how to formulate logical matrix in a loop?

2018 10 24
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更新时间：2018 10 25
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i am having a matrix and want to separate them in the depending on its value
A=[1,2,3,4,5,6,7,8,9,10]
expected result are
idx1=[1,0,0,0,0,0,0,0,0,0] % for 1
...
idx10=[0,0,0,0,0,0,0,0,0,1] % for 10

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Andrei Bobrov
Andrei Bobrov 2018-10-25

1 个投票

A = 1:10;
idx = A(:) == A(:)'

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what should i change to have a 5 rows with a step of 2
Andrei Bobrov
Andrei Bobrov 2018-10-25
what should i change to have a 5 rows with a step of 2
?
A = 1:2:10;
idx = A(:) == A(:)';

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更多回答（1 个）

madhan ravi
madhan ravi 2018-10-24
编辑：madhan ravi 2018-10-24

1 个投票

A=[1,2,3,4,5,6,7,8,9,10]
RESULT = zeros(1,numel(A));
RESULT1= RESULT;
for i = 1:numel(A)
idx(i)=A(i)==1;
idx1(i)=A(i)==10;
end
RESULT(idx) = A(idx)
RESULT1(idx1) = A(idx1)

10 个评论
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madhan ravi
madhan ravi 2018-10-24
No need to use logical indexing , see the above illustration
madhan ravi
madhan ravi 2018-10-24
See the edited answer ,problem solved!
Shubham Mohan Tatpalliwar
编辑：Shubham Mohan Tatpalliwar 2018-10-25
A=[1,2,3,4,5,6,7,8,9,10]
idy=zeros(1,length(A));
for i=1:1:10
idx22=find(A >=(i) & A<(i+1));
idy(i,idx22)=1;
end
madhan ravi
madhan ravi 2018-10-24
编辑：madhan ravi 2018-10-24
hows this related to the question??? what did you do in the above comment?
Shubham Mohan Tatpalliwar
编辑：Shubham Mohan Tatpalliwar 2018-10-25
i have created a array
in which at evry row the numbers are indexed in a step of 1
from this array i can use evry row as a vector
and important is that, its done in a loop so i do not have to repeat the operation 100 times
try the updated code to get a clear idea
madhan ravi
madhan ravi 2018-10-25
I know but you don’t have Use a loop
madhan ravi
madhan ravi 2018-10-25
A=[1,2,3,4,5,6,7,8,9,10]
idy=zeros(1,length(A));
for i=1:1:10
idx22=find(A >=(i) & A<(i+1));
idy(i,idx22)=1;
end
madhan ravi
madhan ravi 2018-10-25
what do you mean by the above code?
madhan ravi
madhan ravi 2018-10-25
What should be the result after the loop?

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