清空筛选条件
清空筛选条件

issues extracting sub matrix

1 次查看（过去 30 天）
B
B 2017-1-23
编辑： John Kelly 2017-2-3
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
显示 3更早的评论
Guillaume
Guillaume 2017-1-26
编辑：Guillaume 2017-1-26
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
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.

请先登录，再进行评论。

请先登录，再回答此问题。

采纳的回答

Guillaume
Guillaume 2017-1-25
编辑：Guillaume 2017-1-25
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
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
<mailto:jgb2012@sky.com jgb2012@sky.com>
Jan
Jan 2017-1-26
编辑：Jan 2017-1-26
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
Guillaume 2017-1-26
Works as long, as a ~= b. Otherwise you get an increment of zero and thus an empty vector.

请先登录，再进行评论。

类别

在 Help Center 和 File Exchange 中查找有关 Creating and Concatenating Matrices 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by