, 4, 5], [3, 5, 6], [4, 0, 1], [4, 1, 2], [4, 2, 3], [4, 5, 6], [5, 0, 1], 
#[5, 1, 2], [5, 2, 3], [5, 3, 4], [6, 0, 1], [6, 1, 2], [6, 2, 3], [6, 3, 4], [6, 4, 5]]

d = dict()
#make your thingy
for i in sets:
    try:
        d[i[0]] = d[i[0]]+i[1:]
    except:
        d[i[0]] = i[1:]

    d[i[0]] = list(set(d[i[0]]))

#output D

{0: [1, 2, 3, 4, 5, 6],
 1: [2, 3, 4, 5, 6],
 2: [0, 1, 3, 4, 5, 6],
 3: [0, 1, 2, 4, 5, 6],
 4: [0, 1, 2, 3, 5, 6],
 5: [0, 1, 2, 3, 4],
 6: [0, 1, 2, 3, 4, 5]}

这是您想要的吗？：D

创建具有给定大小集合的子集列表的列表

2 个答案: