Mahout协同过滤算法源码分析（3-2）--QR分解数据流
Mahout版本：0.7，hadoop版本：1.0.4，jdk：1.7.0_25 64bit。
接上篇，当数据准备完成后，就可以来分析其数据流了。
首先要分析的是new QRDecomposition(Ai)，这个初始化QRDecomposition就做了好多的事情，具体分析如下：
先贴上源码，然后再分析：
public QRDecomposition(Matrix a) {
// Initialize.
qr = a.clone();
originalRows = a.numRows();
originalColumns = a.numCols();
rDiag = new DenseVector(originalColumns);
// precompute and cache some views to avoid regenerating them time and again
Vector[] QRcolumnsPart = new Vector[originalColumns];
for (int k = 0; k < originalColumns; k++) {
QRcolumnsPart[k] = qr.viewColumn(k).viewPart(k, originalRows - k);
}
// Main loop.
for (int k = 0; k < originalColumns; k++) {
//DoubleMatrix1D QRcolk = QR.viewColumn(k).viewPart(k,m-k);
// Compute 2-norm of k-th column without under/overflow.
double nrm = 0;
//if (k<m) nrm = QRcolumnsPart[k].aggregate(hypot,F.identity);
for (int i = k; i < originalRows; i++) { // fixes bug reported by [email protected]
nrm = Algebra.hypot(nrm, qr.getQuick(i, k));
}
if (nrm != 0.0) {
// Form k-th Householder vector.
if (qr.getQuick(k, k) < 0) {
nrm = -nrm;
}
QRcolumnsPart[k].assign(Functions.div(nrm));
/*
for (int i = k; i < m; i++) {
QR[i][k] /= nrm;
}
*/
qr.setQuick(k, k, qr.getQuick(k, k) + 1);
// Apply transformation to remaining columns.
for (int j = k + 1; j < originalColumns; j++) {
Vector QRcolj = qr.viewColumn(j).viewPart(k, originalRows - k);
double s = QRcolumnsPart[k].dot(QRcolj);
/*
// fixes bug reported by John Chambers
DoubleMatrix1D QRcolj = QR.viewColumn(j).viewPart(k,m-k);
double s = QRcolumnsPart[k].zDotProduct(QRcolumns[j]);
double s = 0.0;
for (int i = k; i < m; i++) {
s += QR[i][k]*QR[i][j];
}
*/
s = -s / qr.getQuick(k, k);
//QRcolumnsPart[j].assign(QRcolumns[k], F.plusMult(s));
for (int i = k; i < originalRows; i++) {
qr.setQuick(i, j, qr.getQuick(i, j) + s * qr.getQuick(i, k));
}
}
}
rDiag.setQuick(k, -nrm);
}
}初始化qr矩阵：就是Ai的clone值
[[31.678402777777777, 4.08661209859189, 4.573918596524476], [4.08661209859189, 1.0203966547288652, 0.3987296589988406], [4.573918596524476, 0.3987296589988406, 1.059026647737198]]QRcolumnsPart: 截取qr的半个矩阵（对角线）
[{0:31.678402777777777,1:4.08661209859189,2:4.573918596524476},
{0:1.0203966547288652,1:0.3987296589988406},
{0:1.059026647737198}]
+++++++++++++++++++++++++---------------------------------first time k=0
nrm：
32.26673724322168assign函数：不仅仅是更新了QRcolumnsPart值，同时更新了qr的值，QRcolumnsPart[k].assign(Functions.div(nrm))，这里因为QRcolumnsPart和qr中的部分引用是一样的，如下图：
对于QRcolumnsPart来说，做了如下更新（div_num在程序中的使用的变量是nrm）：
设 div_num=(sqrt(QRcolumnsPart[k][0]^2+QRcolumnsPart[k][1]^2+...QRcolumnsPart[k][size-1]^2))
QRcolumnsPart[k]=QRcolumnsPart[k]/div_num
其中QRcolumnsPart[k][i]是QRcolumnsPart[k]中的第i个元素，size是QRcolumnsPart[k]中一共含有的元素个数；
QRcolumnsPart更新为：
{0:0.9817665337214256,1:0.12665092438034994,2:0.14175336545641365}
