Removing the columns of a matrix
3 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
Hi,
I try to write a code that compares strings in a cell array. In this code, after it compares the strings in the cell array "type", it assigns the corresponding element in lat matrix to lat1 matrix. This situation is same in lat2 and long2 matrices. But in if statement, when it TF≠1, then it assigns 0 element in lat1 matrix. How can I remove the 0 elements from lat1,lat2,long1 and long2 matrices?
clear all
[lat,long,station,type]=textread('latandlong.txt','%f%f%s%s');
lat1=[];
long1=[];
lat2=[];
long2=[];
for i=1:24
TF=strcmpi('down',type{i,:})
if TF==1
lat1(i,:)=lat(i,:);
long1(i,:)=long(i,:);
elseif TF==0
lat2(i,:)=lat(i,:);
long2(i,:)=long(i,:);
end
end
1 Kommentar
Jan
am 19 Okt. 2011
About the useless "clear all" see: http://www.mathworks.com/matlabcentral/answers/16484-good-programming-practice#answer_22301
Akzeptierte Antwort
Jan
am 19 Okt. 2011
The 0 are not written into lat1 in the "if TF==0" block, but lat1(1, :) is filled with zeros automatically if you assign lat1(2, :). Try this:
x = [];
x(2) = 5
Modification of your code:
lat1=[];
long1=[];
lat2=[];
long2=[];
for i=1:24
TF=strcmpi('down', typeC{i,:}) % "type" => "typeC"!
if TF==1
lat1 = cat(1, lat1, lat(i,:));
long1 = cat(1, long1, long(i,:));
elseif TF==0
lat2 = cat(1, lat2, lat(i,:));
long2 = cat(1, long2, long(i,:));
end
end
But letting a matrix grow in each iteration is very inefficient. Faster and nicer:
index = strcmpi('down', typeC); % "type" => "typeC"!
lat1 = lat(index, :);
long1 = long(index, :);
lat2 = lat(~index, :);
long2 = long(~index, :);
Do not use "type" as name of a variable, because this shadows the built-in function with the same name.
0 Kommentare
Weitere Antworten (2)
Siehe auch
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!