如何在python中用黄土平滑时间序列和年度数据

Question

我有一些每年记录的数据如下。

mydata = [0.6619346141815186, 0.7170140147209167, 0.692265510559082, 0.6394098401069641, 0.6030995845794678, 0.6500746607780457, 0.6013327240943909, 0.6273292303085327, 0.5865356922149658, 0.6477396488189697, 0.5827181339263916, 0.6496025323867798, 0.6589270234107971, 0.5498126149177551, 0.48638370633125305, 0.5367399454116821, 0.517595648765564, 0.5171639919281006, 0.47503289580345154, 0.6081966757774353, 0.5808742046356201, 0.5856912136077881, 0.5608134269714355, 0.6400936841964722, 0.6766082644462585]

corresponding_year = [1970,1971,1972,1973,1974,1975,1976,1977,1978,1979,1980,1981,1982,1983,1984,1985,1986,1987,1988,1989,1990,1991,1992,1993,1994]]

我使用statsmodelspython包计算lowess如下。

import statsmodels.api as sm
lowess = sm.nonparametric.lowess

z = lowess(x, y, frac= 1./3, it=3)

我得到的输出如下。

      [[1.96000000e+03, 6.95703548e-01],
       [1.96100000e+03, 6.81750671e-01],
       [1.96200000e+03, 6.68002318e-01],
       [1.96300000e+03, 6.55138324e-01],
       [1.96400000e+03, 6.38960761e-01],
       [1.96500000e+03, 6.25042177e-01],
       [1.96600000e+03, 6.18586936e-01],
       [1.96700000e+03, 6.17026334e-01],
       [1.96800000e+03, 6.14565102e-01],
       [1.96900000e+03, 6.17610340e-01],
       [1.97000000e+03, 6.20404414e-01],
       [1.97100000e+03, 6.10193222e-01],
       [1.97200000e+03, 5.90100648e-01],
       [1.97300000e+03, 5.70935248e-01],
       [1.97400000e+03, 5.47818726e-01],
       [1.97500000e+03, 5.25788570e-01],
       [1.97600000e+03, 5.18661218e-01],
       [1.97700000e+03, 5.28921300e-01],
       [1.97800000e+03, 5.42783400e-01],
       [1.97900000e+03, 5.55425915e-01],
       [1.98000000e+03, 5.71486587e-01],
       [1.98100000e+03, 5.91539778e-01],
       [1.98200000e+03, 6.13021691e-01],
       [1.98300000e+03, 6.34508409e-01],
       [1.98400000e+03, 6.57703989e-01]]

但是，我不清楚我在statsmodel 中得到的两个值是什么。有什么我做错了。另外我也想知道frac和it这两个参数是干什么的？

此外，我还想使用seaborn 绘制平滑时间序列。似乎 seaborn 支持lowess。但是，它没有frac 和it 参数。请参阅下面的代码。

import numpy as np
import seaborn as sns

x = np.arange(0, 10, 0.01)
ytrue = np.exp(-x / 5) + 2 * np.sin(x / 3)
y = ytrue + np.random.normal(size=len(x))

sns.regplot(x, y, lowess=True)

在这种情况下，是否可以使用statmodels 输出在seaborn 中绘制regplot？

如果需要，我很乐意提供更多详细信息。

Answer 1

lowess 结果可以绘制成如下代码所示。请注意，lowess() 第一个参数是 y-value (endog)，第二个参数是 x (exog)。默认结果有z[:,0] 是排序后的x-values，z[:,1] 是相应的估计y-values。

import matplotlib.pyplot as plt
import statsmodels.api as sm
import numpy as np

mydata = [0.6619346141815186, 0.7170140147209167, 0.692265510559082, 0.6394098401069641, 0.6030995845794678, 0.6500746607780457, 0.6013327240943909, 0.6273292303085327, 0.5865356922149658, 0.6477396488189697, 0.5827181339263916, 0.6496025323867798, 0.6589270234107971, 0.5498126149177551, 0.48638370633125305, 0.5367399454116821, 0.517595648765564, 0.5171639919281006, 0.47503289580345154, 0.6081966757774353, 0.5808742046356201, 0.5856912136077881, 0.5608134269714355, 0.6400936841964722, 0.6766082644462585]
corresponding_year = [1970,1971,1972,1973,1974,1975,1976,1977,1978,1979,1980,1981,1982,1983,1984,1985,1986,1987,1988,1989,1990,1991,1992,1993,1994]

x = np.array(corresponding_year)
y = np.array(mydata)
z = sm.nonparametric.lowess(y, x, frac= 1./3, it=3)

plt.plot(x, y, color='dodgerblue')
plt.plot(z[:,0], z[:,1], 'ro-')

plt.show()

PS：要与同一情节上的 seaborn regplot 进行比较，请将其称为：

sns.regplot(x, y, lowess=True, ax=plt.gca())