Splitting a vector up into unequal sections seperated by zeros

15 次查看（过去 30 天）
Bran
Bran 2014-9-29
评论： Adam 2023-7-17
Hi everyone
I was wondering if someone could help me split up a vector into sections seperated by zeros. I have alot of data but if I give you an example with shorter data
if I have A = [ 1 2 3 0 0 0 0 0 2 3 4 0 0 0 0 0 4 5 6 7 0 0 0 0 1 1 1] and I would like section 1 = [ 1 2 3] section 2 = [ 2 3 4] section 3 = [ 4 5 6 7] section 4 = [ 1 1 1] how would I go about this?

请先登录，再回答此问题。

采纳的回答

Star Strider
Star Strider 2014-9-29
This works:
A = [1 2 3 0 0 0 0 0 2 3 4 0 0 0 0 0 4 5 6 7 0 0 0 0 1 1 1];
ne0 = find(A~=0); % Nonzero Elements
ix0 = unique([ne0(1) ne0(diff([0 ne0])>1)]); % Segment Start Indices
ix1 = ne0([find(diff([0 ne0])>1)-1 length(ne0)]); % Segment End Indices
for k1 = 1:length(ix0)
section{k1} = A(ix0(k1):ix1(k1));
end
celldisp(section) % Display Results
  19 个评论
显示 17更早的评论
Star Strider
Star Strider 2014-9-30
编辑：Star Strider 2014-9-30
I haven’t seen the attachment yet. Took a break, then came back to this, adding leading and trailing zeros to ‘A’ as well, since I discovered that those crashed my previous code. They don’t crash this version.
This works, producing ‘pure’ non-zero and zero sections:
A = [0 0 1 2 3 0 0 0 0 0 2 3 4 0 0 0 0 0 4 5 6 7 0 0 0 0 1 1 1 0 0 0];
ne0 = find(A~=0); % Nonzero Elements
ix0 = unique([ne0(1) ne0(diff([0 ne0])>1)]); % Non-Zero Segment Start Indices
eq0 = find(A==0); % Zero Elements
ix1 = unique([eq0(1) eq0(diff([0 eq0])>1)]); % Zero Segment Start Indices
ixv = sort([ix0 ix1 length(A)]); % Consecutive Indices Vector
for k1 = 1:length(ixv)-1
section{k1} = A(ixv(k1):ixv(k1+1)-1);
end
celldisp(section)
I ended up duplicating the index calculations for the non-zero segments to find the zero segments as well. This was an interesting problem!
EDIT — Just now saw ‘matrix A.mat’. My current code processed it without errors. (I displayed a few sections of it and the original matrix to be sure.) The only changes to my code to run it were to replace the original ‘A’ assignment with:
matA = load('matrix A.mat');
A = matA.ABSXX2;
since it is easier than changing the two references to ‘A’.
Adam
Adam 2023-7-17
Hello,
Your code helped me a lot, but it doesn't treat well the last section. I propose this code, which, I believe, covers all possible cases without error.
A = [0 0 1 2 3 0 0 0 0 0 2 3 4 0 0 0 0 0 4 5 6 7 0 0 0 0 1 1 1 0 0 0];
ne0 = find(A~=0); % Nonzero Elements
if ~isempty(ne0)
ix0 = unique([ne0(1) ne0(diff([0 ne0])>1)]); % Non-Zero Segment Start Indices
else
ix0 = [];
end
eq0 = find(A==0); % Zero Elements
if ~isempty(eq0)
ix1 = unique([eq0(1) eq0(diff([0 eq0])>1)]); % Zero Segment Start Indices
else
ix1 = [];
end
ixv = sort([ix0 ix1]); % Consecutive Indices Vector
section = cell(1, length(ixv));
for k1 = 1:length(ixv)
if k1 ~= length(ixv)
section{k1} = A(ixv(k1):ixv(k1+1)-1);
else
section{k1} = A(ixv(k1):length(A));
end
end
celldisp(section)

请先登录，再进行评论。

更多回答（1 个）

Michael Haderlein
Michael Haderlein 2014-9-29
Not sure if this question is finally answered as Bran seems to still struggle with an error. Anyway, I just wanted to mention this little workaround with strings:
c=cellfun(@abs,strsplit(char(A),char(0)),'uniform',false);

类别

在 Help Center 和 File Exchange 中查找有关 Logical 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