{0:1.0203966547288652,1:0.3987296589988406}
{0:1.059026647737198} 对于qr，做了如下修改,因为qr存储的地址和QRcolumnsPart中的matrix中的values地址一样，所以修改了QRcolumnsPart 就会导致qr也被修改
qr[j]=qr[j]/div_num
qr更新为：
[[0.9817665337214256, 4.08661209859189, 4.573918596524476], [0.12665092438034994, 1.0203966547288652, 0.3987296589988406], [0.14175336545641365, 0.3987296589988406, 1.059026647737198]]然后qr的对角线（row=col）自增1：
qr更新为：
[[1.9817665337214256, 4.08661209859189, 4.573918596524476], [0.12665092438034994, 1.0203966547288652, 0.3987296589988406], [0.14175336545641365, 0.3987296589988406, 1.059026647737198]]QRcolumnsPart更新为：
[{0:1.9817665337214256,1:0.12665092438034994,2:0.14175336545641365}, -- 主循环改变的值QRcolumnsPart[k]
{0:1.0203966547288652,1:0.3987296589988406},
{0:1.059026647737198}] 内层for循环 ，循环从k+1开始，终止为Ai的列数，用j来表示次数
------------------------------sub-first time j=k+1=1 ，k=0
QRcolj: 取qr的第j列，且从k行开始
{0:4.08661209859189,1:1.0203966547288652,2:0.3987296589988406} s:就是QRcolumnsPart[k]和QRcolj的点乘，即每项对应相乘
8.28446654391689s重新赋值：s = -s / qr.getQuick(k, k);
-4.180344355881341qr：把qr的第j列第k行开始更新为原始值+s*第k列的相应值
[[1.9817665337214256, -4.197854445325, 4.573918596524476], [0.12665092438034994, 0.4909521778283148, 0.3987296589988406], [0.14175336545641365, -0.19384822221406328, 1.059026647737198]]QRcolumnsPart更新为：QRcolumnsPart截取的是qr的左下部分，和qr一直保持一致
[{0:1.9817665337214256,1:0.12665092438034994,2:0.14175336545641365},
{0:0.4909521778283148,1:-0.19384822221406328}, -- 内循环改变的值QRcolumnsPart[k+1]
{0:1.059026647737198}] --------------------------------sub-second time j=j+1=2，k=0
QRcolj: 取qr的第j列，且从k行开始
{0:4.573918596524476,1:0.3987296589988406,2:1.059026647737198} s:就是QRcolumnsPart[k]和QRcolj的点乘，即每项对应相乘
9.265058873873116s重新赋值：s = -s / qr.getQuick(k, k);
-4.675151545966863qr：把qr的第j列第k行开始更新为原始值+s*第k列的相应值
[[1.9817665337214256, -4.197854445325, -4.69114027734864], [0.12665092438034994, 0.4909521778283148, -0.1933826059160847], [0.14175336545641365, -0.19384822221406328, 0.39630818207763985]]QRcolumnsPart更新为：QRcolumnsPart截取的是qr的左下部分，和qr一直保持一致
[{0:1.9817665337214256,1:0.12665092438034994,2:0.14175336545641365},
{0:0.4909521778283148,1:-0.19384822221406328},
{0:0.39630818207763985}] -- 内循环改变的值QRcolumnsPart[k+2] ---------------------------------------- 内循环结束，可以看到外循环当k=0时设置的是QRcolumnsPart[0]的值
---------------------------------------- 内循环依次设置QRcolumnsPart[1]、...、QRcolumnsPart[rowSize-1]的值；
设置变量rDiag：rDiag.setQuick(k, -nrm);，其中的nrm就是前面的div_num
{0:-32.26673724322168}
+++++++++++++++++++++++++++++++++---------------------------------------------------second time k=1
nrm:
0.527836313803738QRcolumnsPart更新为: 经过assign函数后QRcolumnsPart[k]和qr（qr中和QRcolumnsPart[k]对应的部分）都会改变
[{0:1.9817665337214256,1:0.12665092438034994,2:0.14175336545641365},
{0:0.9301220188705365,1:-0.36725063650346085},
