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将单元分配给单元阵列

[英]assign a cell to a cell array

I have the following matrix : 我有以下矩阵:

 adj =     O1   O2    O3    O4    S1    S2    S3    S4    S5

 O1        0     0     0     0     0     1     1     0     0
 O2        0     0     0     0     0     1     0     1     0
 O3        0     0     0     0     0     1     1     0     0
 O4        0     0     0     0     0     0     0     1     1
 S1        0     0     1     0     0     0     0     0     0
 S2        0     1     0     0     0     0     0     0     0
 S3        0     1     1     0     0     0     0     0     0
 S4        0     1     0     1     0     0     0     0     0
 S5        0     0     0     1     0     0     0     0     0

I have a cellarray nodeNames that contains the labels of the lines and columns of my matrix so : 我有一个cellarray nodeNames,其中包含矩阵行和列的标签,因此:

nodeNames = {O1 O2 O3 O4 S1 S2 S3 S4 S5}

I want to do the following: browse the matirce adj in the lines from the nodes S (i = 5: 9) and for the columns the nodes O (j = 1; 4) if adj (i, j) = 1 then get the label of the node Oj from nodeNames and put it in the cell w {i} of a new cellarray that we will name w . 我要执行以下操作:在节点S (i = 5:9)的行中浏览matirce adj,对于列O (j = 1; 4)的列,如果adj (i, j) = 1则得到从nodeNames删除节点Oj的标签,并将其放入新单元格的单元格w {i}中,我们将其命名为w so I want to have the following: 所以我想要以下内容:

w{1}={O3}.
w{2}={O2}.
w{3}={O2 O3}.
w{4}={O2 O4}.
w{5}={O4}.

I have tried to do that but I get an empty cellarray in my result : 我已经尝试过这样做,但是结果是空单元格:

for i=5 : 9
    k=1;
    for j=1:4
        if adj(i,j)==1;
            w{i}{k}=nodeNames{j};
        end
        k=k+1;
    end
end

I know that the problem is in the assignment to the cellarray w but can't get the right one , Any suggestion ? 我知道问题出在分配给细胞阵列w的问题上,但找不到正确的方法,有什么建议吗?

nodeNames or nodNames , also your k=k+1 must be in the if-statement. nodeNamesnodNames ,您的k = k + 1也必须在if语句中。

adj=[0,0,0,0,0,1,1,0,0;0,0,0,0,0,1,0,1,0;0,0,0,0,0,1,1,0,0;0,0,0,0,0,0,0,1,1;0,0,1,0,0,0,0,0,0;0,1,0,0,0,0,0,0,0;0,1,1,0,0,0,0,0,0;0,1,0,1,0,0,0,0,0;0,0,0,1,0,0,0,0,0]==1;
nodeNames = {'O1','O2','O3','O4','S1','S2','S3','S4','S5'};
w={};
for i = 5:9
    k=1;
    for j=1:4
        if adj(i,j)
        w{i-4,k}=nodeNames{j};
        k=k+1;
        end
    end
end

Result: 结果:

w =

  5×2 cell array

    'O3'      []
    'O2'      []
    'O2'    'O3'
    'O2'    'O4'
    'O4'      []

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