我对将卡尔曼滤波器（KF）应用于以下预测问题所看到的行为有疑问 . 我已经包含了一个简单的代码示例 .
Goal: 我想知道KF是否适合使用现在（t）时获得的测量结果提前一天（t 24小时）改善预测/模拟结果 . 目标是使预测尽可能接近测量
Assumption: 我们假设测量是完美的（即，如果我们能够完全匹配测量结果，我们很高兴） .
我们有一个测量变量（z，实际风速）和一个模拟变量（x，预测风速） .
模拟风速x使用各种气象数据（对我来说是黑盒子）从NWP（数值天气预报）软件产生 . 模拟文件每天生成，每半小时包含一次数据 .
我尝试使用标量卡尔曼滤波器，使用我现在获得的测量和现在的预测数据（t-24小时前生成）来校正t 24hr预测 . 作为参考，我用过：http://www.swarthmore.edu/NatSci/echeeve1/Ref/Kalman/ScalarKalman.html
Code:
#! /usr/bin/python
import numpy as np
import pylab
import os
def main():
# x = 336 data points of simulated wind speed for 7 days * 24 hour * 2 (every half an hour)
# Imagine at time t, we will get a x_t fvalue or t+48 or a 24 hours later.
x = load_x()
# this is a list that will contain 336 data points of our corrected data
x_sample_predict_list = []
# z = 336 data points for 7 days * 24 hour * 2 of actual measured wind speed (every half an hour)
z = load_z()
# Here is the setup of the scalar kalman filter
# reference: http://www.swarthmore.edu/NatSci/echeeve1/Ref/Kalman/ScalarKalman.html
# state transition matrix (we simply have a scalar)
# what you need to multiply the last time's state to get the newest state
# we get the x_t+1 = A * x_t, since we get the x_t+1 directly for simulation
# we will have a = 1
a = 1.0
# observation matrix
# what you need to multiply to the state, convert it to the same form as incoming measurement
# both state and measurements are wind speed, so set h = 1
h = 1.0
Q = 16.0 # expected process variance of predicted Wind Speed
R = 9.0 # expected measurement variance of Wind Speed
p_j = Q # process covariance is equal to the initial process covariance estimate
# Kalman gain is equal to k = hp-_j / (hp-_j + R). With perfect measurement
# R = 0, k reduces to k=1/h which is 1
k = 1.0
# one week data
# original R2 = 0.183
# with delay = 6, R2 = 0.295
# with delay = 12, R2 = 0.147
# with delay = 48, R2 = 0.075
delay = 6
# Kalman loop
for t, x_sample in enumerate(x):
if t <= delay:
# for the first day of the forecast,
# we don't have forecast data and measurement
# from a day before to do correction
x_sample_predict = x_sample
else: # t > 48
# for a priori estimate we take x_sample as is
# x_sample = x^-_j = a x^-_j_1 + b u_j
# Inside the NWP (numerical weather prediction,
# the x_sample should be on x_sample_j-1 (assumption)
x_sample_predict_prior = a * x_sample
# we use the measurement from t-delay (ie. could be a day ago)
# and forecast data from t-delay, to produce a leading residual that can be used to
# correct the forecast.
residual = z[t-delay] - h * x_sample_predict_list[t-delay]
p_j_prior = a**2 * p_j + Q
k = h * p_j_prior / (h**2 * p_j_prior + R)
# we update our prediction based on the residual
x_sample_predict = x_sample_predict_prior + k * residual
p_j = p_j_prior * (1 - h * k)
#print k
#print p_j_prior
#print p_j
#raw_input()
x_sample_predict_list.append(x_sample_predict)
# initial goodness of fit
R2_val_initial = calculate_regression(x,z)
R2_string_initial = "R2 initial: {0:10.3f}, ".format(R2_val_initial)
print R2_string_initial # R2_val_initial = 0.183
# final goodness of fit
R2_val_final = calculate_regression(x_sample_predict_list,z)
R2_string_final = "R2 final: {0:10.3f}, ".format(R2_val_final)
print R2_string_final # R2_val_final = 0.117, which is worse
timesteps = xrange(len(x))
pylab.plot(timesteps,x,'r-', timesteps,z,'b:', timesteps,x_sample_predict_list,'g--')
pylab.xlabel('Time')
pylab.ylabel('Wind Speed')
pylab.title('Simulated Wind Speed vs Actual Wind Speed')
pylab.legend(('predicted','measured','kalman'))
pylab.show()
def calculate_regression(x, y):
R2 = 0
A = np.array( [x, np.ones(len(x))] )
model, resid = np.linalg.lstsq(A.T, y)[:2]
R2_val = 1 - resid[0] / (y.size * y.var())
return R2_val
def load_x():
return np.array([2, 3, 3, 5, 4, 4, 4, 5, 5, 6, 5, 7, 7, 7, 8, 8, 8, 9, 9, 10, 10, 10, 11, 11,
11, 10, 8, 8, 8, 8, 6, 3, 4, 5, 5, 5, 6, 5, 5, 5, 6, 5, 5, 6, 6, 7, 6, 8, 9, 10,
12, 11, 10, 10, 10, 11, 11, 10, 8, 8, 9, 8, 9, 9, 9, 9, 8, 9, 8, 11, 11, 11, 12,
12, 13, 13, 13, 13, 13, 13, 13, 14, 13, 13, 12, 13, 13, 12, 12, 13, 13, 12, 12,
11, 12, 12, 19, 18, 17, 15, 13, 14, 14, 14, 13, 12, 12, 12, 12, 11, 10, 10, 10,
10, 9, 9, 8, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 6, 6, 6, 7, 7, 8, 8, 8, 6, 5, 5,
5, 5, 5, 5, 6, 4, 4, 4, 6, 7, 8, 7, 7, 9, 10, 10, 9, 9, 8, 7, 5, 5, 5, 5, 5, 5,
5, 5, 6, 5, 5, 5, 4, 4, 6, 6, 7, 7, 7, 7, 6, 6, 5, 5, 4, 2, 2, 2, 1, 1, 1, 2, 3,
13, 13, 12, 11, 10, 9, 10, 10, 8, 9, 8, 7, 5, 3, 2, 2, 2, 3, 3, 4, 4, 5, 6, 6,
7, 7, 7, 6, 6, 6, 7, 6, 6, 5, 4, 4, 3, 3, 3, 2, 2, 1, 5, 5, 3, 2, 1, 2, 6, 7,
