logical indexing
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imagine you have a two matrixes:
a=[1 2 3 4 5 6 7 8 9];
b=[1 0 1 0 1];
how do i use the b matrix as a logical index? I'd expect:
a(b)
ans =
[1 3 5]
but instead i get the error: "Subscript indices must either be real positive integers or logicals."
if I try
a(~b)
ans =
[2 4]
now I could use a(~~b) which does what i want but this seems inelegant. Can anyone suggest a better solution?
採用された回答
その他の回答 (4 件)
Aldin
2012 年 3 月 14 日
Here:
for i = 1:5
if b(i) == 1
disp(a(i))
end
end
:)
6 件のコメント
Aldin
2012 年 3 月 14 日
If you want i can put the result 1,3,5 in an array ???
Thijs
2012 年 3 月 14 日
Aldin
2012 年 3 月 14 日
Thanks
Aldin
2012 年 3 月 14 日
index = 0;
for i = 1:length(b)
if b(i) == 1
index = index + 1;
result(index) = a(i);
end
end
Aldin
2012 年 3 月 14 日
result =
1 3 5
Aldin
2012 年 3 月 14 日
Here is another solution:
a(b(1:5)==1)
Onomitra Ghosh
2012 年 3 月 14 日
Your "b" matrix is in double. You need to convert that to logical values for logical indexing:
>> a(boolean(b))
ans =
1 3 5
Aldin
2012 年 3 月 14 日
0 投票
but what if you haven't only '1' and '0' in b array. I think it's better my first solution or second &Onomitra Ghosh his code with logical work correctly
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