Music@life

 

 

 Case Study
http://kevinjc.pixnet.net/album/photo/296640567
 

轉載：Stigmergy!如白蟻建築般的權力構造
 
 

KMeans.java
/**************************************************************************************************************************
(c) Copyright: Kevin Chen , No.79912001, 2010~2012. All rights reserved.
Author:         : Kevin Chen  
Desc            :
Change History  : -
Class used in   :
Change History  :   Level              Changed By              Date                        Change description
                   -------------------------------------------------------------------------------------------------------
                    (Initial)          Kevin Chen             20100112                  (Initial)
'**************************************************************************************************************************/
package e9912001;
public class KMeans {
    double[][] m_TestData;
    int m_NumOfClusters;
    int m_TestCases;
    int m_TestVariables;
    double[][] m_Centroid;
    int[][] m_aryResultClusters;
    int  m_MAXLOOP = 100;
    public int getM_NumOfClusters() {
        return m_NumOfClusters;
    }
    public void setM_NumOfClusters(int mNumOfClusters) {
        m_NumOfClusters = mNumOfClusters;
        m_ClusterNo = new int[m_NumOfClusters];
    }
    public double[][] getM_TestData() {
        return m_TestData;
    }
    public int getM_TestCases() {
        return m_TestCases;
    }
    public int getM_TestVariables() {
        return m_TestVariables;
    }
    public int[] getM_ClusterNo() {
        return m_ClusterNo;
    }
    public double[][] getM_Centroid() {
        return m_Centroid;
    }
    int[] m_ClusterNo ;
    
    public KMeans(double[][] input_data){
        m_TestData = input_data;
        m_TestCases = m_TestData.length;
        m_TestVariables = m_TestData[0].length;
        //m_NumOfGroups = (int)Math.Sqrt(m_TestCases) + 1;
        m_NumOfClusters = 3;
        m_ClusterNo = new int[m_NumOfClusters];
    }
    public double[] getGrandCentroid(){
        double[] ret = new double[this.m_TestVariables];
        int sum;
        for(int i=0; i < this.m_TestData.length;i++){
            sum =0;
            for(int j=0; j < this.m_TestData[i].length;j++){
                sum += this.m_TestData[i][j];
            }
            ret[i] = sum / this.m_TestCases;            
        }        
        return ret;
    }

Exploit vs Explore
Exploit  翻譯成開採比較能理解，開採有在現有地區(地點、市場、方案)挖掘(找尋)，往深度進行(z軸的搜索)。
Explore 翻譯成探勘，有開發新地區、市場，新方案的意思，往廣度進行(x,y軸的搜索)。

 Association 與 Correlation analysis 差異在哪?
查網路有關詞的解釋
association =>
聯結:結合在一起

Algorithm
The following algorithm is an agglomerative scheme that erases rows and columns in a proximity matrix as old clusters are merged into new ones. The proximity matrix D contains all distances d(i,j). The clusterings are assigned sequence numbers 0,1,......, (n − 1) and L(k) is the level of the kth clustering. A cluster with sequence number m is denoted (m) and the proximity between clusters (r) and (s) is denoted d[(r),(s)].

Additional DetailsFormulate and then solve a linear programming model of this problem, to determine how many
containers of each product to produce tomorrow to maximize profits.
The company makes four
juice products using orange, grapefruit, and pineapple juice.
Product Retail Price Per Quart
Orange juice $1.00
Grapefruit juice .90
Pineapple juice .80
All-in-One 1.10
The All-in-One juice has equal parts of orange, grapefruit, and pineapple juice. Each product
is produced in a one- quart size. On hand are 400 gallons of orange juice,300 gallons of grapefruit
juice, and 200 gallons of pineapple juice.
The cost per gallon is $2.00 for orange juice,
$1.60 for grapefruit juice,
and 1.40 for pineapple juice.
In addition, the manager wants grapefruit juice to be used for no more than 30 percent of
the number of containers produced( grapefruit <= 0.3(orange +grapefruit + pineapple + all-in-one). She wants the ratio of the number of containers of orange
juice to the number of containers of pineapple juice to be at least 7 to 5.( orange:pineapple >= 7:5)
1) find how many containers of each product to produce tomorrow to mazimize profits.
2) Find any Surplus and/or Slack.
I've been looking at this for days.. and I can't seem to grasp the idea and work it out.

email regex pattern
emails = re.findall('[a-zA-Z0-9+_\-\.]+@[0-9a-zA-Z][.-0-9a-zA-Z]*.[a-zA-Z]+', page)
 

用看的比用說的容易理解