为什么我的metropolis算法(mcmc)的python实现如此缓慢?
我试图用Python实现该算法(Metropolis Hastings算法的一个简单版本) 以下是我的实现:为什么我的metropolis算法(mcmc)的python实现如此缓慢?,python,performance,machine-learning,random,mcmc,Python,Performance,Machine Learning,Random,Mcmc,我试图用Python实现该算法(Metropolis Hastings算法的一个简单版本) 以下是我的实现: def Metropolis_Gaussian(p, z0, sigma, n_samples=100, burn_in=0, m=1): """ Metropolis Algorithm using a Gaussian proposal distribution. p: distribution that we want to sample from (can
def Metropolis_Gaussian(p, z0, sigma, n_samples=100, burn_in=0, m=1):
"""
Metropolis Algorithm using a Gaussian proposal distribution.
p: distribution that we want to sample from (can be unnormalized)
z0: Initial sample
sigma: standard deviation of the proposal normal distribution.
n_samples: number of final samples that we want to obtain.
burn_in: number of initial samples to discard.
m: this number is used to take every mth sample at the end
"""
# List of samples, check feasibility of first sample and set z to first sample
sample_list = [z0]
_ = p(z0)
z = z0
# set a counter of samples for burn-in
n_sampled = 0
while len(sample_list[::m]) < n_samples:
# Sample a candidate from Normal(mu, sigma), draw a uniform sample, find acceptance probability
cand = np.random.normal(loc=z, scale=sigma)
u = np.random.rand()
try:
prob = min(1, p(cand) / p(z))
except (OverflowError, ValueError) as error:
continue
n_sampled += 1
if prob > u:
z = cand # accept and make candidate the new sample
# do not add burn-in samples
if n_sampled > burn_in:
sample_list.append(z)
# Finally want to take every Mth sample in order to achieve independence
return np.array(sample_list)[::m]
这段代码需要相当长的时间来运行,我不知道为什么。在我的Metropolis_Gaussian代码中,我试图通过
函数
pdf\t
的定义如下
from scipy.stats import t
def pdf_t(x, df=10):
return t.pdf(x, df=df)
我回答了一个问题。我在这里提到的很多东西(不是每次迭代都计算当前的可能性,预先计算随机创新等)都可以在这里使用
实现的其他改进是不使用列表来存储示例。相反,您应该为样本预先分配内存,并将它们存储为数组。类似于samples=np的东西。零(n_samples)
比在每次迭代时附加到列表更有效
您已经提到,您试图通过不记录老化样本来提高效率。这是个好主意。您还可以通过只记录每个第m个样本来实现类似的细化,因为您在返回语句中使用np.array(sample_list)[::m]
丢弃了这些样本。您可以通过更改以下内容来执行此操作:
# do not add burn-in samples
if n_sampled > burn_in:
sample_list.append(z)
到
还值得注意的是,您不需要计算min(1,p(cand)/p(z))
,只需计算p(cand)/p(z)
。我意识到在形式上,min
是必要的(以确保概率在0和1之间)。但是,在计算上,我们不需要最小值,因为如果p(cand)/p(z)>1
,那么p(cand)/p(z)
总是大于u
将这一切加在一起,以及预先计算随机创新、接受概率u
,并仅在您确实需要时计算可能性,我得出:
def my_Metropolis_Gaussian(p, z0, sigma, n_samples=100, burn_in=0, m=1):
# Pre-allocate memory for samples (much more efficient than using append)
samples = np.zeros(n_samples)
# Store initial value
samples[0] = z0
z = z0
# Compute the current likelihood
l_cur = p(z)
# Counter
iter = 0
# Total number of iterations to make to achieve desired number of samples
iters = (n_samples * m) + burn_in
# Sample outside the for loop
innov = np.random.normal(loc=0, scale=sigma, size=iters)
u = np.random.rand(iters)
while iter < iters:
# Random walk innovation on z
cand = z + innov[iter]
# Compute candidate likelihood
l_cand = p(cand)
# Accept or reject candidate
if l_cand / l_cur > u[iter]:
z = cand
l_cur = l_cand
# Only keep iterations after burn-in and for every m-th iteration
if iter > burn_in and iter % m == 0:
samples[(iter - burn_in) // m] = z
iter += 1
return samples
我回答了一个问题。我在这里提到的很多东西(不是每次迭代都计算当前的可能性,预先计算随机创新等)都可以在这里使用
实现的其他改进是不使用列表来存储示例。相反,您应该为样本预先分配内存,并将它们存储为数组。类似于samples=np的东西。零(n_samples)
比在每次迭代时附加到列表更有效
您已经提到,您试图通过不记录老化样本来提高效率。这是个好主意。您还可以通过只记录每个第m个样本来实现类似的细化,因为您在返回语句中使用np.array(sample_list)[::m]
丢弃了这些样本。您可以通过更改以下内容来执行此操作:
# do not add burn-in samples
if n_sampled > burn_in:
sample_list.append(z)
到
还值得注意的是,您不需要计算min(1,p(cand)/p(z))
,只需计算p(cand)/p(z)
。我意识到在形式上,min
是必要的(以确保概率在0和1之间)。但是,在计算上,我们不需要最小值,因为如果p(cand)/p(z)>1
,那么p(cand)/p(z)
总是大于u
将这一切加在一起,以及预先计算随机创新、接受概率u
,并仅在您确实需要时计算可能性,我得出:
def my_Metropolis_Gaussian(p, z0, sigma, n_samples=100, burn_in=0, m=1):
# Pre-allocate memory for samples (much more efficient than using append)
samples = np.zeros(n_samples)
# Store initial value
samples[0] = z0
z = z0
# Compute the current likelihood
l_cur = p(z)
# Counter
iter = 0
# Total number of iterations to make to achieve desired number of samples
iters = (n_samples * m) + burn_in
# Sample outside the for loop
innov = np.random.normal(loc=0, scale=sigma, size=iters)
u = np.random.rand(iters)
while iter < iters:
# Random walk innovation on z
cand = z + innov[iter]
# Compute candidate likelihood
l_cand = p(cand)
# Accept or reject candidate
if l_cand / l_cur > u[iter]:
z = cand
l_cur = l_cand
# Only keep iterations after burn-in and for every m-th iteration
if iter > burn_in and iter % m == 0:
samples[(iter - burn_in) // m] = z
iter += 1
return samples
已在此网站上询问了一个问题。虽然标题听起来可能不是同一个问题,但我给您的答案与此处相同:。我在此再次强调,不包括失败接受的重复是渐进不正确的,并且会导致低可能性样本值的过度代表。本网站已经询问了A的可能重复。虽然标题听起来可能不像是同一个问题,但我给你的答案与此处相同:。我在这里再次强调,不包括失败接受的重复是渐进错误的,并且会导致低可能性样本值的过度代表性
In [1]: %timeit Metropolis_Gaussian(pdf_t, 3, 1, n_samples=100, burn_in=100, m=10)
205 ms ± 2.16 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
In [2]: %timeit my_Metropolis_Gaussian(pdf_t, 3, 1, n_samples=100, burn_in=100, m=10)
102 ms ± 1.12 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)