Go語言的線性迴歸庫

Question

我正在尋找一個使用MLE或LSE實現線性迴歸的Go庫。 有沒有人見過？

有此統計庫，但它似乎並不有我需要什麼： https://github.com/grd/statistics

謝謝！

Answer 1

實現LSE（最小平方誤差）線性迴歸相當簡單。

Here是JavaScript中的一個實現 - 移植到Go應該是微不足道的。

---

Here的一個（未經測試）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() { 
    // ... 
} 

Answer 2

有一個叫gostat項目，有這應該是能夠做到線性迴歸貝葉斯包。

不幸的是，文檔有點缺乏，所以您可能需要閱讀代碼以瞭解如何使用它。我自己嘗試了一下，但沒有觸及bayes包。

Answer 3

我的Gentleman's AS75（在線）線性迴歸算法的端口是用Go（golang）編寫的，它進行了正態OLS普通最小二乘迴歸。在線部分意味着它可以處理不受限制的數據行：如果您習慣於提供（nxp）設計矩陣，則有點不同：您調用includ（）n次（或者如果獲得更多數據，則更多） ，每次給它一個p值的向量。這可以有效地處理n增長較大的情況，並且您可能必須從磁盤中傳輸數據，因爲它不會全部放入內存。

https://github.com/glycerine/zettalm

Answer 4

我已實現了以下使用梯度下降，它不僅賦予了係數，但需要任意數量的解釋變量，是相當準確：

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) 

} 

結果：

Thetas : [6.999959251448524 -0.769216974483968] 
Thetas_2 : [1.5694174539341945 -0.06169183063112409 0.2359981255871977 0.2424327101610395] 

go playground

我檢查我的結果與python.pandas，他們是非常接近：

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---------------------------------