PyMC3 多元混合模型:约束分量为非空



我正在pymc3中实现多元高斯回归的个性化混合,并遇到了空组件的问题。在参考了相关的 PyMC3 混合模型示例后,我尝试使用单变量法线来实现该模型,但我在那里也遇到了一些问题。

我尝试了几种策略来将每个组件约束为非空组件,但每种策略都失败了。这些显示在下面的代码中。我的具体问题是:在使用pymc3的多元高斯混合中将所有分量限制为非空的最佳方法是什么?

请注意,以下代码中的尝试 #1 来自 PyMC3 示例中的混合模型,在此处不起作用。

您可以复制我在此要点中的函数中使用的合成数据。

import pymc3 as pm
import numpy as np
import theano
import theano.tensor as T
from scipy import stats
# Extract problem dimensions.
N = X.shape[0]  # number of samples
F = X.shape[1]  # number of features
pids = I[:, 0].astype(np.int)  # primary entity ids
uniq_pids = np.unique(pids)  # array of unique primary entity ids
n_pe = len(uniq_pids)  # number of primary entities
with pm.Model() as gmreg:
    # Init hyperparameters.
    a0 = 1
    b0 = 1
    mu0 = pm.constant(np.zeros(F))
    alpha = pm.constant(np.ones(K))
    coeff_precisions = pm.constant(1 / X.var(0))
    # Init parameters.
    # Dirichlet shape parameter, prior on indicators.
    pi = pm.Dirichlet(
        'pi', a=alpha, shape=K)
    # ATTEMPT 1: Make probability of membership for each cluter >= 0.1
    # ================================================================
    pi_min_potential = pm.Potential(
        'pi_min_potential', T.switch(T.min(pi) < .1, -np.inf, 0))
    # ================================================================
    # The multinomial (and by extension, the Categorical), is a symmetric
    # distribution. Using this as a prior for the indicator variables Z
    # makes the likelihood invariant under the many possible permutations of
    # the indices. This invariance is inherited in posterior inference.
    # This invariance model implies unidentifiability and induces label
    # switching during inference.
    # Resolve by ordering the components to have increasing weights.
    # This does not deal with the parameter identifiability issue.
    order_pi_potential = pm.Potential(
        'order_pi_potential',
        T.sum([T.switch(pi[k] - pi[k-1] < 0, -np.inf, 0)
               for k in range(1, K)]))
    # Indicators, specifying which cluster each primary entity belongs to.
    # These are draws from Multinomial with 1 trial.
    init_pi = stats.dirichlet.rvs(alpha.eval())[0]
    test_Z = np.random.multinomial(n=1, pvals=init_pi, size=n_pe)
    as_cat = np.nonzero(test_Z)[1]
    Z = pm.Categorical(
        'Z', p=pi, shape=n_pe, testval=as_cat)
    # ATTEMPT 2: Give infinite negative likelihood to the case
    # where any of the clusters have no users assigned.
    # ================================================================
    # sizes = [T.eq(Z, k).nonzero()[0].shape[0] for k in range(K)]
    # nonempty_potential = pm.Potential(
    #     'comp_nonempty_potential',
    #     np.sum([T.switch(sizes[k] < 1, -np.inf, 0) for k in range(K)]))
    # ================================================================
    # ATTEMPT 3: Add same sample to each cluster, each has at least 1.
    # ================================================================
    # shared_X = X.mean(0)[None, :]
    # shared_y = y.mean().reshape(1)
    # X = T.concatenate((shared_X.repeat(K).reshape(K, F), X))
    # y = T.concatenate((shared_y.repeat(K), y))
    # Add range(K) on to the beginning to include shared instance.
    # Z_expanded = Z[pids]
    # Z_with_shared = T.concatenate((range(K), Z_expanded))
    # pid_idx = pm.Deterministic('pid_idx', Z_with_shared)
    # ================================================================
    # Expand user cluster indicators to each observation for each user.
    pid_idx = pm.Deterministic('pid_idx', Z[pids])
    # Construct masks for each component.
    masks = [T.eq(pid_idx, k).nonzero() for k in range(K)]
    comp_sizes = [masks[k][0].shape[0] for k in range(K)]
    # Component regression precision parameters.
    beta = pm.Gamma(
        'beta', alpha=a0, beta=b0, shape=(K,),
        testval=np.random.gamma(a0, b0, size=K))
    # Regression coefficient matrix, with coeffs for each component.
    W = pm.MvNormal(
        'W', mu=mu0, tau=T.diag(coeff_precisions), shape=(K, F),
        testval=np.random.randn(K, F) * std)
    # The mean of the observations is the result of a regression, with
    # coefficients determined by the cluster the sample belongs to.
    # Now we have K different multivariate normal distributions.
    X = T.cast(X, 'float64')
    y = T.cast(y, 'float64')
    comps = []
    for k in range(K):
        mask_k = masks[k]
        X_k = X[mask_k]
        y_k = y[mask_k]
        n_k = comp_sizes[k]
        precision_matrix = beta[k] * T.eye(n_k)
        comp_k = pm.MvNormal(
            'comp_%d' % k,
            mu=T.dot(X_k, W[k]), tau=precision_matrix,
            observed=y_k)
        comps.append(comp_k)

