如何在python中使用lowess平滑具有年度数据的时间序列

我有一些数据每年记录如下

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]]

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

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

      [[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]]

我使用

statsmodels

python包计算lowess，如下所示

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]]

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

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

      [[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]]

我得到的结果如下

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]]

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

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

      [[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，我不清楚在

statsmodel

中得到的两个值是什么。是不是我做错了什么。此外，我还想知道这两个参数

frac

和

it

的作用是什么

此外，我还想使用

seaborn

绘制平滑的时间序列。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。

结果可以按照下面的代码绘制。请注意，

lowess（）

第一个参数是

y

-值（

endog

），第二个参数是

x

（

exog

）。默认结果的

z[：，0]

为排序后的

x

-值，而

z[：，1]

为相应的估计

y

-值

导入matplotlib.pyplot作为plt
将statsmodels.api作为sm导入
将numpy作为np导入
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]
对应年份=[1970197119721973197419751976197719781979198019811982198319841985198619871988198919901991199219931994]
x=np.数组（对应的年）
y=np.数组（mydata）
z=sm.非参数lowess（y，x，分形=1./3，it=3）
plt.plot（x，y，color='dodgerblue'）
plt.绘图（z[：，0]，z[：，1]，'ro-'）
plt.show（）

PS：要与同一绘图上的seaborn

regplot

进行比较，请将其称为：

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

结果可以如下面的代码所示绘制。请注意，

lowess（）

第一个参数是

y

-值（

endog

），第二个参数是

x

（

exog

）。默认结果是

z[：，0]

作为排序后的

x

-值和

z[：，1]

相应的估计值

y

-值

导入matplotlib.pyplot作为plt
将statsmodels.api作为sm导入
将numpy作为np导入
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]
对应年份=[1970197119721973197419751976197719781979198019811982198319841985198619871988198919901991199219931994]
x=np.数组（对应的年）
y=np.数组（mydata）
z=sm.非参数lowess（y，x，分形=1./3，it=3）
plt.plot（x，y，color='dodgerblue'）
plt.绘图（z[：，0]，z[：，1]，'ro-'）
plt.show（）

PS：要与同一绘图上的seaborn

regplot

进行比较，请将其称为：

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

---

谢谢你的回答。请让我知道这两个参数

frac

和

it

的作用是什么？期待你的回复：）

frac=1/3

似乎意味着在评估一个点的估计值时，要考虑所有数据的三分之一（默认值

frac=2/3

）.

it=3

意味着权重提高了3倍。和中的更多信息。感谢您的回答。请让我知道

frac

和

it

这两个参数的作用是什么？期待您的回复：）

frac=1/3

似乎意味着在评估一个点（默认值

frac=2/3

）。

it=3

表示权重提高了3倍。和中的更多信息。