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我在使用时遇到了一个pykalman 0.9.5问题Python 3.6.3

请参阅下面的代码，为什么结果来自kf2和kf3不同，而结果来自kf1和kf3相同？

kf2和之间过程的区别kf3在于我只是将迭代拆分为运行 2 次函数 for kf2。

感谢大家调查它。

>>>pri_mean[:10]
array([ 2827.2222,  2829.6   ,  2831.    ,  2832.1   ,  2833.1   ,  2835.3   ,  2833.9   ,
        2833.8   ,  2833.6   ,  2833.    ])

>>>kf1 = KalmanFilter()
>>>kf1 = kf1.em(pri_mean, 10, em_vars='all')
>>>print(kf1.transition_matrices, kf1.transition_offsets, kf1.transition_covariance)
[[ 0.99741876]] [ 10.04426882] [[ 2896.92752373]]
>>>kf2 = kf1.em(pri_mean, 10, em_vars='all')
>>>print(kf2.transition_matrices, kf2.transition_offsets, kf2.transition_covariance)
[[ 0.99364606]] [ 20.02260806] [[ 2600.94151188]]

>>>kf3 = KalmanFilter()
>>>kf3 = kf3.em(pri_mean, 20, em_vars='all')
>>>print(kf3.transition_matrices, kf3.transition_offsets, kf3.transition_covariance)
[[ 0.99741876]] [ 10.04426882] [[ 2896.92752373]]
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1 回答 1

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我编辑了答案，因为我误解了这个问题。我认为问题在于您缺少关键字n_iter

请参阅此代码：

kf1 = KalmanFilter()
kf1 = kf1.em(pri_mean, n_iter=10, em_vars='all')
print(kf1.transition_matrices, kf1.transition_offsets, kf1.transition_covariance)
kf1 = kf1.em(pri_mean, n_iter=10, em_vars='all')
print(kf1.transition_matrices, kf1.transition_offsets, kf1.transition_covariance)
kf1 = KalmanFilter()
kf1 = kf1.em(pri_mean, n_iter=20, em_vars='all')
print(kf1.transition_matrices, kf1.transition_offsets, kf1.transition_covariance)

我通过创建一个过滤器KalmanFilter()，循环 10 次迭代并打印，然后再循环 10 次并打印。这相当于.em()立即调用 20 次迭代。

将产生以下输出

[[ 0.95500561]] [ 113.29118228] [[ 6431.66262464]]
[[ 0.93636512]] [ 119.32378005] [[ 249.67547612]]
[[ 0.93636512]] [ 119.32378005] [[ 249.67547612]]
于 2017-12-21T11:13:54.357 回答