前两种方法无法确保非空簇;尝试采样会导致LinAlgError

with gmreg:
step1 = pm.Metropolis(vars=[pi, beta, W])
step2 = pm.ElemwiseCategoricalStep(vars=[Z], values=np.arange(K))
tr = pm.sample(100, step=[step1, step2])
...:     
Failed to compute determinant []
---------------------------------------------------------------------------
LinAlgError                               Traceback (most recent call last)
<ipython-input-2-c7df53f4c6a5> in <module>()
      2     step1 = pm.Metropolis(vars=[pi, beta, W])
      3     step2 = pm.ElemwiseCategoricalStep(vars=[Z], values=np.arange(K))
----> 4     tr = pm.sample(100, step=[step1, step2])
      5 
/home/mack/anaconda/lib/python2.7/site-packages/pymc3/sampling.pyc in sample(draws, step, start, trace, chain, njobs, tune, progressbar, model, random_seed)
    155         sample_args = [draws, step, start, trace, chain,
    156                        tune, progressbar, model, random_seed]
--> 157     return sample_func(*sample_args)
    158 
    159 
/home/mack/anaconda/lib/python2.7/site-packages/pymc3/sampling.pyc in _sample(draws, step, start, trace, chain, tune, progressbar, model, random_seed)
    164     progress = progress_bar(draws)
    165     try:
--> 166         for i, strace in enumerate(sampling):
    167             if progressbar:
    168                 progress.update(i)
/home/mack/anaconda/lib/python2.7/site-packages/pymc3/sampling.pyc in _iter_sample(draws, step, start, trace, chain, tune, model, random_seed)
    246         if i == tune:
    247             step = stop_tuning(step)
--> 248         point = step.step(point)
    249         strace.record(point)
    250         yield strace
/home/mack/anaconda/lib/python2.7/site-packages/pymc3/step_methods/compound.pyc in step(self, point)
     12     def step(self, point):
     13         for method in self.methods:
---> 14             point = method.step(point)
     15         return point
/home/mack/anaconda/lib/python2.7/site-packages/pymc3/step_methods/arraystep.pyc in step(self, point)
     87             inputs += [point]
     88 
---> 89         apoint = self.astep(bij.map(point), *inputs)
     90         return bij.rmap(apoint)
     91 
/home/mack/anaconda/lib/python2.7/site-packages/pymc3/step_methods/gibbs.pyc in astep(self, q, logp)
     38 
     39     def astep(self, q, logp):
---> 40         p = array([logp(v * self.sh) for v in self.values])
     41         return categorical(p, self.var.dshape)
     42 
/home/mack/anaconda/lib/python2.7/site-packages/pymc3/blocking.pyc in __call__(self, x)
    117 
    118     def __call__(self, x):
--> 119         return self.fa(self.fb(x))
/home/mack/anaconda/lib/python2.7/site-packages/pymc3/model.pyc in __call__(self, *args, **kwargs)
    423     def __call__(self, *args, **kwargs):
    424         point = Point(model=self.model, *args, **kwargs)
--> 425         return self.f(**point)
    426 
    427 compilef = fastfn
/home/mack/anaconda/lib/python2.7/site-packages/theano/compile/function_module.pyc in __call__(self, *args, **kwargs)
    604                         self.fn.nodes[self.fn.position_of_error],
    605                         self.fn.thunks[self.fn.position_of_error],
--> 606                         storage_map=self.fn.storage_map)
    607                 else:
    608                     # For the c linker We don't have access from
/home/mack/anaconda/lib/python2.7/site-packages/theano/compile/function_module.pyc in __call__(self, *args, **kwargs)
    593         t0_fn = time.time()
    594         try:
--> 595             outputs = self.fn()
    596         except Exception:
    597             if hasattr(self.fn, 'position_of_error'):
/home/mack/anaconda/lib/python2.7/site-packages/theano/gof/op.pyc in rval(p, i, o, n)
    766             # default arguments are stored in the closure of `rval`
    767             def rval(p=p, i=node_input_storage, o=node_output_storage, n=node):
--> 768                 r = p(n, [x[0] for x in i], o)
    769                 for o in node.outputs:
    770                     compute_map[o][0] = True
/home/mack/anaconda/lib/python2.7/site-packages/theano/tensor/nlinalg.pyc in perform(self, node, (x,), (z,))
    267     def perform(self, node, (x,), (z, )):
    268         try:
--> 269             z[0] = numpy.asarray(numpy.linalg.det(x), dtype=x.dtype)
    270         except Exception:
    271             print 'Failed to compute determinant', x
/home/mack/anaconda/lib/python2.7/site-packages/numpy/linalg/linalg.pyc in det(a)
   1769     """
   1770     a = asarray(a)
-> 1771     _assertNoEmpty2d(a)
   1772     _assertRankAtLeast2(a)
   1773     _assertNdSquareness(a)
/home/mack/anaconda/lib/python2.7/site-packages/numpy/linalg/linalg.pyc in _assertNoEmpty2d(*arrays)
    220     for a in arrays:
    221         if a.size == 0 and product(a.shape[-2:]) == 0:
--> 222             raise LinAlgError("Arrays cannot be empty")
    223 
    224 
LinAlgError: Arrays cannot be empty
Apply node that caused the error: Det(Elemwise{Mul}[(0, 1)].0)
Inputs types: [TensorType(float64, matrix)]
Inputs shapes: [(0, 0)]
Inputs strides: [(8, 8)]
Inputs values: [array([], shape=(0, 0), dtype=float64)]
Backtrace when the node is created:
  File "/home/mack/anaconda/lib/python2.7/site-packages/pymc3/distributions/multivariate.py", line 66, in logp
    result = k * T.log(2 * np.pi) + T.log(1./det(tau))
HINT: Use the Theano flag 'exception_verbosity=high' for a debugprint and storage map footprint of this apply node.

