简单的把矩阵分解成一个正交矩阵和一个对角线全为1的上三角矩阵(转载)

转载： https://blog.csdn.net/aimatfuture/article/details/7741892

function [He Xx] = GQR(A)
% param A : 是一个可逆矩阵
% return He : 是一个满足每两个不同列向量的内积为0的矩阵
% return Xx : 是一个主对角线全为1的上三角矩阵
% function : 分解 A = He * Xx
% CopyRight : edu.nust.cs726.JunH(111060881)


[H L] = size(A);
if H ~= L     error('hehehe');
end


if det(A) == 0     error('hahaha');
end


BIJ = eye(H);
B = zeros(H, L);
B(:, 1) = A(:, 1);
for R = 2:L     ar = A(:, R);     br = ar;     for K = 1:R-1
%         disp(ar)
%         disp(B(:, K));
%         disp(' ');
%         disp(' ');         bij = Gb(ar, B(:, K));         BIJ(K, R) = bij;         br = br - bij*B(:, K);     end     B(:, R) = br;
end
He = B;
Xx = BIJ;


function  bij = Gb(aierfa, beita)
if 1 == size(aierfa, 1) && 1 == size(beita, 1)     bij = (aierfa*beita') / (beita*beita');
else     bij = (aierfa'*beita) / (beita'*beita);
end

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import sys, os
import json
import traceback
import time
import csv
import numpy as np
from PIL import Image
import math

# https://blog.csdn.net/aimatfuture/article/details/7741892
# http://blog.sina.com.cn/s/blog_bb81c2230102xe1u.html
def getScalue(a, b, bandSize):
  bb =0.0
  ab = 0.0
  for i in range(bandSize):
    ab = ab + a[i]*b[i]
    bb = bb + b[i]*b[i]
  scale = ab / bb
  return scale


def gramSchmidt( rasterData,  bandCount,  bandSize,  R):
  for i in range(bandCount):
    for j in range(bandCount):
      R[i][j] = 1 if (i == j) else 0
  #
  column =  [0] * bandSize
  for i in range(1,bandCount):
    for j in range(0, bandSize):
      column[j] = rasterData[i][j]
    #
    for j in range(0, i):
      scaleValue = getScalue(column, rasterData[j], bandSize)
      R[j][i] = scaleValue
      for k in range(0, bandSize):
        column[k] = column[k] - scaleValue * rasterData[j][k]
    #
    for j in range(0, bandSize):
      rasterData[i][j] = column[j]


  return rasterData


def GSModify(U):
  d, n = U.shape
  m = min(d, n)
  R = np.eye(m, n)

  Q = np.zeros((d, m), dtype=np.float32)
  D = np.zeros(m, dtype=np.float32)

  for i in range(0, m):
    if i == 0:
      e = U[:, i]
      Q[:, i] = e
      D[i] = np.dot(e, e.T)
    else:
      a = Q[:, 0:i]  # Q[:, 0:i-1]
      b = 1 / D[0:i]  # 1 / D[0:i-1]
      c = np.multiply(a, b)
      d = c.T            # .reshape((height*width, i))
      e = U[:, i]
      f = np.dot(d, e)
      R[0:i, i] = f
      g = np.dot(a, f)
      h = (e-g)
      Q[:, i] = h
      k = np.dot(h, h.T)
      D[i] = k
  return Q, R


rasterData1 = [ [1, 2, -1], [-1,3,1], [4,-1,0]]
R1 = [ [0, 0, 0], [0, 0, 0], [0, 0, 0]]

Q1 = gramSchmidt( rasterData1,  3,  3,  R1)

print(R1)
print(Q1)

print('---------------------------------------------------------------')
panDataset = np.array([1, 2, -1, -1,3,1, 4,-1,0])
rasterData2 = panDataset.reshape((3, 3), order='F')
Q2, R2 = GSModify(rasterData2)

print(rasterData2)
print(R2)
print(Q2)

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来源：https://www.cnblogs.com/gispathfinder/p/11753840.html

标签

矩阵分解

return

column