问题描述
假定大小为1 x n的向量v和函数fun接受长度为L的向量并返回大小为p x 1的向量.
Assuming vector v of size 1 x n and function fun that takes in a vector of length L and returns a vector of size p x 1.
是否有一个MATLAB函数可以接收向量v，并使用fun函数处理长度为L的每个滑动窗口，并返回大小为p x n(或p x (n-L))的矩阵.
Is there a MATLAB function that would take in vector v, process each sliding window of length L with function fun, and return a matrix of size p x n (or p x (n-L)).
我知道这可以通过使用im2col创建窗口矢量矩阵并对其进行处理来实现，但这对于长矢量v来说会占用太多内存.
I am aware this could be achieved with creating a matrix of windowed vectors with im2col and processing each of those, but this takes too much memory for a long vector v.
推荐答案
funsl=@(is) fun(v(is:is+l-1));
cell2mat(arrayfun(funsl,1:length(v)-l+1,'UniformOutput',false))
我在这里所做的是定义一个匿名函数，对于固定的v和l和起始索引参数(is)，获取相应的v切片并将fun应用于该函数
What I did here is define an anonymous function that, for a fixed v and l and a starting index parameter (is), gets the respective slice of v and applies fun to it.
然后通过arrayfun将此函数应用于该起始索引的所有有用值.出于某种原因，我本人目前无法准确命名，每个应用程序都返回一个p x 1向量，但是arrayfun无法将其排列成适当的矩阵，因此UniformOutput = false设置和cell2mat围绕它进行调用.
Then this function is applied, via arrayfun, to all useful values for this starting index. For reasons I myself cannot quite name at the moment, each application returns a p x 1 vector, but arrayfun cannot arrange it into a proper matrix, thus the UniformOutput=false setting and the cell2mat call around it.
编辑:要使用将1×5矢量转换为我使用的4×1矢量的功能进行测试
Edit: To test this with a function that turns 1-by-5 vectors into 4-by-1 vectors I used
l=5;v=1:12; fun=@(x) cumsum(x(2:end))';
并得到以下结果:
ans =
2 3 4 5 6 7 8 9
5 7 9 11 13 15 17 19
9 12 15 18 21 24 27 30
14 18 22 26 30 34 38 42
再次注意，在funsl v的定义中是固定的，要将这种方法应用于不同的v，您可以制作另一个采用v(和l)的函数，如果您不想这样做的话.将其修复为参数，包含上面的两行并返回第二行的结果.
Note again that in the definition of funsl v is fixed, To apply this approach to different v you could make another function that takes v (and l, if you do not want to fix this) as parameter(s), contains the two lines above and returns the result of the second one.
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