答案 0 :(得分：4)
我使用渐变下降实现了以下功能，它只给出了系数，但是它采用了任意数量的解释变量并且相当准确：
package main
import "fmt"
func calc_ols_params(y []float64, x[][]float64, n_iterations int, alpha float64) []float64 {
thetas := make([]float64, len(x))
for i := 0; i < n_iterations; i++ {
my_diffs := calc_diff(thetas, y, x)
my_grad := calc_gradient(my_diffs, x)
for j := 0; j < len(my_grad); j++ {
thetas[j] += alpha * my_grad[j]
}
}
return thetas
}
func calc_diff (thetas []float64, y []float64, x[][]float64) []float64 {
diffs := make([]float64, len(y))
for i := 0; i < len(y); i++ {
prediction := 0.0
for j := 0; j < len(thetas); j++ {
prediction += thetas[j] * x[j][i]
}
diffs[i] = y[i] - prediction
}
return diffs
}
func calc_gradient(diffs[] float64, x[][]float64) []float64 {
gradient := make([]float64, len(x))
for i := 0; i < len(diffs); i++ {
for j := 0; j < len(x); j++ {
gradient[j] += diffs[i] * x[j][i]
}
}
for i := 0; i < len(x); i++ {
gradient[i] = gradient[i] / float64(len(diffs))
}
return gradient
}
func main(){
y := []float64 {3,4,5,6,7}
x := [][]float64 {{1,1,1,1,1}, {4,3,2,1,3}}
thetas := calc_ols_params(y, x, 100000, 0.001)
fmt.Println("Thetas : ", thetas)
y_2 := []float64 {1,2,3,4,3,4,5,4,5,5,4,5,4,5,4,5,6,5,4,5,4,3,4}
x_2 := [][]float64 {{1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1},
{4,2,3,4,5,4,5,6,7,4,8,9,8,8,6,6,5,5,5,5,5,5,5},
{4,1,2,3,4,5,6,7,5,8,7,8,7,8,7,8,7,7,7,7,7,6,5},
{4,1,2,5,6,7,8,9,7,8,7,8,7,7,7,7,7,7,6,6,4,4,4},}
thetas_2 := calc_ols_params(y_2, x_2, 100000, 0.001)
fmt.Println("Thetas_2 : ", thetas_2)
}
结果：
Thetas : [6.999959251448524 -0.769216974483968]
Thetas_2 : [1.5694174539341945 -0.06169183063112409 0.2359981255871977 0.2424327101610395]
我使用python.pandas检查了我的搜索结果，结果非常接近：
In [24]: from pandas.stats.api import ols
In [25]: df = pd.DataFrame(np.array(x).T, columns=['x1','x2','x3','y'])
In [26]: from pandas.stats.api import ols
In [27]: x = [
[4,2,3,4,5,4,5,6,7,4,8,9,8,8,6,6,5,5,5,5,5,5,5],
[4,1,2,3,4,5,6,7,5,8,7,8,7,8,7,8,7,7,7,7,7,6,5],
[4,1,2,5,6,7,8,9,7,8,7,8,7,7,7,7,7,7,6,6,4,4,4]
]
In [28]: y = [1,2,3,4,3,4,5,4,5,5,4,5,4,5,4,5,6,5,4,5,4,3,4]
In [29]: x.append(y)
In [30]: df = pd.DataFrame(np.array(x).T, columns=['x1','x2','x3','y'])
In [31]: ols(y=df['y'], x=df[['x1', 'x2', 'x3']])
Out[31]:
-------------------------Summary of Regression Analysis-------------------------
Formula: Y ~ <x1> + <x2> + <x3> + <intercept>
Number of Observations: 23
Number of Degrees of Freedom: 4
R-squared: 0.5348
Adj R-squared: 0.4614
Rmse: 0.8254
F-stat (3, 19): 7.2813, p-value: 0.0019
Degrees of Freedom: model 3, resid 19
-----------------------Summary of Estimated Coefficients------------------------
Variable Coef Std Err t-stat p-value CI 2.5% CI 97.5%
--------------------------------------------------------------------------------
x1 -0.0618 0.1446 -0.43 0.6741 -0.3453 0.2217
x2 0.2360 0.1487 1.59 0.1290 -0.0554 0.5274
x3 0.2424 0.1394 1.74 0.0983 -0.0309 0.5156
intercept 1.5704 0.6331 2.48 0.0226 0.3296 2.8113
---------------------------------End of Summary---------------------------------
和
In [34]: df_1 = pd.DataFrame(np.array([[3,4,5,6,7], [4,3,2,1,3]]).T, columns=['y', 'x'])
In [35]: df_1
Out[35]:
y x
0 3 4
1 4 3
2 5 2
3 6 1
4 7 3
[5 rows x 2 columns]
In [36]: ols(y=df_1['y'], x=df_1['x'])
Out[36]:
-------------------------Summary of Regression Analysis-------------------------
Formula: Y ~ <x> + <intercept>
Number of Observations: 5
Number of Degrees of Freedom: 2
R-squared: 0.3077
Adj R-squared: 0.0769
Rmse: 1.5191
F-stat (1, 3): 1.3333, p-value: 0.3318
Degrees of Freedom: model 1, resid 3
-----------------------Summary of Estimated Coefficients------------------------
Variable Coef Std Err t-stat p-value CI 2.5% CI 97.5%
--------------------------------------------------------------------------------
x -0.7692 0.6662 -1.15 0.3318 -2.0749 0.5365
intercept 7.0000 1.8605 3.76 0.0328 3.3534 10.6466
---------------------------------End of Summary---------------------------------
In [37]: df_1 = pd.DataFrame(np.array([[3,4,5,6,7], [4,3,2,1,3]]).T, columns=['y', 'x'])
In [38]: ols(y=df_1['y'], x=df_1['x'])
Out[38]:
-------------------------Summary of Regression Analysis-------------------------
Formula: Y ~ <x> + <intercept>
Number of Observations: 5
Number of Degrees of Freedom: 2
R-squared: 0.3077
Adj R-squared: 0.0769
Rmse: 1.5191
F-stat (1, 3): 1.3333, p-value: 0.3318
Degrees of Freedom: model 1, resid 3
-----------------------Summary of Estimated Coefficients------------------------
Variable Coef Std Err t-stat p-value CI 2.5% CI 97.5%
--------------------------------------------------------------------------------
x -0.7692 0.6662 -1.15 0.3318 -2.0749 0.5365
intercept 7.0000 1.8605 3.76 0.0328 3.3534 10.6466
---------------------------------End of Summary---------------------------------
答案 1 :(得分：3)
实现LSE（最小平方误差）线性回归非常简单。
Here是JavaScript中的一个实现 - 移植到Go应该是微不足道的。
Here是一个（未经测试的）端口：
package main
import "fmt"
type Point struct {
X float64
Y float64
}
func linearRegressionLSE(series []Point) []Point {
q := len(series)
if q == 0 {
return make([]Point, 0, 0)
}
p := float64(q)
sum_x, sum_y, sum_xx, sum_xy := 0.0, 0.0, 0.0, 0.0
for _, p := range series {
sum_x += p.X
sum_y += p.Y
sum_xx += p.X * p.X
sum_xy += p.X * p.Y
}
m := (p*sum_xy - sum_x*sum_y) / (p*sum_xx - sum_x*sum_x)
b := (sum_y / p) - (m * sum_x / p)
r := make([]Point, q, q)
for i, p := range series {
r[i] = Point{p.X, (p.X*m + b)}
}
return r
}
func main() {
// ...
}
答案 2 :(得分：1)
有一个名为gostat的项目，它有一个贝叶斯软件包，可以进行线性回归。
不幸的是文档有点缺乏，因此您可能必须阅读代码以了解如何使用它。我自己稍微涉足了它，但没有触及贝叶斯包装。
答案 3 :(得分：1)
我的绅士AS75（在线）线性回归算法的端口用Go（golang）编写。它做正常的OLS普通最小二乘回归。在线部分意味着它可以处理无限数据行：如果您习惯于提供（nxp）设计矩阵，这有点不同：您调用Includ（）n次，（或者如果您获得更多数据则更多） ，每次给它一个p值的向量。这有效地处理了n变大的情况，你可能不得不从磁盘中传输数据，因为它不会全部适合内存。