问题描述
我有一个矩阵A，其中包含一年中每一天的24个值(每小时一个值). A的每一列都是不同的一天，每天都有24行数据(A是24-by-365).我想通过比较每天的小时数据来相互比较每一天.为此，我获取一列数据并将其与下一列进行比较.我将两列中每个小时的数据之差取平方，然后求和并求和得出一个单一值，该值指示两天的相似程度.然后，我用可能的几天组合进行此操作，创建一个365 x 365矩阵d，该矩阵表示每一天之间的相似度.例如，元素d(20,100)包含一个值，该值指示一年中的第20天与第100天有多相似.该代码正在运行，但是速度很慢，我希望能够对其进行向量化.帮助将不胜感激.
I have a matrix A that contains 24 values for each day of the year (one value for each hour). Each column of A is a different day and each day has 24 rows worth of data (A is 24-by-365). I want to compare each day to each other by comparing the hour data of each day. To do this, I take one column of data and compare it to the next column. I take the difference of each hour's data in the two columns and then square and sum them to get a single value indicating how similar the two days are. I then do this with every possible combination of days, creating a 365-by-365 matrix, d, indicating how similar each day is to each other day. For example, element d(20,100) contains a value indicating how similar the 20th day of the year is to the 100th. The code is working, but it is quite slow and I would like to be able to vectorize it. Help would be greatly appreciated.
for j=1:365
for k=1:365
d(j,k)=sqrt(sum((A(:,j)-A(:,k)).^2));
end
end
推荐答案
使用 pdist ，它在C语言中起着举足轻重的作用，并且> c6> 创建距离矩阵:
d = squareform(pdist(A.'));
如果您需要更快的速度(虽然365 x 365并不是很大)，请在此处或尝试使用此文件交换程序
If you need this to be even faster (365-by-365 is not very big though), see my answer here or try this File Exchange program.
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