issues extracting sub matrix
显示 更早的评论
matrix x=[1:10]. I need to extract the submatrix s=[6, 4, 2] from x. every time i try to use the command:
x=[1:10]
s=[(min(x)+5:2:max(x)-6)]
i get this error, when the solution should be s= 6, 4:*
s =
1×0 empty double row vector
any help as to why this function will not extract a submatrix would be appreciated!
5 个评论
Guillaume
2017-1-25
The square brackets are not need for s either. It's all unnecessary clutter that slows the code down (square brackets are translated to a call to horzcat) and make it more difficult to read.
John BG
2017-1-26
编辑：John Kelly
2017-2-3
feel free to use the brackets the best you you consider.
MATLAB allows the omission of brackets when defining ranges in the way you have, yet it helps readers, it's ok, syntactically correct, it's ok there is more than one way of defining ranges.
Well, if you want extra useless character why not
x = [[[((((1)'*(((1)'')''+(0)')'')'':-(-1):10^1))]]]
It produces the same result as
x = 1:10
Matlab allows the addition (there's no omission, no bracket is the default state) of plenty of characters that may or may not do anything depending on context. Useless clutter does not make the code more readable.
Jan
2017-1-26
编辑：John Kelly
2017-2-3
It seems like this operator for concatenation is confused with a constructor of vectors, but this is a misunderstanding. It is not just my opinion or a question of taste, but even Matlab's code analyser suggests to omit this clutter by showing an MLint warning in the editor. This might not convince you, but it convinces me. And therefore you are correct: I will keep telling users of this forum how to improve Matlab code.
See Answers: why-not-use-square-brackets for more details and an example of speeding up indexing by a factor of 4 by avoiding the brackets around an indexing vector.
采纳的回答
The only safe way to make your expression work for any arbitrary x where min(x)+5 may or may not be greater than max(x)-6 is to make the sign of the increment dependent on the sign of the difference between the bounds. Therefore:
x = 1:10; %brackets not needed!
startbound = min(x)+5;
endbound = max(x)-6;
if endbound > startbound
s = startbound:2:endbound;
else
s = startbound:-2:endbound;
end
更多回答（2 个）
John BG
2017-1-23
Because
(min(x)+5<max(x)-6)
=
0
With a positive step, unless you provide the initial element of the range smaller than the end element MATLAB will not attempt to guess it.
s=[max(x)-6:2:min(x)+5]
=
4 6
or you provide a negative step
s=[(min(x)+5:-2:max(x)-6)]
s =
6 4
.
if you find this answer useful would you please be so kind to mark my answer as Accepted Answer?
To any other reader, please if you find this answer of any help solving your question,
please click on the thumbs-up vote link,
thanks in advance
John BG
Another solution:
x = 1:10;
a = min(x)+5;
b = max(x)-6;
s = x(a:(2 * sign(b-a)):b); % Fails for a==b, thanks Guillaume
w = 2 * (-1 + 2 * (b > a)); % Not cute
s = x(a:w:b);
1 个评论
Guillaume
2017-1-26
Works as long, as a ~= b. Otherwise you get an increment of zero and thus an empty vector.
类别
在 帮助中心 和 File Exchange 中查找有关 Creating and Concatenating Matrices 的更多信息
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
Translated by 
