基于质心的算法 代表 K-means
from sklearn.cluster import KMeans
from sklearn.preprocessing import MinMaxScaler
from sklearn.datasets import load_iris
from sklearn.manifold import TSNE
df=pd.read_excel("lip_cluster.xlsx")
df1=df[['price','number_pay','amount','adress']]#选中数据框某几列
print(np.isnan(df1).any())
df1nona=df1[df1['amount'].notna()]#空值删除整行
print(np.isnan(df1nona).any())#检查是否含有空行
scale=MinMaxScaler().fit(df1nona)
df1_scale=scale.transform(df1nona)
plt.scatter(df1_scale[:, 2], df1_scale[:, 3],##原始数据散点图
c = "red", marker='o', label='see')
plt.xlabel('lip number_pay')
plt.ylabel('lip amount')
plt.legend(loc=2)
plt.show()
kmeans=KMeans(n_clusters=3).fit(df1_scale)#构造聚类器,estimator初始化Kmeans聚类；estimator.fit聚类内容拟合；
label_pred = kmeans.labels_ #获取聚类标签
centroids=kmeans.cluster_centers_#获取聚类中心
inertia = kmeans.inertia_ # 获取聚类准则的总和
ssa=kmeans.inertia_#组内平方和
y_kmeans2=kmeans.predict(df1_scale)
plt.scatter(df1_scale[:, 2], df1_scale[:, 3],
c=y_kmeans2, marker='p', cmap='rainbow', linewidths=4)
plt.show()
from mpl_toolkits.mplot3d import Axes3D
fig = plt.figure()
ax1 = plt.axes(projection='3d')
ax1.scatter3D(df1nona.values[:, 0], df1nona.values[:, 1],
df1nona.values[:, 3], cmap='Blues') #绘制散点图
plt.show()
import numpy as np
class HiddenMarkov:
def forward(self, Q, V, A, B, O, PI): # 使用前向算法
N = len(Q) #可能存在的状态数量
M = len(O) # 观测序列的大小
alphas = np.zeros((N, M)) # alpha值
T = M # 有几个时刻，有几个观测序列，就有几个时刻
for t in range(T): # 遍历每一时刻，算出alpha值
indexOfO = V.index(O[t]) # 找出序列对应的索引
for i in range(N):
if t == 0: # 计算初值
alphas[i][t] = PI[t][i] * B[i][indexOfO] # P176（10.15）
print(
'alpha1(%d)=p%db%db(o1)=%f' % (i, i, i, alphas[i][t]))
else:
alphas[i][t] = np.dot(
[alpha[t - 1] for alpha in alphas],
[a[i] for a in A]) * B[i][indexOfO] # 对应P176（10.16）
print('alpha%d(%d)=[sigma alpha%d(i)ai%d]b%d(o%d)=%f' %
(t, i, t - 1, i, i, t, alphas[i][t]))
print(alphas)
P = np.sum([alpha[M - 1] for alpha in alphas]) # P176(10.17)
def backward(self, Q, V, A, B, O, PI): # 后向算法
N = len(Q) # 可能存在的状态数量
M = len(O) # 观测序列的大小
betas = np.ones((N, M)) # beta
for i in range(N):
print('beta%d(%d)=1' % (M, i))
for t in range(M - 2, -1, -1):
indexOfO = V.index(O[t + 1]) # 找出序列对应的索引
for i in range(N):
betas[i][t] = np.dot(
np.multiply(A[i], [b[indexOfO] for b in B]),
[beta[t + 1] for beta in betas])
realT = t + 1
realI = i + 1
print(
'beta%d(%d)=[sigma a%djbj(o%d)]beta%d(j)=(' %
(realT, realI, realI, realT + 1, realT + 1),
end='')
for j in range(N):
print(
"%.2f*%.2f*%.2f+" % (A[i][j], B[j][indexOfO],
betas[j][t + 1]),
end='')
print("0)=%.3f" % betas[i][t])
# print(betas)
indexOfO = V.index(O[0])
P = np.dot(
np.multiply(PI, [b[indexOfO] for b in B]),
[beta[0] for beta in betas])
print("P(O|lambda)=", end="")
for i in range(N):
print(
"%.1f*%.1f*%.5f+" % (PI[0][i], B[i][indexOfO], betas[i][0]),
end="")
print("0=%f" % P)
