我是PYMC3的新手。也许这是一个天真的问题,但我搜索了很多,并没有找到任何解释这个问题。 基本上,我想在PYMC3中进行线性回归,但训练非常缓慢,训练集的模型性能也非常差。以下是我的代码:
this.input.getImages()
.throttleTime(500)
.switchMap(image =>
new Observable<{}>(observer => {
// Some operation
})
.map(i => ({ image, i }))
).filter(({ i }) => {
// some filtering
})
.map(({ image }) => image)
.take(6)
.bufferCount(6)
.map(images => // switch map??
Observable.fromPromise(this.server.validate(images))
)
.retry(2) // This only retrys the request, I want it to retry the whole chain (to get valid images)
.subscribe(images => {
console.log('All done')
},
err => {console.log(err)}
)
在这里,我使用PYMC3来训练模型:
X_Tr = np.array([ 13.99802212, 13.8512075 , 13.9531636 , 13.97432944,
13.89211468, 13.91357953, 13.95987483, 13.86476587,
13.9501789 , 13.92698143, 13.9653932 , 14.06663115,
13.91697969, 13.99629862, 14.01392784, 13.96495713,
13.98697998, 13.97516973, 14.01048397, 14.05918188,
14.08342002, 13.89350606, 13.81768849, 13.94942447,
13.90465027, 13.93969029, 14.18771189, 14.08631113,
14.03718829, 14.01836206, 14.06758363, 14.05243539,
13.96287123, 13.93011351, 14.01616973, 14.01923812,
13.97424024, 13.9587175 , 13.85669845, 13.97778302,
14.04192138, 13.93775494, 13.86693585, 13.79985956,
13.82679677, 14.06474544, 13.90821822, 13.71648423,
13.78899668, 13.76857337, 13.87201756, 13.86152949,
13.80447525, 13.99609891, 14.0210165 , 13.986906 ,
13.97479211, 14.04562055, 14.03293095, 14.15178043,
14.32413197, 14.2330354 , 13.99247751, 13.92962912,
13.95394525, 13.87888254, 13.82743111, 14.10724699,
14.23638905, 14.15731881, 14.13239278, 14.13386722,
13.91442452, 14.01056255, 14.19378649, 14.22233852,
14.30405399, 14.25880108, 14.23985258, 14.21184303,
14.4443183 , 14.55710331, 14.42102092, 14.29047616,
14.43712609, 14.58666212])
y_tr = np.array([ 13.704, 13.763, 13.654, 13.677, 13.66 , 13.735, 13.845,
13.747, 13.747, 13.606, 13.819, 13.867, 13.817, 13.68 ,
13.823, 13.779, 13.814, 13.936, 13.956, 13.912, 13.982,
13.979, 13.919, 13.944, 14.094, 13.983, 13.887, 13.902,
13.899, 13.881, 13.784, 13.909, 13.99 , 14.06 , 13.834,
13.778, 13.703, 13.965, 14.02 , 13.992, 13.927, 14.009,
13.988, 14.022, 13.754, 13.837, 13.91 , 13.907, 13.867,
14.014, 13.952, 13.796, 13.92 , 14.051, 13.773, 13.837,
13.745, 14.034, 13.923, 14.041, 14.077, 14.125, 13.989,
14.174, 13.967, 13.952, 14.024, 14.171, 14.175, 14.091,
14.267, 14.22 , 14.071, 14.112, 14.174, 14.289, 14.146,
14.356, 14.5 , 14.265, 14.259, 14.406, 14.463, 14.473,
14.413, 14.507])
sns.regplot(x=X_tr, y=y_tr.flatten());
然后检查性能:
shA_X = shared(X_tr)
with pm.Model() as linear_model:
alpha = pm.Normal("alpha", mu=14,sd=100)
betas = pm.Normal("betas", mu=0, sd=100, shape=1)
sigma = pm.HalfCauchy('sigma', beta=10, testval=1.)
mu = alpha + betas * shA_X
forecast = pm.Normal("forecast", mu=mu, sd=sigma, observed=y_tr)
step = pm.NUTS()
trace=pm.sample(3000, tune=1000)
为什么训练集的预测如此之差? 任何建议和想法都表示赞赏。
答案 0 :(得分:3)
这个模型做得很好 - 我认为混淆是如何处理PyMC3
对象(感谢您使用示例轻松实现!)。通常,PyMC3
将用于量化模型中的不确定性。
例如,trace['betas'].mean()
约为0.83(这将取决于您的随机种子),而最小二乘估计,例如,sklearn
给出的将为0.826。同样,trace['alpha'].mean()
给出2.34,而“真实”值为2.38。
您还可以使用迹线为您的最佳拟合线绘制许多不同的合理绘图:
for draw in trace[::100]:
pred = draw['betas'] * X_tr + draw['alpha']
plt.plot(X_tr, pred, '--', alpha=0.2, color='grey')
plt.plot(X_tr, y_tr, 'o');
请注意,这些是从您的数据的“最佳匹配”分布中提取的。您还使用sigma
对噪点进行建模,您也可以绘制此值:
for draw in trace[::100]:
pred = draw['betas'] * X_tr + draw['alpha']
plt.plot(X_tr, pred, '-', alpha=0.2, color='grey')
plt.plot(X_tr, pred + draw['sigma'], '-', alpha=0.05, color='red')
plt.plot(X_tr, pred - draw['sigma'], '-', alpha=0.05, color='red');
plt.plot(X_tr, y_tr, 'o');
使用sample_ppc
从后验分布中获取观察到的值,因此ppc_w['forecast']
的每一行都是“下次”生成数据的合理方式。您可以这样使用该对象:
ppc_w = pm.sample_ppc(trace, 1000, linear_model,
progressbar=False)
for draw in ppc_w['forecast'][::5]:
sns.regplot(X_tr, draw, scatter_kws={'alpha': 0.005, 'color': 'r'}, fit_reg=False)
sns.regplot(X_tr, y_tr, color='k', fit_reg=False);