HMM 模型输入数据处理的优雅做法 来自实际项目

 

实际项目我是这样做的：

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def mining_ue_procedures_behavior(seq, lengths, imsi_list):      print("seq 3:", seq[:3], "lengths 3:", lengths[:3])     # model.fit(seq, lengths)     fitter = LabelEncoder().fit(seq)       import sys     n_components=[5, 10, 20, 30][int(sys.argv[1])]     n_iter=[10, 30, 50, 100][int(sys.argv[2])]       model_file = 'hmm_model_{}_{}.pkl'.format(n_components, n_iter)     if os.path.exists(model_file):         input_file = open(model_file, 'rb')         model = pickle.load(input_file)         input_file.close()     else:         model = hmm.MultinomialHMM(n_components=n_components, n_iter=n_iter)         seq2 = fitter.transform(seq)         model.fit(np.array([seq2]).T, lengths)         output_file = open(model_file, 'wb')         pickle.dump(model, output_file)         output_file.close()     print("model.startprob_:", model.startprob_)     print("model.transmat_:", model.transmat_)     print("model.emissionprob_:", model.emissionprob_)     ## [[  1.11111111e-01   2.22222222e-01   6.66666667e-01]     ##  [  5.55555556e-01   4.44444444e-01   6.27814351e-28]]     start = 0     ans = []     for i,l in enumerate(lengths):         s = seq[start: start+l]         score = model.score(np.array([[d] for d in fitter.transform(s)]))         ans.append([score, imsi_list[i], s])         # print("score:", model.score(np.array([[d] for d in fitter.transform(s)])), s)         start += l     ans.sort(key=lambda x: x[0])     score_index = 0     malicious_ue = []     for i,item in enumerate(ans):         if item[score_index] < Config.HMMBaseScore:             malicious_ue.append(item)         print(item)     # print(ans)

　　

 

输入数据参考了下面的优雅做法：

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# predict a sequence of hidden states based on visible states seq = [] lengths = [] for _ in range(100):     length = random.randint(5, 10)     lengths.append(length)     for _ in range(length):         r = random.random()         if r < .2:             seq.append(0)         elif r < .6:             seq.append(1)         else:             seq.append(2) seq = np.array([seq]).T model = model.fit(seq, lengths)

 

此外，HMM模型的持续增量训练：

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# 解决问题3，学习问题，仅给出X，估计模型参数,鲍姆-韦尔奇算法，其实就是基于EM算法的求解 # 解决这个问题需要X的有一定的数据量，然后通过model.fit(X, lengths=None)来进行训练然后自己生成一个模型 # 并不需要设置model.startprob_,model.transmat_,model.emissionprob_ # 例如:   import numpy as np from hmmlearn import hmm   states = ["Rainy", "Sunny"]##隐藏状态 n_states = len(states)##隐藏状态长度   observations = ["walk", "shop", "clean"]##可观察的状态 n_observations = len(observations)##可观察序列的长度   model = hmm.MultinomialHMM(n_components=n_states, n_iter=1000, tol=0.01)   X = np.array([[2, 0, 1, 1, 2, 0],[0, 0, 1, 1, 2, 0],[2, 1, 2, 1, 2, 0]]) model.fit(X) print model.startprob_ print model.transmat_ print model.emissionprob_ # [[  1.11111111e-01   2.22222222e-01   6.66666667e-01] #  [  5.55555556e-01   4.44444444e-01   6.27814351e-28]] print model.score(X) model.fit(X) print model.startprob_ print model.transmat_ print model.emissionprob_ 和第一次fit(X)得到的行顺序不一样 # [[  5.55555556e-01   4.44444444e-01   9.29759770e-28] #  [  1.11111111e-01   2.22222222e-01   6.66666667e-01]] print model.score(X) model.fit(X) print model.startprob_ print model.transmat_ print model.emissionprob_ print model.score(X) # 可以进行多次fit,然后拿评分最高的模型，就可以预测了 print model.predict(bob_Actions, lengths=None) # 预测最可能的隐藏状态 # 例如: # [0 1 0 0 0 1] print model.predict_proba(bob_Actions, lengths=None)# 预测各个隐藏状态的概率 # 例如: # [[ 0.82770645  0.17229355] #  [ 0.27361913  0.72638087] #  [ 0.58700959  0.41299041] #  [ 0.69861348  0.30138652] #  [ 0.81799813  0.18200187] #  [ 0.24723966  0.75276034]] # 在生成的模型中，可以随机生成随机生成一个模型的Z和X X,Z = model.sample(n_samples=5, random_state=None) print "Bob Actions:", ", ".join(map(lambda x: observations[x], X)) print "weathers:", ", ".join(map(lambda x: states[x], Z))     # 保存模型 import pickle output = open('D:\\xxx\\data1111.pkl', 'wb') s = pickle.dump(model, output) output.close() # 调用模型 input = open('D:\\xxx\\data.pkl', 'rb') model = pickle.load(model) input.close() model.predict(X)