gpt4 book ai didi

python - PyMC3 多元混合模型 : Constraining components to be non-empty

转载 作者:太空宇宙 更新时间:2023-11-04 05:36:27 25 4
gpt4 key购买 nike

我正在 pymc3 中实现多元高斯回归的个性化混合,并遇到空组件问题。引用后related PyMC3 mixture model example ,我尝试使用单变量法线来实现模型,但我有 some issues there as well .

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

请注意,下面代码中的尝试 #1 来自 Mixture Model in PyMC3 Example并且在这里不起作用。

您可以使用 function in this gist 复制我正在使用的合成数据.

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;尝试在 LinAlgError 中采样结果:

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.

...表示组件为空,因为精度矩阵的形状为 (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。我选择了一个基于轨迹图的老化,并稀释到每 10 个样本。样本仍然高度自相关,但比没有细化要好得多。此时,我对样本中的 Z 值求和,这就是我得到的结果:

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集群。

最佳答案

我遇到过类似的问题。这样的事情适用于多元高斯模型的混合。至于它是否是最好的,它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))

此处的关键是,您需要考虑每个 低于临界值的潜力。此外,您应该调整 pi 分布的 shape,如评论中所述。这将影响您在 T.switch 调用中的索引(在 pi[i,0] 上)。

关于python - PyMC3 多元混合模型 : Constraining components to be non-empty,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/35438967/

25 4 0
Copyright 2021 - 2024 cfsdn All Rights Reserved 蜀ICP备2022000587号
广告合作:1813099741@qq.com 6ren.com