{0:0.39630818207763985}]qr：
[[1.9817665337214256, -4.197854445325, -4.69114027734864], [0.12665092438034994, 0.9301220188705365, -0.1933826059160847], [0.14175336545641365, -0.36725063650346085, 0.39630818207763985]]更新qr，对角线（row=col）加1
qr：
[[1.9817665337214256, -4.197854445325, -4.69114027734864], [0.12665092438034994, 1.9301220188705366, -0.1933826059160847], [0.14175336545641365, -0.36725063650346085, 0.39630818207763985]]QRcolumnsPart：
[{0:1.9817665337214256,1:0.12665092438034994,2:0.14175336545641365},
{0:1.9301220188705366,1:-0.36725063650346085},
{0:0.39630818207763985}
内层for循环 ，循环从k+1开始，终止为Ai的列数，用j来表示次数
------------------------------sub-first time j=k+1=2 ，k=1
QRcolj: 取qr的第j列，且从k行开始
{0:-0.1933826059160847,1:0.39630818207763985} s:就是QRcolumnsPart[k]和QRcolj的点乘，即每项对应相乘
-0.5187964578647415s重新赋值：s = -s / qr.getQuick(k, k);
0.2687894613876947qr：把qr的第j列第k行开始更新为原始值+s*第k列的相应值
[[1.9817665337214256, -4.197854445325, -4.69114027734864], [0.12665092438034994, 1.9301220188705366, 0.3254138519486568], [0.14175336545641365, -0.36725063650346085, 0.29759508129758655]]QRcolumnsPart更新为：QRcolumnsPart截取的是qr的左下部分，和qr一直保持一致
[{0:1.9817665337214256,1:0.12665092438034994,2:0.14175336545641365},
{0:1.9301220188705366,1:-0.36725063650346085},
{0:0.29759508129758655}] --------------------------------------------内层循环结束
--------------------------------------------
设置变量rDiag：rDiag.setQuick(k, -nrm);，其中的nrm就是前面的div_num
{0:-32.26673724322168,1:-0.527836313803738}++++++++++++++++++++++++++++++++---------------------------------------------------third time k=2
nrm:0.29759508129758655
QRcolumnsPart更新为: 经过assign函数后QRcolumnsPart[k]和qr（qr中和QRcolumnsPart[k]对应的部分）都会改变
qr：
[[1.9817665337214256, -4.197854445325, -4.69114027734864], [0.12665092438034994, 1.9301220188705366, 0.3254138519486568], [0.14175336545641365, -0.36725063650346085, 0.9999999999999999]]QRcolumnsPart：
[{0:1.9817665337214256,1:0.12665092438034994,2:0.14175336545641365},
{0:1.9301220188705366,1:-0.36725063650346085},
{0:0.9999999999999999}]更新qr，对角线（row=col）加1
qr：
[[1.9817665337214256, -4.197854445325, -4.69114027734864], [0.12665092438034994, 1.9301220188705366, 0.3254138519486568], [0.14175336545641365, -0.36725063650346085, 2.0]]QRcolumnsPart：
[{0:1.9817665337214256,1:0.12665092438034994,2:0.14175336545641365},
{0:1.9301220188705366,1:-0.36725063650346085},
{0:2.0}] 内层for循环 ，循环从k+1开始，终止为Ai的列数，用j来表示次数
------------------------------sub-first time j=k+1=3 ，k=2
-------------------------------- 直接退出，内层循环结束
设置变量rDiag：rDiag.setQuick(k, -nrm);，其中的nrm就是前面的div_num
{0:-32.26673724322168,1:-0.527836313803738,2:-0.29759508129758655}
至此，new QRDecomposition(Ai)算是分析完毕。下篇分析new QRDecomposition(Ai).solve(Vi).viewColumn(0)。好吧，我也快吐了。。。
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