这种方法假设样本点在光滑的流形上,这一方法的计算数据的低维表达,局部近邻信息被最优的保存。以这种方式,可以得到一个能反映流形的几何结构的解。
步骤一:构建一个图G=(V,E),其中V={vi,i=1,2,3…n}是顶点的集合,E={eij}是连接顶点的vi和vj边,图的每一个节点vi与样本集X中的一个点xi相关。如果xi,xj相距较近,我们就连接vi,vj。也就是说在各自节点插入一个边eij,如果Xj在xi的k领域中,k是定义参数。
步骤二:每个边都与一个权值Wij相对应,没有连接点之间的权值为0,连接点之间的权值:
使是最小的m+1个本征值。忽略与=0相关的本征向量,选取另外m个本征向量即为降维后的向量。
1、python实现拉普拉斯降维
- def laplaEigen(dataMat,k,t):
- m,n=shape(dataMat)
- W=mat(zeros([m,m]))
- D=mat(zeros([m,m]))
- for i in range(m):
- k_index=knn(dataMat[i,:],dataMat,k)
- for j in range(k):
- sqDiffVector = dataMat[i,:]-dataMat[k_index[j],:]
- sqDiffVector=array(sqDiffVector)**2
- sqDistances = sqDiffVector.sum()
- W[i,k_index[j]]=math.exp(-sqDistances/t)
- D[i,i]+=W[i,k_index[j]]
- L=D-W
- Dinv=np.linalg.inv(D)
- X=np.dot(D.I,L)
- lamda,f=np.linalg.eig(X)
- return lamda,f
- def knn(inX, dataSet, k):
- dataSetSize = dataSet.shape[0]
- diffMat = tile(inX, (dataSetSize,1)) - dataSet
- sqDiffMat = array(diffMat)**2
- sqDistances = sqDiffMat.sum(axis=1)
- distances = sqDistances**0.5
- sortedDistIndicies = distances.argsort()
- return sortedDistIndicies[0:k]
- dataMat, color = make_swiss_roll(n_samples=2000)
- lamda,f=laplaEigen(dataMat,11,5.0)
- fm,fn =shape(f)
- print 'fm,fn:',fm,fn
- lamdaIndicies = argsort(lamda)
- first=0
- second=0
- print lamdaIndicies[0], lamdaIndicies[1]
- for i in range(fm):
- if lamda[lamdaIndicies[i]].real>1e-5:
- print lamda[lamdaIndicies[i]]
- first=lamdaIndicies[i]
- second=lamdaIndicies[i+1]
- break
- print first, second
- redEigVects = f[:,lamdaIndicies]
- fig=plt.figure('origin')
- ax1 = fig.add_subplot(111, projection='3d')
- ax1.scatter(dataMat[:, 0], dataMat[:, 1], dataMat[:, 2], c=color,cmap=plt.cm.Spectral)
- fig=plt.figure('lowdata')
- ax2 = fig.add_subplot(111)
- ax2.scatter(f[:,first], f[:,second], c=color, cmap=plt.cm.Spectral)
- plt.show()
2、拉普拉斯降维实验
用如下参数生成实验数据存在swissdata.dat里面:
- def make_swiss_roll(n_samples=100, noise=0.0, random_state=None):
- #Generate a swiss roll dataset.
- t = 1.5 * np.pi * (1 + 2 * random.rand(1, n_samples))
- x = t * np.cos(t)
- y = 83 * random.rand(1, n_samples)
- z = t * np.sin(t)
- X = np.concatenate((x, y, z))
- X += noise * random.randn(3, n_samples)
- X = X.T
- t = np.squeeze(t)
- return X, t
实验结果如下:
N=5,t=15: N=7,t=15: N=9,t=15:
N=11,t=15: N=13,t=15: N=15,t=15:
N=17,t=15: N=19,t=15: N=21,t=15:
N=23,t=15: N=25,t=15: N=27,t=15:
N=29,t=15: N=31,t=15: N=33,t=15:
N=25,t=5: N=25,t=8: N=25,t=10:
N=25,t=12: N=25,t=14: N=25,t=50:
N=25,t=Inf: