Table Variable Name-dot indexing

6 次查看（过去 30 天）

显示更早的评论

Mos_bad 2021-2-14

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/744862-table-variable-name-dot-indexing

评论： Walter Roberson 2021-2-15

在 MATLAB Online 中打开

I need to creat a table with the following variable names :

A1 B1 C1 A2 B2 C2 A3 B3 C3 , ... An Bn Cn X Y Z

A1 B1 C1 are Table's variable names that their values come from seprate arrays A, B, and C. Also, the last three column of the table, X, Y, AND Z, have nothing to do with aformentioned arrays. Any thoughts how to define variable names as above?

Here are what I wrote. Besides getting an error "Unable to perform assignment because dot indexing is not supported for variables of this type" , I believe there is definately a way more efficient way to do that.

    % Assume A, B, and C  as Location %d–Building Loss, Location %d–Time Element Loss, and Location %d–Total Loss 
    for n=1:Nbu
        VarName{1 , (n-1)*3+1} = sprintf('Location %d–Building Loss', n);
        VarName{1 , (n-1)*3+2} = sprintf('Location %d–Time Element Loss', n);
        VarName{1 , (n-1)*3+3} = sprintf('Location %d–Total Loss', n);
    end
    
    % Assume Building Loss, Time Element Loss, and Total Loss(Building + Time Element Loss) as X,Y, and Z; VARIABLESS' NAMES OF THE LAST THREE COLUMNS 
    VarName{1 , Nbu*3+1} = sprintf('Building Loss', n);
    VarName{1 , Nbu*3+2} = sprintf('Time Element Loss', n);
    VarName{1 , Nbu*3+3} = sprintf('Total Loss(Building + Time Element Loss)', n);
    Table1.Properties.VariableNames = VarName(1,:); 

7 个评论
显示 5更早的评论隐藏 5更早的评论

Mos_bad 2021-2-14

编辑：Walter Roberson 2021-2-15

Thanks, That worked and "compose" reduced several line of codes.

Last question, Any thoughts on rewriting the nested for loop in my previous comment?

(here is the direct link)

https://www.mathworks.com/matlabcentral/answers/744862-table-variable-name-dot-indexing#comment_1327452

Walter Roberson 2021-2-15

在 MATLAB Online 中打开

        for j=1:Nbu
            Table1{i,(j-1)*3+1} = BuildingLoss{i}(:,j) ; 
            Table1{i,(j-1)*3+2} = BuildingTimeElementLoss{i}(:,j); 
            Table1{i,(j-1)*3+3} = BuildTotLoss{i}(:,j) ;
        end

could be:

       Table1(i,:,1) = num2cell(BuildingLoss{i}, 1);
       Table1(i,:,2) = num2cell(BuildingTimeElementLoss{i}, 1);
       Table1(i,:,3) = num2cell(BuildingTotLoss{i}, 1);
       
       Table1(i,:,4) = {CoverageA(i,1)};  %same value for each j
       Table1(i,:,5) = {CoverageD(i,1)};  %same value for each j
       Table1(i,:,6) = {TotalLoss(i,1)};  %same value for each j