单调递增的，不平稳
单位根检验下
from statsmodels.tsa.stattools import adfuller as ADFADF(data['销量'])>>>(1.8137710150945274, 0.9983759421514264, 10, 26, {'1%': -3.7112123008648155, '10%': -2.6300945562130176, '5%': -2.981246804733728}, 299.46989866024177)元组第二个值为p值，p>0.05，接受原假设，存在单位根
差分（1阶）
d_data = data.diff().dropna()d_data.columns = ['一阶差分']d_data.plot()再次单位根检验下
ADF(d_data['一阶差分'])>>>(-3.1560562366723532, 0.02267343544004886, 0, 35, {'1%': -3.6327426647230316, '10%': -2.6130173469387756, '5%': -2.9485102040816327}, 287.5909090780334)p<0.05,拒绝原假设，不存在单位根了。
随机性检验（白噪声检验）
from statsmodels.stats.diagnostic import acorr_ljungboxacorr_ljungbox(d_data,lags =1)>>>(array([11.30402222]), array([0.00077339]))p=0.00077339<0.05，拒绝原假设，所以一阶差分后的序列不是随机的。
确定阶数p,q建模
一、根据自相关系数和偏自相关系数目测
模型自相关系数（ACF）偏自相关系数（PACF）AR（P）拖尾p阶截尾MA（q）q阶截尾拖尾ARMA（p，q）p阶拖尾q阶拖尾p看pacf，q看acf
from statsmodels.graphics.tsaplots import plot_acffrom statsmodels.graphics.tsaplots import plot_pacfplot_acf(d_data)plot_pacf(d_data)由上图可知，p = 0，q=1
二、根据bic遍历p，q值，取bic最小时对应的p，q
from statsmodels.tsa.arima_model import ARIMAtmp = []for p in range(4): for q in range(4): try: tmp.append([ARIMA(data,(p,1,q)).fit().bic,p,q]) except: tmp.append([None,p,q])tmp = pd.DataFrame(tmp,columns = ['bic','p','q'])tmp[tmp['bic'] ==tmp['bic'].min()]>>> bic pq422.51008201与目测一样，综上p = 0，q = 1
建模预测
model = ARIMA(data,(0,1,1)).fit()model.summary()>>>ARIMA Model ResultsDep. Variable:D.销量No. Observations:36Model:ARIMA(0, 1, 1)Log Likelihood-205.880Method:css-mleS.D. of innovations73.086Date:Tue, 31 Jul 2018AIC417.760Time:15:12:13BIC422.510Sample:01-02-2015HQIC419.418- 02-06-2015coefstd errzP>|z|[0.0250.975]const49.956120.1392.4810.01810.48489.428ma.L1.D.销量0.67100.1654.0710.0000.3480.994RootsRealImaginaryModulusFrequencyMA.1-1.4902+0.0000j1.49020.5000yp = model.forecast(5)#预测未来5年yp[0]>>>array([4873.9665477 , 4923.92261622, 4973.87868474, 5023.83475326, 5073.79082178])完！