Question

有没有一种很好的方法来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的所有密钥组合，其中n可能是fx 3

所以在上面的例子中，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，但我的字典相当大，我正在重新设计整个算法，因为它在n > 3时开始疯狂交换，从而极大地降低了效率

tl; dr 有没有办法从没有collect字典的字典中创建组合迭代器？

Answer 1

以下是一个直接的实现，它试图通过字典最小化一点。此外，它使用OrderedDict，因此保持关键索引是有意义的（因为Dicts不承诺每次都进行一致的密钥迭代，因此有意义的密钥索引）。

using Iterators
using DataStructures

od = OrderedDict([1] => [1,2], [2,3] => [15], [3] => [6,7,8], [4,9,11] => [3])

sv = map(length,keys(od))        # store length of keys for quicker calculations
maxmaxlen = sum(sv)              # maximum total elements in good key
for maxlen=1:maxmaxlen           # replace maxmaxlen with lower value if too slow
  @show maxlen
  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->println(x),gkeys)
end

这比你现在拥有的更有效吗？将代码放在函数中或将其转换为Julia任务也会更好，每次迭代都会生成下一个键集。

---更新---

使用https://stackoverflow.com/a/41074729/3580870

中任务中的迭代器答案

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迭代所有密钥子集，直到组合大小为4（在运行其他StackOverflow答案的代码之后）：

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]]

（从链接的StackOverflow答案中定义NewTask是必要的）。

Julia - 迭代字典中的键组合

1 个答案: