【统计学习方法】支持向量机(SVM) Python实现

「爱情、让人受尽委屈。」 2022-06-09 03:39 185阅读 0赞

## 前言 ##

代码可在Github上下载:[代码下载][Link 1]  
支持向量机，一个理论先于实践的产物，由Corinna Cortes和Vapnik等在1995年首次提出的，取得了非常好的效果。  
SVM分为线性可分（又可分硬间隔与软间隔）和线性不可分的问题，线性不可分可通过核函数进行映射到空间里，从而线性可分。

## 线性可分支持向量机 ##

模型： f ( x ) = s i g n ( w ⋅ x + b ) f\\left( x \\right) = sign\\left( \{w \\cdot x + b\} \\right) f(x)=sign(w⋅x\+b)  
函数间隔（关于样本点）： γ ^ i = y i ( w ⋅ x i + b ) \{ \{\\hat \\gamma \}\_i\} = \{y\_i\}\\left( \{w \\cdot \{x\_\{\\rm\{i\}\}\} + b\} \\right) γ^i=yi(w⋅xi\+b)  
函数间隔（关于数据集T）: γ ^ = min ⁡ i = 1 , . . . , N γ i \\hat \\gamma = \\mathop \{\\min \}\\limits\_\{i = 1,...,N\} \{\\gamma \_i\} γ^=i=1,...,Nminγi  
函数间隔表示一个确信度  
几何间隔： γ i = y i ( w ∥ w ∥ ⋅ x i + b ∥ w ∥ ) \{\\mathop \\gamma \\limits^\{\} \_i\} = \{y\_i\}\\left( \{\\frac\{w\}\{ \{\\left\\| w \\right\\|\}\} \\cdot \{x\_i\} + \\frac\{b\}\{ \{\\left\\| w \\right\\|\}\}\} \\right) iγ=yi(∥w∥w⋅xi\+∥w∥b)  
其实这里的几何间隔的就是点直线的距离。  
由此我们可以得到 max ⁡ w , b γ ^ ∥ w ∥ \\mathop \{\\max \}\\limits\_\{w,b\} \\frac\{ \{\\hat \\gamma \}\}\{ \{\\left\\| w \\right\\|\}\} w,bmax∥w∥γ^  
 s . t . y i ( w ⋅ x i + b ) − 1 ≥ 0 , i = 1 , 2 , . . . , N s.t.\{\\rm\{ \}\}\{y\_i\}\\left( \{w \\cdot \{x\_i\} + b\} \\right) - 1 \\ge 0,i = 1,2,...,N s.t.yi(w⋅xi\+b)−1≥0,i=1,2,...,N  
同时等价于  
 min ⁡ w , b 1 2 ∥ w ∥ 2 \\mathop \{\\min \}\\limits\_\{w,b\} \\frac\{1\}\{2\}\{\\left\\| w \\right\\|^2\} w,bmin21∥w∥2  
 s . t . y i ( w ⋅ x i + b ) − 1 ≥ 0 , i = 1 , 2 , . . . , N s.t.\{\\rm\{ \}\}\{y\_i\}\\left( \{w \\cdot \{x\_i\} + b\} \\right) - 1 \\ge 0,i = 1,2,...,N s.t.yi(w⋅xi\+b)−1≥0,i=1,2,...,N  
构造拉格朗日乘子 L ( w , b , α ) = 1 2 ∥ w ∥ 2 − ∑ i = 1 N α i y i ( w ⋅ x i + b ) + ∑ i = 1 N α i L\\left( \{w,b,\\alpha \} \\right) = \\frac\{1\}\{2\}\{\\left\\| w \\right\\|^2\} - \\sum\\limits\_\{i = 1\}^N \{ \{\\alpha \_i\}\{y\_i\}\\left( \{w \\cdot \{x\_i\} + b\} \\right) + \\sum\\limits\_\{i = 1\}^N \{ \{\\alpha \_i\}\} \} L(w,b,α)=21∥w∥2−i=1∑Nαiyi(w⋅xi\+b)\+i=1∑Nαi  
通过求导等于0课得 w ∗ = ∑ i = 1 N α i y i x i \{w^\*\} = \\sum\\limits\_\{i = 1\}^N \{ \{\\alpha \_i\}\{y\_i\}\{x\_i\}\} w∗=i=1∑Nαiyixi  
将 w w w代入到 y j ( w ∗ ⋅ x j + b ∗ ) − 1 = 0 \{y\_j\}\\left( \{ \{w^\*\} \\cdot \{x\_j\} + \{b^\*\}\} \\right) - 1 = 0 yj(w∗⋅xj\+b∗)−1=0  
可得 b ∗ = 1 y j − ∑ i = 1 N α i y i x i x j = y j − ∑ i = 1 N α i y i x i x j \{b^\*\} = \\frac\{1\}\{ \{ \{y\_j\}\}\} - \\sum\\limits\_\{i = 1\}^N \{ \{\\alpha \_i\}\{y\_i\}\{x\_i\}\{x\_j\}\} = \{y\_j\} - \\sum\\limits\_\{i = 1\}^N \{ \{\\alpha \_i\}\{y\_i\}\{x\_i\}\{x\_j\}\} b∗=yj1−i=1∑Nαiyixixj=yj−i=1∑Nαiyixixj  
我们的目标是 max ⁡ α min ⁡ w , b L ( w , b , α ) \\mathop \{\\max \}\\limits\_\\alpha \\mathop \{\\min \}\\limits\_\{w,b\} L\\left( \{w,b,\\alpha \} \\right) αmaxw,bminL(w,b,α)  
然后用对偶来表示，最终的对偶形式  
 min ⁡ α 1 2 ∑ i = 1 N ∑ j = 1 N α i α j y i y j ( x i ⋅ x j ) − ∑ i = 1 N α i \\mathop \{\\min \}\\limits\_\\alpha \\frac\{1\}\{2\}\\sum\\limits\_\{i = 1\}^N \{\\sum\\limits\_\{j = 1\}^N \{ \{\\alpha \_i\}\{\\alpha \_j\}\{y\_i\}\{y\_j\}\\left( \{ \{x\_i\} \\cdot \{x\_j\}\} \\right)\} - \\sum\\limits\_\{i = 1\}^N \{ \{\\alpha \_i\}\} \} αmin21i=1∑Nj=1∑Nαiαjyiyj(xi⋅xj)−i=1∑Nαi  
 s . t . ∑ i = 1 N α i y i = 0 s.t.\{\\rm\{ \}\}\\sum\\limits\_\{i = 1\}^N \{ \{\\alpha \_i\}\{y\_i\} = 0\} s.t.i=1∑Nαiyi=0  
 α i ≥ 0 , i = 1 , 2 , . . . , N \{\\alpha \_i\} \\ge 0,i = 1,2,...,N αi≥0,i=1,2,...,N  
但是有时数据难免有偏差（比如数据点不允许在间隔内），如果不允许这些数据点在间隔内，我们很难找到理想的间隔来划分数据，于是就有了软间隔最大化，C作为一个惩罚参数，C值大则对误分类的惩罚增大，反之同理。  
 min ⁡ α 1 2 ∑ i = 1 N ∑ j = 1 N α i α j y i y j ( x i ⋅ x j ) − ∑ i = 1 N α i \{\\mathop \{\\min \}\\limits\_\\alpha \{1 \\over 2\}\\sum\\limits\_\{i = 1\}^N \{\\sum\\limits\_\{j = 1\}^N \{ \{\\alpha \_i\}\} \} \{\\alpha \_j\}\{y\_i\}\{y\_j\}\\left( \{ \{x\_i\} \\cdot \{x\_j\}\} \\right) - \\sum\\limits\_\{i = 1\}^N \{ \{\\alpha \_i\}\} \} αmin21i=1∑Nj=1∑Nαiαjyiyj(xi⋅xj)−i=1∑Nαi  
 s . t . ∑ i = 1 N α i y i = 0 \{s.t.\\sum\\limits\_\{i = 1\}^N \{ \{\\alpha \_i\}\} \{y\_i\} = 0\} s.t.i=1∑Nαiyi=0  
 0 ≤ α i ≤ C , i = 1 , 2 , . . . , N \{0 \\le \{\\alpha \_i\} \\le C,i = 1,2,...,N\} 0≤αi≤C,i=1,2,...,N  
可以看到软间隔最大化的公式主要差别在于 α \\alpha α的取值大小。  
到这步，我们需要求出 α \\alpha α来得到 w w w，如何求出 α \\alpha α通常是选择SMO算法。  
《统计学习方法》P124 7.4序列最小化优化算法开始专门为这部分进行了一个介绍。  
首先这个算法最主要的是想求出 α \\alpha α，有了 α \\alpha α我们就能得到 w w w和 b b b。  
那SMO的算法主要是分两步，P128有详细介绍，这里简单描述下，第一步，找出第一个 α \\alpha α，这个可以找到一个不满足KKT条件的，进行优化，第二步，找出能使得 ∣ E 1 − E 2 ∣ % \\left| \{ \{E\_1\} - \{E\_\{\\rm\{2\}\}\}\} \\right| ∣E1−E2∣最大的数，这里可以通过对 E 1 \{E\_1\} E1进行判断，如果 E 1 \{E\_1\} E1是正的，就选个最小的 E i \{E\_i\} Ei作为 E 2 \{E\_2\} E2，反之同理。

for i in range(N):  #外循环
                    alpha1old = alpha[i].copy()   #未更新的alpha,也就是alpha_old
                    y1 = labels[i]  #y1
                    data1 = dataSet[i]
                    g1 = self.calcG(alpha, labels, data1, dataSet, b)
                    alphaIndex1 = -1    #存储不满足KKT条件的alphaIndex1
                    if alpha1old == 0:   #判断是否满足KKT条件 (7.111)
                        if y1 * g1 < 1:
                            alphaIndex1 = i
                            flag = True
                    if alpha1old > 0 and alpha[i] < C:   #(7.112)
                        if y1 * g1 != 1:
                            alphaIndex1 = i
                            flag = True
                    if alpha1old == C:   #(7.1132)
                        if y1 * g1 <= 1:
                            alphaIndex1 = i
                            flag = True
                    if alphaIndex1 == -1:   #说明满足KKT条件，继续下一次循环来找alpha1
                        continue
                    E1 = g1 - y1    #(7.105)

那首先有个外循环，来遍历每一个 α \\alpha α来找第一个 α 1 o l d \{\\alpha\_1^\{old\}\} α1old，书本说的是选取违反KKT最严重的样本点，那这里做了简单处理，只要是违反的就行，并且算出 E 1 \{E\_1\} E1。  
其中E的计算方式为 E i = g ( x i ) − y i = ( ( ∑ j = 1 N α j y j K ( x j , x i ) ) + b ) − y i \{E\_i\} = g\\left( \{ \{x\_i\}\} \\right) - \{y\_i\} = \\left( \{\\left( \{\\sum\\limits\_\{j = 1\}^N \{ \{\\alpha \_j\}\{y\_j\}K\\left( \{ \{x\_j\},\{x\_i\}\} \\right)\} \} \\right) + b\} \\right) - \{y\_i\} Ei=g(xi)−yi=((j=1∑NαjyjK(xj,xi))\+b)−yi。

for j in range(N): #内循环
                        if i != j:  #相等就没法选了
                            yj = labels[j]
                            gj = self.calcG(alpha, labels, dataSet[j], dataSet, b)
                            Ej = gj - yj
                            if E1 > 0:  #说明要选最小的E2
                                if Ej < selectedE2:
                                    selectedE2 = Ej
                                    alphaIndex2 = j
                            else:
                                if Ej > selectedE2:
                                    selectedE2 = Ej
                                    alphaIndex2 = j

以上的代码就是寻找出第二个 α 2 o l d \{\\alpha\_2^\{old\}\} α2old。

L = 0  # P126末尾两段
                    H = 0
                    y2 = labels[alphaIndex2]
                    alpha2old = alpha[alphaIndex2].copy()
                    data2 = dataSet[alphaIndex2]
                    E2 = selectedE2
                    if (y1 == y2):  #alpha2取值范围必须限制在L<alpha2<H
                        L = np.maximum(0, alpha2old - alpha1old)  # L
                        H = np.minimum(C, C + alpha2old - alpha1old)  #H
                    else:
                        L = np.maximum(0, alpha2old + alpha1old - C)  # L
                        H = np.minimum(C, C + alpha2old + alpha1old)    #H
                    eta = self.calcK(data1, data1) + self.calcK(data2, data2) - 2 * self.calcK(data1, data2)
                    if eta == 0:    #没法选
                        continue
                    alpha2new = alpha2old + (y2 * (E1 - E2)) / eta
                    if (alpha2new > H):  # (7.108)
                        alpha[alphaIndex2] = H
                    elif (alpha2new < L):
                        alpha[alphaIndex2] = L
                    else:
                        alpha[alphaIndex2] = alpha2new
                    alpha1new = alpha1old * y1 * y2 * (alpha2old - alpha2new)   #(7.109)
                    alpha[alphaIndex1] = alpha1new

