%?LLE?ALGORITHM?（using?K?nearest?neighbors）
%
%?[Y]?=?lle（XKdmax）
%
%?X?=?data?as?D?x?N?matrix?（D?=?dimensionality?N?=?#points）
%?K?=?number?of?neighbors
%?dmax?=?max?embedding?dimensionality
%?Y?=?embedding?as?dmax?x?N?matrix
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%?function?[Y]?=?lle（XKd）
clc;clear
load?MAIN_data1
feature=MAIN_data;
%4種故障模式
X=[feature（1:180:）;feature（541:720:）;feature（1081:1260:）;feature（1621:1800:）]‘;????%取5層信號(hào)小波包分解后的能量特征和9個(gè)時(shí)域特征
K=9;?????????????????????????????????????????????%近鄰個(gè)數(shù)
%?d=3;????????????????????????????????????????????%降維個(gè)數(shù)
[DN]?=?size（X）;??????????????????????????????????%D行N列，一列代表一個(gè)樣本，共有N個(gè)樣本
fprintf（1‘LLE?running?on?%d?points?in?%d?dimensions\n‘ND）;
%?STEP1:?COMPUTE?PAIRWISE?DISTANCES?&?FIND?NEIGHBORS?計(jì)算距離，尋找近鄰
fprintf（1‘-->Finding?%d?nearest?neighbours.\n‘K）;
%?X2?=?sum（X.^21）;???????????????????????????????%對(duì)每列求和，其中一列代表一個(gè)樣本，所以X2大小為1xN
%?distance?=?repmat（X2N1）+repmat（X2‘1N）-2*X‘*X;
distance?=?squareform（pdist（X‘?‘euclidean‘））;?????%計(jì)算歐式距離
[sortedindex]?=?sort（distance）;??????????????????%對(duì)樣本間的距離按從小到大排序
neighborhood?=?index（2:（1+K）:）;??????????????????%選取前K個(gè)近鄰



%?STEP2:?SOLVE?FOR?RECONSTRUCTION?WEIGHTS?計(jì)算重構(gòu)參數(shù)
fprintf（1‘-->Solving?for?reconstruction?weights.\n‘）;
if（K>D）?
??fprintf（1‘???[note:?K>D;?regularization?will?be?used]\n‘）;?
??tol=1e-3;?%?regularlizer?in?case?constrained?fits?are?ill?conditioned
else
??tol=0;
end

W?=?zeros（