def viterbi(self, Q, V, A, B, O, PI):
N = len(Q) # 可能存在的状态数量
M = len(O) # 观测序列的大小
deltas = np.zeros((N, M))
psis = np.zeros((N, M))
I = np.zeros((1, M))
for t in range(M):
realT = t + 1
indexOfO = V.index(O[t]) # 找出序列对应的索引
for i in range(N):
realI = i + 1
if t == 0:
deltas[i][t] = PI[0][i] * B[i][indexOfO]
psis[i][t] = 0
print('delta1(%d)=pi%d * b%d(o1)=%.2f * %.2f=%.2f' %
(realI, realI, realI, PI[0][i], B[i][indexOfO],
deltas[i][t]))
print('psis1(%d)=0' % (realI))
else:
deltas[i][t] = np.max(
np.multiply([delta[t - 1] for delta in deltas],
[a[i] for a in A])) * B[i][indexOfO]
print(
'delta%d(%d)=max[delta%d(j)aj%d]b%d(o%d)=%.2f*%.2f=%.5f'
% (realT, realI, realT - 1, realI, realI, realT,
np.max(
np.multiply([delta[t - 1] for delta in deltas],
[a[i] for a in A])), B[i][indexOfO],
deltas[i][t]))
psis[i][t] = np.argmax(
np.multiply(
[delta[t - 1] for delta in deltas],
[a[i]
for a in A])) + 1 # 由于其返回的是索引，因此应+1才能和正常的下标值相符合。
print('psis%d(%d)=argmax[delta%d(j)aj%d]=%d' %
(realT, realI, realT - 1, realI, psis[i][t]))
print(deltas)
print(psis)
I[0][M - 1] = np.argmax([delta[M - 1] for delta in deltas
]) + 1 # 由于其返回的是索引，因此应+1才能和正常的下标值相符合。
print('i%d=argmax[deltaT(i)]=%d' % (M, I[0][M - 1]))
for t in range(M - 2, -1, -1):
I[0][t] = psis[int(I[0][t + 1]) - 1][t + 1]
print('i%d=psis%d(i%d)=%d' % (t + 1, t + 2, t + 2, I[0][t]))
print("状态序列I：", I)
例子
Q = [1, 2, 3]
V = ['红', '白']
A = [[0.5, 0.2, 0.3], [0.3, 0.5, 0.2], [0.2, 0.3, 0.5]]
B = [[0.5, 0.5], [0.4, 0.6], [0.7, 0.3]]
# O = ['红', '白', '红', '红', '白', '红', '白', '白']
O = ['红', '白', '红', '白'] #习题10.1的例子
PI = [[0.2, 0.4, 0.4]]
HMM = HiddenMarkov()
HMM.viterbi(Q, V, A, B, O, PI)
delta1(1)=pi1 * b1(o1)=0.20 * 0.50=0.10
psis1(1)=0
delta1(2)=pi2 * b2(o1)=0.40 * 0.40=0.16
psis1(2)=0
delta1(3)=pi3 * b3(o1)=0.40 * 0.70=0.28
psis1(3)=0
delta2(1)=max[delta1(j)aj1]b1(o2)=0.06*0.50=0.02800
psis2(1)=argmax[delta1(j)aj1]=3
delta2(2)=max[delta1(j)aj2]b2(o2)=0.08*0.60=0.05040
psis2(2)=argmax[delta1(j)aj2]=3
delta2(3)=max[delta1(j)aj3]b3(o2)=0.14*0.30=0.04200
psis2(3)=argmax[delta1(j)aj3]=3
delta3(1)=max[delta2(j)aj1]b1(o3)=0.02*0.50=0.00756
psis3(1)=argmax[delta2(j)aj1]=2
delta3(2)=max[delta2(j)aj2]b2(o3)=0.03*0.40=0.01008
psis3(2)=argmax[delta2(j)aj2]=2
delta3(3)=max[delta2(j)aj3]b3(o3)=0.02*0.70=0.01470
psis3(3)=argmax[delta2(j)aj3]=3
delta4(1)=max[delta3(j)aj1]b1(o4)=0.00*0.50=0.00189
psis4(1)=argmax[delta3(j)aj1]=1
delta4(2)=max[delta3(j)aj2]b2(o4)=0.01*0.60=0.00302
psis4(2)=argmax[delta3(j)aj2]=2
delta4(3)=max[delta3(j)aj3]b3(o4)=0.01*0.30=0.00220
psis4(3)=argmax[delta3(j)aj3]=3
[[0.1 0.028 0.00756 0.00189 ]
[0.16 0.0504 0.01008 0.003024]
[0.28 0.042 0.0147 0.002205]]
[[0. 3. 2. 1.]
[0. 3. 2. 2.]
[0. 3. 3. 3.]]
i4=argmax[deltaT(i)]=2
i3=psis4(i4)=2
i2=psis3(i3)=2
i1=psis2(i2)=3
状态序列I： [[3. 2. 2. 2.]]
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