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r - auto arima : r and python suggest different arima models for same data, 为什么?

转载 作者:行者123 更新时间:2023-12-05 06:25:41 27 4
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我曾尝试在 python 中同时在 R 上使用 auto arima 获取相同的数据,但得到了不同的 ARIMA 模型选择是具有不同 AIC 的最佳模型。你能告诉我为什么我从两种语言中得到具有不同 AIC 的不同最佳模型吗?

R 的数据和代码

wineind <- c(15136., 16733., 20016., 17708., 18019., 19227., 22893., 23739.,
21133., 22591., 26786., 29740., 15028., 17977., 20008., 21354.,
19498., 22125., 25817., 28779., 20960., 22254., 27392., 29945.,
16933., 17892., 20533., 23569., 22417., 22084., 26580., 27454.,
24081., 23451., 28991., 31386., 16896., 20045., 23471., 21747.,
25621., 23859., 25500., 30998., 24475., 23145., 29701., 34365.,
17556., 22077., 25702., 22214., 26886., 23191., 27831., 35406.,
23195., 25110., 30009., 36242., 18450., 21845., 26488., 22394.,
28057., 25451., 24872., 33424., 24052., 28449., 33533., 37351.,
19969., 21701., 26249., 24493., 24603., 26485., 30723., 34569.,
26689., 26157., 32064., 38870., 21337., 19419., 23166., 28286.,
24570., 24001., 33151., 24878., 26804., 28967., 33311., 40226.,
20504., 23060., 23562., 27562., 23940., 24584., 34303., 25517.,
23494., 29095., 32903., 34379., 16991., 21109., 23740., 25552.,
21752., 20294., 29009., 25500., 24166., 26960., 31222., 38641.,
14672., 17543., 25453., 32683., 22449., 22316., 27595., 25451.,
25421., 25288., 32568., 35110., 16052., 22146., 21198., 19543.,
22084., 23816., 29961., 26773., 26635., 26972., 30207., 38687.,
16974., 21697., 24179., 23757., 25013., 24019., 30345., 24488.,
25156., 25650., 30923., 37240., 17466., 19463., 24352., 26805.,
25236., 24735., 29356., 31234., 22724., 28496., 32857., 37198.,
13652., 22784., 23565., 26323., 23779., 27549., 29660., 23356.)
tswineind<-ts(wineind, start=c(1985,1), frequency=12)
library(forecast)
tswineindbest<-auto.arima(tswineind,approximation = FALSE)
tswineindbest

R 的结果

ARIMA(0,1,3)(0,1,1)[12]

Python 的数据和代码

import numpy as np
import pmdarima as pm
from pmdarima.datasets import load_wineind

# this is a dataset from R
wineind = load_wineind().astype(np.float64)

# fit stepwise auto-ARIMA
stepwise_fit = pm.auto_arima(wineind, start_p=1, start_q=1,
max_p=3, max_q=3, m=12,
start_P=0, seasonal=True,
d=1, D=1, trace=True,
error_action='ignore', # don't want to know if an order does not work
suppress_warnings=True, # don't want convergence warnings
stepwise=True) # set to stepwise
stepwise_fit.summary()

Python 结果

    SARIMAX(1, 1, 2)x(0, 1, 1, 12)  AIC 3066.742

我希望 RPython 具有相同的最佳模型和相同的 AIC

最佳答案

我在网上转了一圈,发现这个 python 代码非常有用

# import package
import itertools

# Define the p, d and q parameters to take any value between 0 and 2
p = d = q = range(0, 3)

# Generate all different combinations of p, q and q triplets
pdq = list(itertools.product(p, d, q))

# Generate all different combinations of seasonal p, q and q
triplets
seasonal_pdq = [(x[0], x[1], x[2], 12) for x in
list(itertools.product(p, d, q))]

print('Examples of parameter combinations for Seasonal ARIMA...')
print('SARIMAX: {} x {}'.format(pdq[1], seasonal_pdq[1]))
print('SARIMAX: {} x {}'.format(pdq[1], seasonal_pdq[2]))
print('SARIMAX: {} x {}'.format(pdq[2], seasonal_pdq[3]))
print('SARIMAX: {} x {}'.format(pdq[2], seasonal_pdq[4]))

warnings.filterwarnings("ignore") # specify to ignore warning messages

for param in pdq:
for param_seasonal in seasonal_pdq:
try:
mod = sm.tsa.statespace.SARIMAX(ts,
order=param,
seasonal_order=param_seasonal,
enforce_stationarity=False,
enforce_invertibility=False)

results = mod.fit()

print('ARIMA{}x{}12 - AIC:{}'.format(param, param_seasonal, results.aic))
except:
continue

这里输入的变量tp是我在python代码中用ts表示的单维时间序列数据。结果与 R 中的 auto.arima 相同。

关于r - auto arima : r and python suggest different arima models for same data, 为什么?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/56802974/

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