OK，有了两个 α \\alpha α，我们可以根据公式 α 2 n e w , u n c = α 2 o l d + y 2 ( E 1 − E 2 ) η \\alpha \_\{\\rm\{2\}\}^\{new,unc\} = \\alpha \_2^\{old\} + \\frac\{ \{ \{y\_2\}\\left( \{ \{E\_1\} - \{E\_2\}\} \\right)\}\}\{\\eta \} α2new,unc=α2old\+ηy2(E1−E2)(7.106)，其中来得出一个未经处理的候选 α \\alpha α，这个 α \\alpha α由于有限制范围，需要处理下。  
具体处理方法为(7.108)： α 2 n e w = \{ H , α 2 n e w , u n c > H α 2 n e w , u n c , L ≤ α 2 n e w , u n c ≤ H L , α 2 n e w , u n c < L \\alpha \_2^\{new\} = \\left\\\{ \\begin\{array\}\{l\} H,\{\\rm\{ \}\}\\alpha \_2^\{new,unc\} > H\\\\ \\alpha \_2^\{new,unc\},L \\le \\alpha \_2^\{new,unc\} \\le H\\\\ L,\{\\rm\{ \}\}\\alpha \_2^\{new,unc\} < L \\end\{array\} \\right. α2new=⎩⎨⎧H,α2new,unc>Hα2new,unc,L≤α2new,unc≤HL,α2new,unc<L  
有了 α 2 n e w \\alpha \_2^\{new\} α2new之后，, α 1 n e w = α 1 o l d + y 1 y 2 ( α 2 o l d − α 2 n e w ) \\alpha \_1^\{new\} = \\alpha \_1^\{old\} + \{y\_1\}\{y\_2\}\\left( \{\\alpha \_2^\{old\} - \\alpha \_2^\{new\}\} \\right) α1new=α1old\+y1y2(α2old−α2new)

b1new = -E1 - y1 * self.calcK(data1, data1) * (alpha1new - alpha1old) - y2 * self.calcK(data2, data1) * (alpha2new - alpha2old) + b #(7.115)
                    b2new = -E2 - y1 * self.calcK(data1, data1) * (alpha1new - alpha1old) - y2 * self.calcK(data2, data2) * (alpha2new - alpha2old) + b #(7.116)
                    if (alpha1new > 0 and alpha1new < C):
                        b = b1new
                    else:
                        b = (b1new + b2new) / 2

同时，由于计算过程中需要用到 b b b，所以需要在此过程中更新 b b b  
公式在P130页(7.115)和(7.116)， b 1 n e w = − E 1 − y 1 K 11 ( α 1 n e w − α 1 o l d ) − y 2 K 21 ( α 2 n e w − α 2 o l d ) + b o l d b 2 n e w = − E 2 − y 1 K 12 ( α 1 n e w − α 1 o l d ) − y 2 K 21 ( α 2 n e w − α 2 o l d ) + b o l d \\begin\{array\}\{l\} b\_1^\{new\} = - \{E\_1\} - \{y\_1\}\{K\_\{11\}\}\\left( \{\\alpha \_\{\\rm\{1\}\}^\{new\} - \\alpha \_\{\\rm\{1\}\}^\{old\}\} \\right) - \{y\_2\}\{K\_\{21\}\}\\left( \{\\alpha \_2^\{new\} - \\alpha \_2^\{old\}\} \\right) + \{b^\{old\}\}\\\\ b\_2^\{new\} = - \{E\_2\} - \{y\_1\}\{K\_\{12\}\}\\left( \{\\alpha \_\{\\rm\{1\}\}^\{new\} - \\alpha \_\{\\rm\{1\}\}^\{old\}\} \\right) - \{y\_2\}\{K\_\{21\}\}\\left( \{\\alpha \_2^\{new\} - \\alpha \_2^\{old\}\} \\right) + \{b^\{old\}\}\\\\\\end\{array\} b1new=−E1−y1K11(α1new−α1old)−y2K21(α2new−α2old)\+boldb2new=−E2−y1K12(α1new−α1old)−y2K21(α2new−α2old)\+bold  
然后如果 α 1 n e w \\alpha\_1^\{new\} α1new和 α 2 n e w \\alpha\_2^\{new\} α2new同时满足 0 < α 1 n e w < C 0<\{\\alpha\_1^\{new\}\}<C 0<α1new<C，那么 b = b 1 n e w b=\{b\_1^\{new\}\} b=b1new或者 b = b 2 n e w b=\{b\_2^\{new\}\} b=b2new，如果不符合，则选这两个值的中值。

