arrays - In octave, what gives with literal structs?

Question

when I do this:

pedigree = struct('parents', [0,0;1,3;0,0]);
pedigree.names = {'Ira','James','Robin'};

I get a different thing to when I do this:

pedigree = struct('parents', [0,0;1,3;0,0], 'names', {'Ira','James','Robin'});

What are the two things, how are they different, is there a literal form for whatever is made by the first thing, and more generally, what are the words for whatever is going on?

Answer 1

问题是结构也是一个数组。struct当您使用命令创建结构时

pedigree = struct('parents', [0,0;1,3;0,0], 'names', {'Ira','James','Robin'});

each 的值应该是一个元胞数组，一个元胞数组的每个元素都被分配。这样做是将矩阵分别设置[0 0; 1 3; 0 0]为 all pedigree.parents、和Ira、James和Robin、pedigree(1).names和。要做你想做的，只需传递一个带有单个元素的单元格数组：pedigree(2).namespedigree(3).names

pedigree = struct ("parents", [0,0;1,3;0,0],
                   "names", {{'Ira','James','Robin'}});

请注意元胞数组本身如何包含元胞数组，因此在这种情况下，您有一个具有单个元素的元胞数组（也是一个元胞数组）。

Answer 2

在这里可以找到很多启示：

https://www.gnu.org/software/octave/doc/interpreter/Data-Containers.html#Data-Containers

Answer 3

在这里可以找到很多启示：

http://abandonmatlab.wordpress.com/2009/08/12/programatically-creating-a-struct/