Tricky matrix conversion question

2 次查看（过去 30 天）
Niels de Vries
Niels de Vries 2018-8-7
编辑： Adam Danz 2018-8-7
Hey all,
I am trying to convert a matrix A into a new matrix B, by looking at each element in matrix A and whenever the next element (in a row) is different, make a pair and add it to the new matrix B. Also elements with number 0 should be ignored.
As an example: Suppose we have a matrix A, and we want the resulted matrix B.
A = [1 1 1 3 5 B = [1 3
2 2 0 4 5 3 5
2 2 0 0 5] 2 4
4 5
2 5]
This problem could be solved by using a for-loop over all elements in the matrix, however for computational time reasons i am looking for a faster way to convert this, any ideas are welcome :D

请先登录，再回答此问题。

采纳的回答

Guillaume
Guillaume 2018-8-7
Actually, it can be done fairly easily without a loop. Whether or not it's faster than Adam's loop, I haven't tested.
A = [1 1 1 3 5;2 2 0 4 5; 2 2 0 0 5];
rows = repmat(1:size(A, 1), size(A, 2), 1);
At = A';
mask = At ~= 0;
Anz = At(mask);
pair1 = diff(Anz) ~= 0 & ~diff(rows(mask));
B = [Anz(pair1), Anz([false; pair1])]
  1 个评论
Adam Danz
Adam Danz 2018-8-7
编辑：Adam Danz 2018-8-7
Slow clap! (+1)
@Niels de Vries, this version is 10x faster than mine when tested with my 1 million randi() data. Use this one.

请先登录，再进行评论。

更多回答（1 个）

Adam Danz
Adam Danz 2018-8-7
编辑：Adam Danz 2018-8-7
Sure, it's probably possible to avoid a loop entirely. But that doesn't mean it's practical.
This proposal loops through rows of A, removes 0s, differentiates to find changes, then extracts pairs and stores them in B to produce your desired results.
B = [];
% Loop over rows of A
for i = 1:size(A,1)
% remove 0s
tempRow = A(i,A(i,:)~=0);
% differentiate
d = diff(tempRow);
% Extract non=0 pairs
pairedIdx = find(d~=0);
% Detect pairs that differ and store in B
pairedNums = [tempRow(pairedIdx)', tempRow(pairedIdx+1)'];
B(end + 1 : end + size(pairedNums,1),:) = pairedNums;
end
In fact, when I change A to
A = randi(10, 1000, 1000)-1;
which contains 1 million elements in 1000 rows, it only takes ~1 sec using tic/toc.

类别

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

标签

产品


版本

R2017b

Community Treasure Hunt

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

Start Hunting!

Translated by