How can I resolve Error Using max function
    4 visualizzazioni (ultimi 30 giorni)
  
       Mostra commenti meno recenti
    
    Sumit Saha
 il 8 Gen 2021
  
    
    
    
    
    Commentato: Star Strider
      
      
 il 9 Gen 2021
            clc
clear all;
close all;
for kk =1:44
for k=1:50
 % here drift values are in the form of drift ratio
 eval(['load IDAOutput/EQ_',num2str(kk),'/Scale_',num2str(k),'/StoryDrifts/','Roof.txt']);
 Roof_Drift_max(kk,k,:)= max(abs(Roof(:,2)));
 eval(['load IDAOutput/EQ_',num2str(kk),'/Scale_',num2str(k),'/StoryDrifts/','Story1.txt']);
 Story1_Drift_max(kk,k,:)= max(abs(Story1(:,2)));
 eval(['load IDAOutput/EQ_',num2str(kk),'/Scale_',num2str(k),'/StoryDrifts/','Story2.txt']);
 Story2_Drift_max(kk,k,:)= max(abs(Story2(:,2)));
 eval(['load IDAOutput/EQ_',num2str(kk),'/Scale_',num2str(k),'/StoryDrifts/','Story3.txt']);
 Story3_Drift_max(kk,k,:)= max(abs(Story3(:,2)));
 eval(['load IDAOutput/EQ_',num2str(kk),'/Scale_',num2str(k),'/StoryDrifts/','Story4.txt']);
 Story4_Drift_max(kk,k,:)= max(abs(Story4(:,2)));
 eval(['load IDAOutput/EQ_',num2str(kk),'/Scale_',num2str(k),'/StoryDrifts/','Story5.txt']);
 Story5_Drift_max(kk,k,:)= max(abs(Story5(:,2)));
 eval(['load IDAOutput/EQ_',num2str(kk),'/Scale_',num2str(k),'/StoryDrifts/','Story6.txt']);
 Story6_Drift_max(kk,k,:)= max(abs(Story6(:,2)));
 eval(['load IDAOutput/EQ_',num2str(kk),'/Scale_',num2str(k),'/StoryDrifts/','Story7.txt']);
 Story7_Drift_max(kk,k,:)= max(abs(Story7(:,2)));
 eval(['load IDAOutput/EQ_',num2str(kk),'/Scale_',num2str(k),'/StoryDrifts/','Story8.txt']);
 Story8_Drift_max(kk,k,:)= max(abs(Story8(:,2)));
 eval(['load IDAOutput/EQ_',num2str(kk),'/Scale_',num2str(k),'/StoryDrifts/','Story9.txt']);
 Story9_Drift_max(kk,k,:)= max(abs(Story9(:,2)));
% Maximum Interstorey Drift Ratio
 MIDR(kk,k,:) =max(Roof_Drift_max(kk,k,:),Story1_Drift_max(kk,k,:),Story2_Drift_max(kk,k,:),Story3_Drift_max(kk,k,:),Story4_Drift_max(kk,k,:),...
                   Story5_Drift_max(kk,k,:),Story6_Drift_max(kk,k,:),Story7_Drift_max(kk,k,:),Story8_Drift_max(kk,k,:),Story9_Drift_max(kk,k,:));
end
end
0 Commenti
Risposta accettata
  Star Strider
      
      
 il 8 Gen 2021
        I am not certain what you want.  
One option is to enclose all the arguments within square brackets [], effectively concatenating them: 
 MIDR(kk,k,:) =max([Roof_Drift_max(kk,k,:),Story1_Drift_max(kk,k,:),Story2_Drift_max(kk,k,:),Story3_Drift_max(kk,k,:),Story4_Drift_max(kk,k,:),...
                   Story5_Drift_max(kk,k,:),Story6_Drift_max(kk,k,:),Story7_Drift_max(kk,k,:),Story8_Drift_max(kk,k,:),Story9_Drift_max(kk,k,:)]);
This will produce an ‘MIDR’ matrix with the same dimensions as the argument matrices.  
Experiment with other approaches to get different results.  
16 Commenti
  Star Strider
      
      
 il 9 Gen 2021
				Try this to extract the variables: 
uz = unzip('IDAOutput.zip');
k2 = 0;
for k1 = 1:numel(uz)
    if ~isdir(uz(k1))
        [~,txtname] = fileparts(uz{k1});
        k2 = k2+1;
        LD{k2,1} = txtname;
        LD{k2,2} = load(uz{k1});
    end
end
I have to unzip it, so I included tahat as part of my code.  Change my code to work with the .zip file or it was created from.  My code extracts the contents of the .zip file to a series of cell arrays, with the first element of the cell array being the name of the file it came from, and the second cell array the contents of that file as a double matrix.  
So to get the information for the first 5 elements of ‘LD(1)’: 
BldgPart_1 = [LD{1,1}]
BldgPrtData_1 = [LD{1,2}(1:5,:)]
produces: 
BldgPart_1 =
    'Roof'
BldgPrtData_1 =
        0.002   6.9378e-08
        0.004   6.9378e-08
        0.006   6.9378e-08
        0.008   6.9378e-08
         0.01   6.9378e-08
and for ‘LD(40)’: 
BldgPart_40 = [LD{40,1}]
BldgPrtData_40 = [LD{40,2}(1:5,:)]
produces: 
BldgPart_40 =
    'Story9'
BldgPrtData_40 =
        0.002   5.5276e-07
        0.004   5.5276e-07
        0.006   5.5276e-07
        0.008   5.5276e-07
         0.01   5.5276e-07
I leave the rest to you.  
Più risposte (1)
Vedere anche
Categorie
				Scopri di più su Get Started with MATLAB in Help Center e File Exchange
			
	Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!