。这表示分量为空,因为精度矩阵的形状(0, 0)

第三种方法实际上解决了空组件问题,但给出了非常奇怪的推理行为。我根据迹线图选择了一个老化图,并细化到每 10 个样本。这些样本仍然是高度自相关的,但比没有变薄要好得多。此时,我对样本中的 Z 值求和,这就是我得到的:

In [3]: with gmreg:
    step1 = pm.Metropolis(vars=[pi, beta, W])
    step2 = pm.ElemwiseCategoricalStep(vars=[Z], values=np.arange(K))
    tr = pm.sample(1000, step=[step1, step2])
   ...:     
 [-----------------100%-----------------] 1000 of 1000 complete in 258.8 sec
...
In [24]: zvals = tr[300::10]['Z']
In [25]: np.array([np.bincount(zvals[:, n]) for n in range(nusers)])
Out[25]: 
array([[ 0,  0, 70],
       [ 0,  0, 70],
       [ 0,  0, 70],
       [ 0,  0, 70],
       [ 0,  0, 70],
       [ 0,  0, 70],
       [ 0,  0, 70],
       [ 0,  0, 70],
       [ 0,  0, 70],
       [ 0,  0, 70],
       [ 0,  0, 70],
       [ 0,  0, 70],
       [ 0,  0, 70],
       [ 0,  0, 70],
       [ 0,  0, 70],
       [ 0,  0, 70],
       [ 0,  0, 70],
       [ 0,  0, 70],
       [ 0,  0, 70],
       [ 0,  0, 70]])

因此,出于某种原因,所有用户都被分配到每个样本的最后一个聚类。

我遇到了类似的问题。这样的东西适用于多元高斯模型的混合。至于它是否是最好的,这肯定是我找到的最好的解决方案。

pm.Potential('pi_min_potential', T.switch(
                                           T.all(
                                                [pi[i, 0] < 0.1 for i in range(K)]), -np.inf, 0))

这里的关键是,您需要考虑低于截止值的每个潜力。此外,您应该调整pi分布的shape,如注释中所述。这将影响您在T.switch调用(在pi[i,0]上(中的索引。

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