weights = np.dot(np.multiply(alpha, labels), dataSet)   #权重

至此， α \\alpha α全部都求出来了，我们通过 w ∗ = ∑ i = 1 N α i y i x i \{w^\*\} = \\sum\\limits\_\{i = 1\}^N \{ \{\\alpha \_i\}\{y\_i\}\{x\_i\}\} w∗=i=1∑Nαiyixi可以得到 w w w。 b b b就是刚才那个 b n e w b^\{new\} bnew了。  
这里有个小插曲， b b b值也可以在每次得到两个 α \\alpha α值后，通过 b ∗ = y j − ∑ i = 1 N α i y i x i x j \{b^\*\} = \{y\_j\} - \\sum\\limits\_\{i = 1\}^N \{ \{\\alpha \_i\}\{y\_i\}\{x\_i\}\{x\_j\}\} b∗=yj−i=1∑Nαiyixixj。  
这个代码在python3环境下运行。

def loadDataSet():  #加载文件
        data = list()
        labels = list()
        with open('testSet.txt') as f:
            lines = f.readlines()
            for line in lines:
                line = line.rstrip().split('\t')
                data.append([float(line[0]), float(line[1])])
                labels.append(float(line[-1]))
        return data, labels
    
    def sign(x):    #符号函数
        if x >= 0:
            return 1
        else:
            return -1
    
    class SVM:
        def train(self, dataSet, labels):   #训练并返回权重和偏置
            b = 0   #偏置
            C = 1 #惩罚系数
            flag = True    #检验是否全部都满足KKT条件
            maxIter = 100 #最大循环次数
            iter = 0
            N = len(dataSet)   #数据的行数
            M = len(dataSet[0])    #数据的列数，维数
            alpha = np.zeros(N)
            while iter < maxIter:
                print(iter)
                iter += 1
                flag = False
                for i in range(N):  #外循环
                    alpha1old = alpha[i].copy()   #未更新的alpha,也就是alpha_old
                    y1 = labels[i]  #y1
                    data1 = dataSet[i]
                    g1 = self.calcG(alpha, labels, data1, dataSet, b)
                    alphaIndex1 = -1    #存储不满足KKT条件的alphaIndex1
                    if alpha1old == 0:   #判断是否满足KKT条件 (7.111)
                        if y1 * g1 < 1:
                            alphaIndex1 = i
                            flag = True
                    if alpha1old > 0 and alpha[i] < C:   #(7.112)
                        if y1 * g1 != 1:
                            alphaIndex1 = i
                            flag = True
                    if alpha1old == C:   #(7.1132)
                        if y1 * g1 <= 1:
                            alphaIndex1 = i
                            flag = True
                    if alphaIndex1 == -1:   #说明满足KKT条件，继续下一次循环来找alpha1
                        continue
                    E1 = g1 - y1    #(7.105)
                    alphaIndex2 = -1
                    if E1 > 0: #正的话要找E2的最小值，反之同理
                        selectedE2 = np.inf
                    else:
                        selectedE2 = -np.inf
                    for j in range(N): #内循环
                        if i != j:  #相等就没法选了
                            yj = labels[j]
                            gj = self.calcG(alpha, labels, dataSet[j], dataSet, b)
                            Ej = gj - yj
                            if E1 > 0:  #说明要选最小的E2
                                if Ej < selectedE2:
                                    selectedE2 = Ej
                                    alphaIndex2 = j
                            else:
                                if Ej > selectedE2:
                                    selectedE2 = Ej
                                    alphaIndex2 = j
                    '''
                       此时应该选到了alpha2了
                   '''
                    L = 0  # P126末尾两段
                    H = 0
                    y2 = labels[alphaIndex2]
                    alpha2old = alpha[alphaIndex2].copy()
                    data2 = dataSet[alphaIndex2]
                    E2 = selectedE2
                    if (y1 == y2):  #alpha2取值范围必须限制在L<alpha2<H
                        L = np.maximum(0, alpha2old - alpha1old)  # L
                        H = np.minimum(C, C + alpha2old - alpha1old)  #H
                    else:
                        L = np.maximum(0, alpha2old + alpha1old - C)  # L
                        H = np.minimum(C, C + alpha2old + alpha1old)    #H
                    eta = self.calcK(data1, data1) + self.calcK(data2, data2) - 2 * self.calcK(data1, data2)
                    if eta == 0:    #没法选
                        continue
                    alpha2new = alpha2old + (y2 * (E1 - E2)) / eta
                    if (alpha2new > H):  # (7.108)
                        alpha[alphaIndex2] = H
                    elif (alpha2new < L):
                        alpha[alphaIndex2] = L
                    else:
                        alpha[alphaIndex2] = alpha2new
                    alpha1new = alpha1old * y1 * y2 * (alpha2old - alpha2new)   #(7.109)
                    alpha[alphaIndex1] = alpha1new
                    b1new = -E1 - y1 * self.calcK(data1, data1) * (alpha1new - alpha1old) - y2 * self.calcK(data2, data1) * (alpha2new - alpha2old) + b #(7.115)
                    b2new = -E2 - y1 * self.calcK(data1, data1) * (alpha1new - alpha1old) - y2 * self.calcK(data2, data2) * (alpha2new - alpha2old) + b #(7.116)
                    if (alpha1new > 0 and alpha1new < C):
                        b = b1new
                    else:
                        b = (b1new + b2new) / 2
            print(alpha)
            weights = np.dot(np.multiply(alpha, labels), dataSet)   #权重
            return weights, b
    