7, 8, 8, 9, 9, 9, 9, 10, 10, 10, 10, 10, 10, 9, 9, 9, 9, 9, 8, 8, 8, 8, 7, 7,
7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 7, 11, 11, 11, 11, 10, 10, 9, 10, 10, 10, 2, 2,
2, 3, 1, 1, 3, 4, 5, 8, 9, 9, 9, 9, 8, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 7,
7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 7, 5, 5, 5, 5, 5, 6, 5])
def load_z():
return np.array([3, 2, 1, 1, 1, 1, 3, 3, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 2, 1, 1, 2, 2, 2,
2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 3, 4, 4, 4, 4, 5, 4, 4, 5, 5, 5, 6, 6,
6, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 7, 8, 8, 8, 8, 8, 8, 9, 10, 9, 9, 10, 10, 9,
9, 10, 9, 9, 10, 9, 8, 9, 9, 7, 7, 6, 7, 6, 6, 7, 7, 8, 8, 8, 8, 8, 8, 7, 6, 7,
8, 8, 7, 8, 9, 9, 9, 9, 10, 9, 9, 9, 8, 8, 10, 9, 10, 10, 9, 9, 9, 10, 9, 8, 7,
7, 7, 7, 8, 7, 6, 5, 4, 3, 5, 3, 5, 4, 4, 4, 2, 4, 3, 2, 1, 1, 2, 1, 2, 1, 4, 4,
4, 4, 4, 3, 3, 3, 1, 1, 1, 1, 2, 3, 3, 2, 3, 3, 3, 2, 2, 5, 4, 2, 5, 4, 1, 1, 1,
1, 1, 1, 1, 2, 2, 1, 1, 3, 3, 3, 3, 3, 4, 3, 4, 3, 4, 4, 4, 4, 3, 3, 4, 4, 4, 4,
4, 4, 5, 5, 5, 4, 3, 3, 3, 3, 3, 3, 3, 3, 1, 2, 2, 3, 3, 1, 2, 1, 1, 2, 4, 3, 1,
1, 2, 0, 0, 0, 2, 1, 0, 0, 2, 3, 2, 4, 4, 3, 3, 4, 5, 5, 5, 4, 5, 4, 4, 4, 5, 5,
4, 3, 3, 4, 4, 4, 3, 3, 3, 4, 4, 4, 5, 5, 5, 4, 5, 5, 5, 5, 6, 5, 5, 8, 9, 8, 9,
9, 9, 9, 9, 10, 10, 10, 10, 10, 10, 10, 9, 10, 9, 8, 8, 9, 8, 9, 9, 10, 9, 9, 9,
7, 7, 9, 8, 7, 6, 6, 5, 5, 5, 5, 3, 3, 3, 4, 6, 5, 5, 6, 5])
if __name__ == '__main__': main() # this avoids executing main on import your_module
-------------------------
Observations:
1）如果昨天的预测是过度预测（正偏差），那么今天，我会通过减去偏差来进行修正 . 在实践中，如果今天我碰巧预测不足，那么减去正偏差会导致更糟糕的预测 . 实际上，我观察到更广泛的数据，整体拟合度较差 . 我的例子出了什么问题？
2）大多数卡尔曼滤波器资源指示卡尔曼滤波器最小化后验协方差p_j = E {（x_j-x ^ _j）}，并且具有证明选择K以最小化p_j . 但有人可以解释如何最小化后验协方差实际上最小化过程白噪声的影响吗？在实时情况下，假设实际风速和测得的风速为5米/秒 . 预测风速为6米/秒，即 . 有一个w = 1米/秒的噪音 . 残留量为5-6 = -1m / s . 您通过从预测中获取1 m / s来恢复5 m / s . 那是如何最小化过程噪声的影响？
3）这篇论文提到应用KF来平滑天气预报 . http://hal.archives-ouvertes.fr/docs/00/50/59/93/PDF/Louka_etal_jweia2008.pdf . 有趣的一点是在第9页eq（7）中，“一旦新的观测值y_t已知，时间t的x估计变为x_t = x_t / t-1 = K（y_t - H_t * x_t / t- 1）“ . 如果我参考实际时间来解释它，那么“一旦现在知道新的观测值，估计现在变成x_t ......” . “我了解KF如何将您的数据实时接近测量结果 . 但是，如果您正在t = now校正预测数据，而t = now的测量数据，那么预测又是怎样的呢？
谢谢！
UPDATE1：
4）如果我们希望卡尔曼处理的数据与测量数据时间序列之间的R2从未处理的数据中得到改善，我在代码中添加了一个延迟来调查预测的时间比当前测量的当前偏差晚多少与测量数据 . 在这个例子中，如果测量用于改进预测6时间步骤（从现在起3小时），它仍然有用（R2从0.183变为0.295） . 但如果测量用于改善从现在起1天的预测，那么它会破坏相关性（R2下降到0.075） .
2 回答
我更新了我的测试标量实现，没有假设完美测量R为1，这就是将卡尔曼增益降低到1的恒定值 . 现在我看到时间序列的改进，降低了RMSE误差 .
这条线不尊重Scalar Kalman Filter：
在我看来你应该做的：