我们可以创建一个递归函数,它会探索您的元胞数组并创建一个树指针数组(如docs) 到每个节点的父节点。
该函数采用一个元胞数组(如您问题中的元胞数组),其中包含标量或嵌套元胞数组。
treebuilder
logic:
- 如果某项是标量,则为其分配父节点编号,并将节点编号加 1
- 如果某个项目是元胞数组,请运行
treebuilder
在该单元上,返回达到的最大节点数(以及生成的子树)。
- 由于步骤 2 的原因,递归函数,因此重复直到完成每个元素
功能:
function treearray = getTreeArray(cellarray)
% initialise the array construction from node 0
treearray = [0, treebuilder(cellarray, 1)];
% recursive tree building function, pass it a cell array and root node
function [out, node] = treebuilder(cellarray, rnode)
% Set up variables to be populated whilst looping
out = [];
% Start node off at root node
node = rnode;
% Loop over cell array elements, either recurse or add node
for ii = 1:numel(cellarray)
tb = []; node = node + 1;
if iscell(cellarray{ii})
[tb, node] = treebuilder(cellarray{ii}, node);
end
out = [out, rnode, tb];
end
end
end
简单示例的用法
这是一个比您的更简单的示例,因此我们可以轻松检查逻辑是否有效。
myCellArray = {1 1 {1 1 1 {1 1 1}}};
% This cell array has 3 levels:
% - 3 child nodes (2,3,4) of the root node (1)
% - Last node on the first level (4) has 4 children:
% - 4 child nodes on second level (5,6,7,8)
% - Last node on the first level (8) has 3 children:
% - 3 child nodes on third level (9,10,11)
myTreeArray = getTreeArray(myCellArray);
% Output, we see the corresponding nodes as listed above:
% [0 1 1 1 4 4 4 4 8 8 8]
treeplot(myTreeArray)
你的元胞数组
我认为这按预期工作,请注意您不必定义myCellArray
or myTreeArray
变量:
treeplot(getTreeArray({1,1,1,{1,1,1,{1,1,{1,{1 1 1 1 1 1 1 1}, 1,1}, 1,1},1,1,1},1,1,1,{1,1,1,1}}))
这里是输出图像,表明该算法可以应对更复杂的树。速度似乎也不太差,尽管显示极其无论如何,复杂的树将是相当多余的!
编辑:标记节点
您可以通过使用获取节点的位置来标记节点treelayout并在构建树形数组时跟踪遇到的值。该函数应该针对“跟踪”进行调整,如下所示:
function [treearray, nodevals] = getTreeArray(cellarray)
% initialise the array construction from node 0
[nodes, ~, nodevals] = treebuilder(cellarray, 1);
treearray = [0, nodes];
% recursive tree building function, pass it a cell array and root node
function [out, node, nodevals] = treebuilder(cellarray, rnode)
% Set up variables to be populated whilst looping
out = []; nodevals = {};
% Start node off at root node
node = rnode;
% Loop over cell array elements, either recurse or add node
for ii = 1:numel(cellarray)
node = node + 1;
if iscell(cellarray{ii})
[tb, node, nv] = treebuilder(cellarray{ii}, node);
out = [out, rnode, tb];
nodevals = [nodevals, nv];
else
out = [out, rnode];
nodevals = [nodevals, {node; cellarray{ii}}];
end
end
end
end
Note:您可以使用类似的改编来跟踪节点number而不是节点value如果您想对绘图上的每个节点进行编号。
我在这里使用了元胞数组,以便每个节点上可以有文本或数值。如果您只想要数值,它可能会缩短存储后的格式nodevals
而是在矩阵中。
然后绘制这个你可以使用
% Run the tree building script above
[treearray, nodevals] = getTreeArray(myCellArray);
% Plot
treeplot(treearray);
% Get the position of each node on the plot
[x,y] = treelayout(treearray);
% Get the indices of the nodes which have values stored
nodeidx = cell2mat(nodevals(1,:));
% Get the labels (values) corresponding to those nodes. Must be strings in cell array
labels = cellfun(@num2str, nodevals(2,:), 'uniformoutput', 0);
% Add labels, with a vertical offset to the y coords so that labels don't sit on nodes
text(x(nodeidx), y(nodeidx) - 0.03, labels);
单元格的输出示例myCellArray = {{17, 99.9}, 50}
,我选择这些数字是为了明确它们不是实际的“节点号”!