Question

我正在使用Matlab，我在过滤矩阵方面遇到了问题。矩阵填充了数字上升，我想删除矩阵中的所有条目，这些条目比给定的intervall更接近于（n-1）之前的条目。

我已经使用for循环完成了这项工作。问题是，计算了五个小时！

我的代码：

for i=2:y   
    if ((arrMaster1(i-1)+deadtime)<arrMaster1(i))
        j=j+1;
        arrMaster1cut(j,3)=int64(arrMaster1(i));
    else
    end   
end

Answer 1

首先，您可以一次完成所有减法操作，而不是每个循环中的单个元素。使用diff就可以更轻松地实现这一点

diffs = diff(arrMaster1(1:y));
% equivalent to: diffs = arrMaster1(2:y) - arrMaster1(1:y-1);

然后创建一个逻辑数组，以测试您的差异是否大于deadtime：

keepthese = [true, diffs > deadtime]; 
% outputs vector like [1 0 1 1 1 0 ...], of length y, where 1 is true and 0 is false
% diffs is of length y-1, so extra "true" is needed.
% There is a "true" in the place of every element you want to keep.

最后，创建输出：

arrMaster1cut = int64(arrMaster1(keepthese));
% This is called logical indexing, you're choosing all elements of
% arrMaster1 where keepthese has the value 1 / true

一起：

diffs = diff(arrMaster1(1:y));                 % Get differences for tolerencing
keepthese = [true, diffs > deadtime];          % Get indices to keep
arrMaster1cut = int64(arrMaster1(keepthese));  % Create output

修改记事

keepthese向量是行向量，因此与逗号连接。如果arrMaster1是列向量，那么diffs也是如此，所以改为使用分号

keepthese = [true; diffs > deadtime];          % Get indices to keep (column vector)

有关索引的说明

逻辑索引与正常索引。假设我们想从5元素矩阵中拉出第1，第2和第4个元素。我们可以使用两种方法：

 elements = mymatrix([1,2,4]);
 elements = mymatrix(logical([1,1,0,1,0])); % logical converts 1/0 to true/false

显然，第二种方法（逻辑索引）在这个例子中并不清楚。但是，当在上面使用时，它允许我们避免调用find函数，这会将逻辑索引转换为编号向量：

find([1,1,0,1,0]); % = [1,2,4]

因此，我们获得了速度优势。

Answer 2

我没有你所有的代码，所以这只是猜测，但你的代码可能运行得很慢，因为你没有为arrMaster1cut预先分配内存，而是在每次迭代中改变它的大小。

此外，您可以使用diff来对整个代码进行矢量化：

% random arrMaster1 array
y = 100;
arrMaster1 = randi(100,[1 y]);
deadtime = 5;
% differences vector
d = diff(arrMaster1);
% differences greater than deadtime
idxs = find(d > deadtime);
% produce arrMaster1cut
arrMaster1cut1 = int64(arrMaster1(idxs+1));
% your method
j = 0;
arrMaster1cut2 = [];
for i=2:y
    if ((arrMaster1(i-1)+deadtime)<arrMaster1(i))
        j=j+1;
        arrMaster1cut2(j)=int64(arrMaster1(i));
    else
    end
end
% check for equal results
isequal(arrMaster1cut2, arrMaster1cut1)

条目关闭时过滤矩阵

2 个答案: