Python 线性回归-减少自由度

我有一个熊猫数据框，其中的列如下

Order     Balance     Profit cum (%)

我在做线性回归

model_profit_tr = pd.ols(y=df_closed['Profit cum (%)'], x=df_closed['Order'])

问题是标准模型类似于（不通过原点的直线方程）

有两个自由度（a和b）

坡度（a）：

及截取(二):

我想减少我的模型的自由度（从2到1），我想有一个像这样的模型

y = a * x

---

使用

intercept

关键字参数：

model_profit_tr = pd.ols(y=df_closed['Profit cum (%)'], 
                         x=df_closed['Order'], 
                         intercept=False)

从文档：

In [65]: help(pandas.ols) 
Help on function ols in module pandas.stats.interface:

ols(**kwargs)

    [snip]

    Parameters
    ----------
    y: Series or DataFrame
        See above for types
    x: Series, DataFrame, dict of Series, dict of DataFrame, Panel
    weights : Series or ndarray
        The weights are presumed to be (proportional to) the inverse of the
        variance of the observations.  That is, if the variables are to be
        transformed by 1/sqrt(W) you must supply weights = 1/W
    intercept: bool
        True if you want an intercept.  Defaults to True.
    nw_lags: None or int
        Number of Newey-West lags.  Defaults to None.

    [snip]

---

以下是显示解决方案的示例：

#!/usr/bin/env python

import pandas as pd
import matplotlib.pylab as plt
import numpy as np

data = [
(0.2, 1.3),
(1.3, 3.9),
(2.1, 4.8),
(2.9,5.5),
(3.3,6.9)
]

df = pd.DataFrame(data, columns=['X', 'Y'])

print(df)

# 2 degrees of freedom : slope / intercept
model_with_intercept = pd.ols(y=df['Y'], x=df['X'], intercept=True)
df['Y_fit_with_intercept'] = model_with_intercept.y_fitted

# 1 degree of freedom : slope ; intersept=0
model_no_intercept = pd.ols(y=df['Y'], x=df['X'], intercept=False)
df['Y_fit_no_intercept'] = model_no_intercept.y_fitted

# 1 degree of freedom : slope ; intersept=offset
offset = -1
df['Yoffset'] = df['Y'] - offset
model_with_offset = pd.ols(y=df['Yoffset'], x=df['X'], intercept=False)
df['Y_fit_offset'] = model_with_offset.y_fitted + offset

print(model_with_intercept)
print(model_no_intercept)
print(model_with_offset)

df.plot(x='X', y=['Y', 'Y_fit_with_intercept', 'Y_fit_no_intercept', 'Y_fit_offset'])
plt.show()

---

非常感谢（为解决方案和帮助提示）！我还有一个问题，但我不知道是否可以在这里问。。。如果我想将intercept设置为给定值（而不是0），我应该怎么做。（我还将把自由度从2减少到1）@FemtoTrader：我认为

ols

没有这个功能。但是，考虑到它是最小二乘法，您可以从

y

中减去该截距，然后使用

ols

和

intercept=False

。应该是一样的。不，那不一样！如果强制直线通过给定的坡度，则坡度不同point@FemtoTrader：否，在最小二乘法中，使用固定的

B

将

a*x+B

拟合到

y

与将

a*x

拟合到

y-B

相同。

model_profit_tr = pd.ols(y=df_closed['Profit cum (%)'], 
                         x=df_closed['Order'], 
                         intercept=False)

In [65]: help(pandas.ols) 
Help on function ols in module pandas.stats.interface:

ols(**kwargs)

    [snip]

    Parameters
    ----------
    y: Series or DataFrame
        See above for types
    x: Series, DataFrame, dict of Series, dict of DataFrame, Panel
    weights : Series or ndarray
        The weights are presumed to be (proportional to) the inverse of the
        variance of the observations.  That is, if the variables are to be
        transformed by 1/sqrt(W) you must supply weights = 1/W
    intercept: bool
        True if you want an intercept.  Defaults to True.
    nw_lags: None or int
        Number of Newey-West lags.  Defaults to None.

    [snip]

#!/usr/bin/env python

import pandas as pd
import matplotlib.pylab as plt
import numpy as np

data = [
(0.2, 1.3),
(1.3, 3.9),
(2.1, 4.8),
(2.9,5.5),
(3.3,6.9)
]

df = pd.DataFrame(data, columns=['X', 'Y'])

print(df)

# 2 degrees of freedom : slope / intercept
model_with_intercept = pd.ols(y=df['Y'], x=df['X'], intercept=True)
df['Y_fit_with_intercept'] = model_with_intercept.y_fitted

# 1 degree of freedom : slope ; intersept=0
model_no_intercept = pd.ols(y=df['Y'], x=df['X'], intercept=False)
df['Y_fit_no_intercept'] = model_no_intercept.y_fitted

# 1 degree of freedom : slope ; intersept=offset
offset = -1
df['Yoffset'] = df['Y'] - offset
model_with_offset = pd.ols(y=df['Yoffset'], x=df['X'], intercept=False)
df['Y_fit_offset'] = model_with_offset.y_fitted + offset

print(model_with_intercept)
print(model_no_intercept)
print(model_with_offset)

df.plot(x='X', y=['Y', 'Y_fit_with_intercept', 'Y_fit_no_intercept', 'Y_fit_offset'])
plt.show()