推 xavierqqqq:非常感謝 04/02 22:43
※ 引述《xavierqqqq (Eye煙霧瀰漫)》之銘言：
: 假如 X 是 data matrix
: - T -1 -
: 要得出 dj = (xj-x) s (xj-x) j=1....n
: 我只會一個一個算
: 請問有什麼辦法可以很快就把n個算出來?
: 因為n很大時不知道該怎辦 3Q
Hi
將以下的latex code貼至
http://www.sciencesoft.at/index.jsp?link=latex&lang=en
大框內. Format改成pdfwrite會出現papersize,再選a4
完成以上設定後按Start latex開始執行程式
程式執行後下方有個link,點下去可以看見pdf檔
%%%%%%%%%%%%%% The below: latex code %%%%%%%%%%%%%%%%%%%
\documentclass[11pt]{letter}
\usepackage{amsmath}
\begin{document}
Hello, you can try it as follows.
Given the $ N \times P $ data matrix $ {\bf X} $ and $ {\bf
1}^T=[\underbrace{1,1,\dots,1}_{N}] $.
Let $ \overline{{\bf x}} = \left( \frac{1}{N} \right) {\bf X}^T{\bf 1} $,
and then you can calculate deviation matrix
$ {\bf D_X}={\bf X}- {\bf 1}\overline{{\bf x}}^T$.
The diagonal part of $ \left( {\bf D_X} \right) {\bf S}^{-1} \left(
{\bf D_X} \right)^T$ is what you want.
\end{document}
%%%%%%%%%%%%%% The above: latex code %%%%%%%%%%%%%%%%%%%
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