我可以计算包含特定值的 Prophet 模型的置信区间吗?

Can I calculate the confidence bound of a Prophet model that would contain a certain value?

我可以使用预测数据框中的 y 值方差、界限和点估计来计算包含给定值的置信度吗?

我已经看到了 I can change my interval level prior to fitting,但从编程的角度来看,这感觉像是大量昂贵的反复试验。 有没有一种方法可以仅使用来自模型参数和预测数据框的信息来估计置信区间?

类似于:

for level in [.05, .1, .15, ... , .95]:
  if value_in_question in (yhat - Z_{level}*yhat_variance/N, yhat + Z_{level}*yhat_variance/N):
        print 'im in the bound level {level}'
# This is sudo code not meant to run in console 

编辑:工作先知示例

# csv from fbprohets working examples https://github.com/facebook/prophet/blob/master/examples/example_wp_log_peyton_manning.csv

import pandas as pd
from fbprophet import Prophet
import os
df = pd.read_csv('example_wp_log_peyton_manning.csv')
m = Prophet()
m.fit(df)
future = m.make_future_dataframe(periods=30)
forecast = m.predict(future)

# the smallest confidence level s.t. the confidence interval of the 30th prediction contains 9

## My current approach
def __probability_calculation(estimate, forecast, j = 30):
    sd_residuals = (forecast.yhat_lower[j] - forecast.yhat[j])/(-1.28)
    for alpha in np.arange(.5, .95, .01):
        z_val = st.norm.ppf(alpha)
        if (forecast.yhat[j]-z_val*sd_residuals < estimate < forecast.yhat[j]+z_val*sd_residuals):
           return alpha

prob = __probability_calculation(9, forecast)

fbprophet 使用 numpy.percentile 方法来估计百分位数,您可以在此处的源代码中看到: https://github.com/facebook/prophet/blob/0616bfb5daa6888e9665bba1f95d9d67e91fed66/python/prophet/forecaster.py#L1448

这里已经回答了如何反算值的百分位数: Map each list value to its corresponding percentile

根据您的代码示例组合所有内容:

import pandas as pd
import numpy as np
import scipy.stats as st
from fbprophet import Prophet

url = 'https://raw.githubusercontent.com/facebook/prophet/master/examples/example_wp_log_peyton_manning.csv'
df = pd.read_csv(url)
   
# put the amount of uncertainty samples in a variable so we can use it later.
uncertainty_samples = 1000 # 1000 is the default
m = Prophet(uncertainty_samples=uncertainty_samples)
m.fit(df)
future = m.make_future_dataframe(periods=30)

# You need to replicate some of the preparation steps which are part of the predict() call internals
tmpdf = m.setup_dataframe(future)
tmpdf['trend'] = m.predict_trend(tmpdf)
sim_values = m.sample_posterior_predictive(tmpdf)

sim_values 对象包含每个数据点的 1000 个模拟,置信区间基于这些模拟。

现在您可以使用任何目标值调用 scipy.stats.percentileofscore 方法

target_value = 8
st.percentileofscore(sim_values['yhat'], target_value, 'weak') / uncertainty_samples
# returns 44.26

为了证明这一点前后都有效,您可以获取 np.percentile 方法的输出并将其放入 scipy.stats.percentileofscore method 中。 这适用于 4 位小数的精度:

ACCURACY = 4
for test_percentile in np.arange(0, 100, 0.5):
    target_value = np.percentile(sim_values['yhat'], test_percentile)
    if not np.round(st.percentileofscore(sim_values['yhat'], target_value, 'weak') / uncertainty_samples, ACCURACY) == np.round(test_percentile, ACCURACY):
        print(test_percentile)
        raise ValueError('This doesnt work')