使用Pandas OLS我可以按照以下方式使用和使用模型:
ols_test = pd.ols(y=merged2[:-1].Units, x=merged2[:-1].lastqu) #to exclude current year, then do forecast method
yrahead=(ols_test.beta['x'] * merged2.lastqu[-1:]) + ols_test.beta['intercept']
我需要切换到statsmodels以获得一些额外的功能(主要是残差图见(question here)
所以现在我有:
def fit_line2(x, y):
X = sm.add_constant(x, prepend=True) #Add a column of ones to allow the calculation of the intercept
model = sm.OLS(y, X,missing='drop').fit()
"""Return slope, intercept of best fit line."""
X = sm.add_constant(x)
return model
和
model=fit_line2(merged2[:-1].lastqu,merged2[:-1].Units)
print fit.summary()
但我无法得到
yrahead2=model.predict(merged2.lastqu[-1:])
或任何变体给我预测?请注意,pd.ols使用相同的merged2.lastqu [-1:]来获取我想要“预测”的数据,无论我放入()预测我没有任何喜悦。似乎statsmodels想要除了熊猫DF单元之外的某些特定的东西我甚至试图在那里放一个数字,例如2696,但仍然没有...... 我当前的错误是
----> 3 yrahead2=model.predict(merged2.lastqu[-1:])
/usr/lib/pymodules/python2.7/statsmodels/base/model.pyc in predict(self, exog, transform, *args, **kwargs)
1004 exog = np.atleast_2d(exog) # needed in count model shape[1]
1005
-> 1006 return self.model.predict(self.params, exog, *args, **kwargs)
1007
1008
/usr/lib/pymodules/python2.7/statsmodels/regression/linear_model.pyc in predict(self, params, exog)
253 if exog is None:
254 exog = self.exog
--> 255 return np.dot(exog, params)
256
257 class GLS(RegressionModel):
ValueError: objects are not aligned
> /usr/lib/pymodules/python2.7/statsmodels/regression/linear_model.py(255)predict()
254 exog = self.exog
--> 255 return np.dot(exog, params)
256
答案 0 :(得分:3)
我更喜欢statsmodels的公式api。至少为此,model.fit().predict
想要一个DataFrame,其中列与预测变量具有相同的名称。这是一个例子:
In [2]: df = pd.DataFrame({'X': np.arange(10), 'Y': np.arange(10) + np.random.randn(10)})
In [3]: mod = sm.OLS.from_formula("Y ~ X", df)
In [4]: res = mod.fit()
In [5]: exog = pd.DataFrame({"X": np.linspace(0, 10, 100)})
In [6]: res.predict(exog)
Out[6]:
array([ 0.99817045, 1.07854804, 1.15892563, 1.23930322, 1.31968081,
1.40005839, 1.48043598, 1.56081357, 1.64119116, 1.72156875,
1.80194634, 1.88232393, 1.96270152, 2.04307911, 2.1234567 ,
2.20383429, 2.28421188, 2.36458947, 2.44496706, 2.52534465,
2.60572224, 2.68609983, 2.76647742, 2.84685501, 2.92723259,
3.00761018, 3.08798777, 3.16836536, 3.24874295, 3.32912054,
3.40949813, 3.48987572, 3.57025331, 3.6506309 , 3.73100849,
3.81138608, 3.89176367, 3.97214126, 4.05251885, 4.13289644,
4.21327403, 4.29365162, 4.3740292 , 4.45440679, 4.53478438,
4.61516197, 4.69553956, 4.77591715, 4.85629474, 4.93667233,
5.01704992, 5.09742751, 5.1778051 , 5.25818269, 5.33856028,
5.41893787, 5.49931546, 5.57969305, 5.66007064, 5.74044823,
5.82082582, 5.9012034 , 5.98158099, 6.06195858, 6.14233617,
6.22271376, 6.30309135, 6.38346894, 6.46384653, 6.54422412,
6.62460171, 6.7049793 , 6.78535689, 6.86573448, 6.94611207,
7.02648966, 7.10686725, 7.18724484, 7.26762243, 7.34800002,
7.4283776 , 7.50875519, 7.58913278, 7.66951037, 7.74988796,
7.83026555, 7.91064314, 7.99102073, 8.07139832, 8.15177591,
8.2321535 , 8.31253109, 8.39290868, 8.47328627, 8.55366386,
8.63404145, 8.71441904, 8.79479663, 8.87517421, 8.9555518 ])
答案 1 :(得分:2)
您的merged2.lastqu[-1:]
不包含常量
yrahead2=model.predict(sm.add_constant(merged2.lastqu[-1:], prepend=True))
应该有用。
另一种方法是以与模型中X相同的方式将常量添加到数据框中,并使用数据框的相应列df[['const', my_other_X]]