我对将卡尔曼滤波器 (KF) 应用于以下预测问题时所看到的行为有疑问。我已经包含了一个简单的代码示例。
目标：我想知道 KF 是否适合使用现在（在 t 时）获得的测量结果来改进前一天（在 t+24 小时）的预测/模拟结果。目标是使预测尽可能接近测量值
假设： 我们假设测量是完美的（即，如果我们能得到预测与测量完美匹配，我们很高兴）。
我们有一个测量变量（z，真实风速）和一个模拟变量（x，预测风速）。
模拟的风速 x 是由 NWP（数值天气预报）软件使用各种气象数据（对我来说是黑匣子）产生的。模拟文件每天生成，每半小时包含一次数据。
我尝试使用我现在获得的测量值和现在的预测数据（在 t-24 小时前生成）使用标量卡尔曼滤波器来纠正 t+24 小时的预测。作为参考，我使用了： http ://www.swarthmore.edu/NatSci/echeeve1/Ref/Kalman/ScalarKalman.html
代码：
#! /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
-------------------------
观察：
1）如果昨天的预测过度预测（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。我实际上观察到更广泛的数据波动，整体拟合较差。我的例子有什么问题？
2) 大多数卡尔曼滤波器资源表明卡尔曼滤波器最小化后验协方差 p_j = E{(x_j – x^_j)}，并且有证据选择 K 来最小化 p_j。但是有人可以解释最小化后验协方差实际上如何最小化过程白噪声 w 的影响吗？在实时情况下，假设实际风速和测量风速为 5 m/s。预测风速为 6 m/s，即。有一个 w = 1 m/s 的噪音。残差为 5 – 6 = -1 m/s。您通过从您的预测中取 1 m/s 来纠正 5 m/s。这就是过程噪声的影响最小化的方式吗？
3）这是一篇提到将KF应用于平滑天气预报的论文。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，在时间 t 的 x 估计值变为 x_t = x_t/t-1 = K(y_t – H_t * x_t/t- 1）”。如果我要根据实际时间来解释它，那么“一旦现在知道新的观察值，现在的估计就变成了 x_t ...。” 我了解 KF 如何让您的数据实时接近测量值。但是，如果您要使用 t=now 的测量数据来修正 t=now 的预测数据，那又如何成为预测？
谢谢！
更新1：
4) 如果我们希望卡尔曼处理数据与测量数据时间序列之间的 R2 从未处理数据中改进，我在代码中添加了一个延迟，以研究预测比从当前测量计算的当前偏差晚多少与测量数据。在此示例中，如果测量用于改进预测 6 时间步（从现在起 3 小时），它仍然有用（R2 从 0.183 变为 0.295）。但是，如果使用测量来改进 1 天后的预测，那么它会破坏相关性（R2 下降到 0.075）。