Anyone know why I am recieving this error?
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z=1;
m=1;
at = 1;
data = rand(7004829, 1);
for k = 2 : at : length(data)
if k ~= (k+1)
index1=k;
y(m,1) = data(index1:z);
z = k;
m = m+1;
end
end
1 件のコメント
採用された回答
Image Analyst
2018 年 12 月 28 日
This is also useless:
if k ~= (k+1)
Can you tell me when k will EVER be equal to k+1? Answer: never. So the condition is always true. That wouldn't cause an error though - it's just bad programming.
Also, id, when you do get around to defining it. Must be a 2-D matrix.
And you might want to preallocate y in advance, though it's not necessary.
And finally, there are no comments. All professional programmers put comments in their code to help them understand later what they did, or to help others understand what they did. I don't really know what you're trying to do. Maybe there is a function or one vectorized line of code to do it, but since there are no comments, I don't know.
3 件のコメント
Walter Roberson
2018 年 12 月 28 日
image analyst: redefine length as uint8(255). when k reaches uint(255) then k~=k+1 would be false.
... it could happen !
Image Analyst
2018 年 12 月 28 日
Walter - true. Cem, if you just want to extract matrices that have only the ID number and no other numbers except zero where that number is not the desired number, you can see this example:
m = randi(3, 4, 6) % Sample ID array
m1 = double(ismember(m, 1)) % Get just the 1's
m2 = 2 * double(ismember(m, 2)) % Get just the 2's
m3 = 3 * double(ismember(m, 3)) % Get just the 3's
You'll see
m =
1 1 2 1 3 2
1 1 3 2 3 2
1 2 3 3 2 1
1 3 2 3 3 2
m1 =
1 1 0 1 0 0
1 1 0 0 0 0
1 0 0 0 0 1
1 0 0 0 0 0
m2 =
0 0 2 0 0 2
0 0 0 2 0 2
0 2 0 0 2 0
0 0 2 0 0 2
m3 =
0 0 0 0 3 0
0 0 3 0 3 0
0 0 3 3 0 0
0 3 0 3 3 0
Is that what you're hoping to do?
その他の回答 (1 件)
madhan ravi
2018 年 12 月 28 日
編集済み: madhan ravi
2018 年 12 月 28 日
- id is not defined
- if condition is superfluos though , always k is not equal to k + 1
- y needs to be preallocated
- y can be preallocate as a cell to further avoid error ( unable to perform ....due to size)
2 件のコメント
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