How can I resolve Error Using max function
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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
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채택된 답변
Star Strider
2021년 1월 8일
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.
댓글 수: 16
Star Strider
2021년 1월 9일
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.
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