Dictionary Julia-迭代字典中的键组合

有没有一种巧妙的方法可以迭代字典中的键组合

我的字典的值如下：

[1] => [1,2], [2,3] => [15], [3] => [6,7,8], [4,9,11] => [3], ... 

我需要做的是获取长度为

1:n

的所有键组合，其中

n

可能是fx 3

就像上面的例子一样，我想迭代

[[1], [3], [2,3], [[1],[1,2]], [[3],[2,3]], [4,9,11]]

我知道我可以收集密钥，但是我的字典相当大，我正在重新设计整个算法，因为当代码> n＞3 < /代码>时，它开始疯狂地交换，大大降低了效率

。

---

tl；dr有没有一种方法可以从字典中创建组合迭代器，而不必收集字典？

下面是一个简单的实现，它尝试在浏览字典时最小化位。此外，它还使用OrderedDict，因此保持键索引是有意义的（因为DICT不能保证每次都进行一致的键迭代，因此键索引是有意义的）

这比你目前拥有的更有效吗？最好将代码放在函数中，或者将其转换为Julia任务，每次迭代生成下一个键集

---更新---

使用中任务中有关迭代器的答案

改进的迭代器版本是：

function keysubsets(n,d)
  Task() do
    od = OrderedDict(d)
    sv = map(length,keys(od))        # store length of keys for quicker calculations
    maxmaxlen = sum(sv)              # maximum total elements in good key
    for maxlen=1:min(n,maxmaxlen)    # replace maxmaxlen with lower value if too slow
      gsets = Vector{Vector{Int}}()  # hold good sets of key _indices_
      for curlen=1:maxlen
        foreach(x->push!(gsets,x),(x for x in subsets(collect(1:n),curlen) if sum(sv[x])==maxlen))
      end
      # indmatrix is necessary to run through keys once in next loop
      indmatrix = zeros(Bool,length(od),length(gsets))
      for i=1:length(gsets)              for e in gsets[i]
          indmatrix[e,i] = true
        end
      end
      # gkeys is the vector of vecotrs of keys i.e. what we wanted to calculate
      gkeys = [Vector{Vector{Int}}() for i=1:length(gsets)]
      for (i,k) in enumerate(keys(od))
        for j=1:length(gsets)
          if indmatrix[i,j]
            push!(gkeys[j],k)
          end
        end
      end
      # do something with each set of good keys
      foreach(x->produce(x),gkeys)
    end
  end
end

现在可以通过这种方式（在运行另一个StackOverflow应答中的代码后）在组合大小为4的所有密钥子集上进行迭代：

（需要从链接的堆栈溢出答案中定义NewTask）

function keysubsets(n,d)
  Task() do
    od = OrderedDict(d)
    sv = map(length,keys(od))        # store length of keys for quicker calculations
    maxmaxlen = sum(sv)              # maximum total elements in good key
    for maxlen=1:min(n,maxmaxlen)    # replace maxmaxlen with lower value if too slow
      gsets = Vector{Vector{Int}}()  # hold good sets of key _indices_
      for curlen=1:maxlen
        foreach(x->push!(gsets,x),(x for x in subsets(collect(1:n),curlen) if sum(sv[x])==maxlen))
      end
      # indmatrix is necessary to run through keys once in next loop
      indmatrix = zeros(Bool,length(od),length(gsets))
      for i=1:length(gsets)              for e in gsets[i]
          indmatrix[e,i] = true
        end
      end
      # gkeys is the vector of vecotrs of keys i.e. what we wanted to calculate
      gkeys = [Vector{Vector{Int}}() for i=1:length(gsets)]
      for (i,k) in enumerate(keys(od))
        for j=1:length(gsets)
          if indmatrix[i,j]
            push!(gkeys[j],k)
          end
        end
      end
      # do something with each set of good keys
      foreach(x->produce(x),gkeys)
    end
  end
end

julia> nt2 = NewTask(keysubsets(4,od))
julia> collect(nt2)
10-element Array{Array{Array{Int64,1},1},1}:
 Array{Int64,1}[[1]]          
 Array{Int64,1}[[3]]          
 Array{Int64,1}[[2,3]]        
 Array{Int64,1}[[1],[3]]      
 Array{Int64,1}[[4,9,11]]     
 Array{Int64,1}[[1],[2,3]]    
 Array{Int64,1}[[2,3],[3]]    
 Array{Int64,1}[[1],[4,9,11]] 
 Array{Int64,1}[[3],[4,9,11]] 
 Array{Int64,1}[[1],[2,3],[3]]