Study Log---linear regression implementation

The calculation of the partial derivative can be obtained by calculating the formula of W:

Assuming that the input data is stored in matrix x, the regression coefficients are stored in the vector W. So for a given data 650) this.width=650; "src="/e/u261/themes/default/images/spacer.gif "style=" Background:url ("/e/u261/lang /zh-cn/images/localimage.png ") no-repeat center;border:1px solid #ddd;" alt= "Spacer.gif"/>650) this.width=650; " Src= "http://img.blog.csdn.net/20150204222354655?watermark/2/text/aHR0cDovL2Jsb2cuY3Nkbi5uZXQvc2hhbmRpYW5rZQ==/ Font/5a6l5l2t/fontsize/400/fill/i0jbqkfcma==/dissolve/70/gravity/center "/> The forecast results will be this.width=650 by 650);" src= "Http://img.blog.csdn.net/20150204222359803?watermark/2/text/aHR0cDovL2Jsb2cuY3Nkbi5uZXQvc2hhbmRpYW5rZQ==/font /5a6l5l2t/fontsize/400/fill/i0jbqkfcma==/dissolve/70/gravity/center "/>650) this.width=650;" Src= "/e/u261/ Themes/default/images/spacer.gif "style=" Background:url ("/e/u261/lang/zh-cn/images/localimage.png") no-repeat center;border:1px solid #ddd; "alt=" spacer.gif "/> Given. For x and Y, how do I find w? The common method is to find the W with the least squared error.

The squared error can be written:

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650) this.width=650; "src="/e/u261/themes/default/images/spacer.gif "style=" Background:url ("/e/u261/lang/zh-cn/ Images/localimage.png ") no-repeat center;border:1px solid #ddd;" alt= "Spacer.gif"/>

With a matrix representation can also be written 650) this.width=650; "Src=" http://img.blog.csdn.net/20150204222409694?watermark/2/text/ ahr0cdovl2jsb2cuy3nkbi5uzxqvc2hhbmrpyw5rzq==/font/5a6l5l2t/fontsize/400/fill/i0jbqkfcma==/dissolve/70/gravity/ Center "/>650" this.width=650; "src="/e/u261/themes/default/images/spacer.gif "style=" Background:url ("/e/u261/ Lang/zh-cn/images/localimage.png ") no-repeat center;border:1px solid #ddd;" alt= "Spacer.gif"/>. To the W derivative, the solution is as follows:

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The data used is a data set that is loaded on the UCI to return the car's msg performance;

Because the data format of the download is not standard, so here wrote a piece of Java code to the format of the data set has been re-structured, the code is as follows:

import java.io.bufferedreader;import java.io.bufferedwriter;import java.io.file;import  Java.io.fileinputstream;import java.io.fileoutputstream;import java.io.inputstreamreader;import  java.io.OutputStreamWriter;public class MyMaze {         public static void main (String[] args)  throws Exception {         FileInputStream fileInputStream = new  FileInputStream (New file ("E:\\dataregression.txt"));         Bufferedreader bufferedreader = new bufferedreader (New inputstreamreader ( FileInputStream));         file file = new file ("E : \\result.txt ");         fileoutputstream fileoutputstream =  new fileoutputstream (file);   &nBsp;     bufferedwriter bufferedwriter = new bufferedwriter (New  outputstreamwriter (FileOutputStream));        string line;         String newline = null;         while ((Line = bufferedreader.readline ())!=null)          {            if (line  == null)             {                 break;             }             int length = line.length ();             for (int  i = 0; i<length; i++)              {                 while (Line.charat (i) = = '   ')                  {                     if (Line.charat (i+1)! = '   ')                      {                          newline = newline +  " ";                         break;                     }                     i++;                 }                 newline = newline +  line.charat (i);            }             newline = newline +  "\ r \ n";             newline = newline.substring ( 4);             bufferedwriter.write (newline);                         newline = null;        }                 bufferedwriter.close ();    }     }

The output file is a dataset with two spaces between each variable, where the first item is the dependent variable, or MSG.

The following is a linear regression of a dataset using the Python method:

Import numpy as npimport matplotlib.pyplot as pltnumfeat = len (Open (' Result.txt '). ReadLine (). Split ('    ')) datamat = []; labelmat = []fr =  open (' result.txt ')//The data in each row is segmented, extracting the data for each row for line in fr.readlines ():     linearr=[]    curline = line.split ('    ')      For i in range (1,numfeat):         linearr.append (Float ( Curline[i])     datamat.append (Linearr)     labelmat.append (float ( CURLINE[0])//convert the sequence to matrix Xmat = np.mat (Datamat) Ymat = np.mat (Labelmat). txtx = xmat.t*xmat/determines whether the determinant value is 0if np.linalg.det (xTx)  == 0.0:     print  "wrong"//Use the formula to find the parameters ws = xtx.i* (Xmat.t*ymat)//Use matplotlib drawing, developed in Fig fig =  Plt.figure () Ax = fig.add_subplot (111) xcopy  = xmat.copy () xcopy.sort (0) yhat = xcopy*ws//here is the relationship between an item in the X matrix and Yhat, as this is the second ax.plot (Xcopy[:,1],yhat) Display Image plt.show ()//Here is the function to find the correlation coefficient, the closer the 1 the better Yhat = xmat*wsprint yhat.t.shapeprint ymat.shapeprint  np.corrcoef (YHAT.T, YMAT.T)

Study Log---linear regression implementation