solving logic , array and condition
1 回表示 (過去 30 日間)
古いコメントを表示
Hope you are doing well :)
I Have code below. I want to create an array with elements p = [5; 10; 10; 15; 15; 20; 20; 25; 25; 30; ]; and I want replace if statement if (stateCount(n)> 5 && stateCount(n)<= 10) with value from an array 5 and 10. How can I do that. Any tips will gladly welcome
% My previous code value comes from a csv file and store in a matrix name value
value =
2.2034000000000
5.9016000000000
24.0833000000000
25.1000000000000
10.3115000000000
15.3934000000000
20.8525000000000
30.7288000000000
21.4918000000000
6.7833000000000
7.3667000000000
16.7541000000000
34.2034000000000
StateCount = value;
stateHit = zeros(length(stateCount),20);
for n = 1:length(stateCount)
if (stateCount(n)<= 5)
stateHit(n,1) = 1;
else
if (stateCount(n)> 5 && stateCount(n)<= 10)
stateHit(n,2) = 1;
else
if (stateCount(n)>10 && stateCount(n)<= 15)
stateHit(n,3) = 1;
else
if (stateCount(n)> 15 && stateCount(n)<= 20)
stateHit(n,4) = 1;
else
if (stateCount(n)> 20 && stateCount(n)<= 25)
stateHit(n,5) = 1;
else
if (stateCount(n)> 25 && stateCount(n)<= 30)
stateHit(n,6) = 1;
end
end
end
end
end
end
end
I have change the code like that below but it doesn't work
% My modified code
value =
2.2034000000000
5.9016000000000
24.0833000000000
25.1000000000000
10.3115000000000
15.3934000000000
20.8525000000000
30.7288000000000
21.4918000000000
6.7833000000000
7.3667000000000
16.7541000000000
34.2034000000000
p = [5; 10; 10; 15; 15; 20; 20; 25; 25; 30; 30; 35;];
StateCount = value;
stateHit = zeros(length(stateCount),6);
for n = 1:length(stateCount)
for i=1:size(p,1)
if (stateCount(n)<= p(i,:) )
stateHit(n,1) = 1;
else
end
end
end
I can't get result in stateHit, matrix is all zero. Could please help me to find the problem in the code
Many thanks in advance
0 件のコメント
回答 (1 件)
Andrei Bobrov
2012 年 5 月 29 日
on first code
value =[...
2.2034000000000
5.9016000000000
24.0833000000000
25.1000000000000
10.3115000000000
15.3934000000000
20.8525000000000
30.7288000000000
21.4918000000000
6.7833000000000
7.3667000000000
16.7541000000000
34.2034000000000];
p0 = [0,5,10,15,20,25,30,35];
[f,f] = histc(value,p0+eps(100));
t = f~=0;
n = numel(value);
s = (1:n)';
stateHit = accumarray([s(t),f(t)],ones(nnz(t),1),[n,numel(p0)-1]);
on second code [EDIT]
stateCount = value;
p = [0;5;10; 15; 20; 25; 30; 35;];
stateHit2 = zeros(numel(stateCount),numel(p)-1);
for n = 1:length(stateCount)
for ii = 1:numel(p)-1
if stateCount(n) > p(ii) & stateCount(n)<= p(ii+1)
stateHit2(n,ii) = 1;
end
end
end
MORE variant
p1 = [p0(1:end-1);p0(2:end)];
stateHit3 = bsxfun(@le,value,p1(2,:)) & bsxfun(@gt,value,p1(1,:));
2 件のコメント
参考
カテゴリ
Help Center および File Exchange で Resizing and Reshaping Matrices についてさらに検索
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!