        def calcK(self, data1, data2):  #线性核函数，返回内积
            return np.dot(data1, data2)
    
        def calcG(self, alpha, labels, data, dataSet, b):   #计算g
            sum = 0
            for j in range(len(alpha)):
                sum += alpha[j] * labels[j] * self.calcK(data, dataSet[j]) #g(x)的计算
            return sum + b
    
    if __name__ == '__main__':
        dataSet, labels = loadDataSet()
        svm = SVM()
        weights, b = svm.train(dataSet, labels)
        print(weights, b)
        x = [1, 2]
        f = sign(np.dot(weights, x) + b)
        print(f)

以下是数据集，请保存为testSet.txt文件放在同目录下。

3.542485	1.977398	-1
    3.018896	2.556416	-1
    7.551510	-1.580030	1
    2.114999	-0.004466	-1
    8.127113	1.274372	1
    7.108772	-0.986906	1
    8.610639	2.046708	1
    2.326297	0.265213	-1
    3.634009	1.730537	-1
    0.341367	-0.894998	-1
    3.125951	0.293251	-1
    2.123252	-0.783563	-1
    0.887835	-2.797792	-1
    7.139979	-2.329896	1
    1.696414	-1.212496	-1
    8.117032	0.623493	1
    8.497162	-0.266649	1
    4.658191	3.507396	-1
    8.197181	1.545132	1
    1.208047	0.213100	-1
    1.928486	-0.321870	-1
    2.175808	-0.014527	-1
    7.886608	0.461755	1
    3.223038	-0.552392	-1
    3.628502	2.190585	-1
    7.407860	-0.121961	1
    7.286357	0.251077	1
    2.301095	-0.533988	-1
    -0.232542	-0.547690	-1
    3.457096	-0.082216	-1
    3.023938	-0.057392	-1
    8.015003	0.885325	1
    8.991748	0.923154	1
    7.916831	-1.781735	1
    7.616862	-0.217958	1
    2.450939	0.744967	-1
    7.270337	-2.507834	1
    1.749721	-0.961902	-1
    1.803111	-0.176349	-1
    8.804461	3.044301	1
    1.231257	-0.568573	-1
    2.074915	1.410550	-1
    -0.743036	-1.736103	-1
    3.536555	3.964960	-1
    8.410143	0.025606	1
    7.382988	-0.478764	1
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    2.387003	1.573131	-1

[Link 1]: https://github.com/FlameCharmander/MachineLearning