逻辑回归的梯度下降法的向量化（详细过程）

小咪咪 2022-08-29 05:49 164阅读 0赞

# 逻辑回归的梯度下降公式 #

逻辑回归的梯度下降公式：

θ j : = θ j − α 1 m ∑ i = 1 m ( h θ ( x ( i ) ) − y ( i ) ) x j ( i ) \\theta\_\{j\}:=\\theta\_\{j\}-\\alpha \\frac\{1\}\{m\} \\sum\_\{i=1\}^\{m\}\\left(h\_\{\\theta\}\\left(x^\{(i)\}\\right)-y^\{(i)\}\\right) x\_\{j\}^\{(i)\} θj:=θj−αm1i=1∑m(hθ(x(i))−y(i))xj(i)

其中：  
 h θ ( x ( i ) ) = g ( θ T x ( i ) ) = 1 1 + e − θ T x ( i ) h\_\{\\theta\}(x^\{(i)\})=g\\left(\\theta^T x^\{(i)\}\\right)=\\frac\{1\}\{1+e^\{-\\theta^\{T\} x^\{(i)\}\}\} hθ(x(i))=g(θTx(i))=1\+e−θTx(i)1

向量化后的公式为：

θ : = θ − α m X T ( g ( X θ ) − y ⃗ ) \\theta:=\\theta-\\frac\{\\alpha\}\{m\} X^\{T\}(g(X \\theta)-\\vec\{y\}) θ:=θ−mαXT(g(Xθ)−y)

其中：

y ⃗ = ( y ( 1 ) y ( 2 ) ⋮ y ( m ) )         θ = ( θ 0 θ 1 ⋮ θ n )        X = \[ x 0 ( 1 ) x 1 ( 1 ) ⋯ x n ( 1 ) x 0 ( 2 ) x 1 ( 2 ) ⋯ x n ( 2 ) ⋮ ⋮ x 0 ( m ) x 1 ( m ) ⋯ x n ( m ) \] m × ( n + 1 ) \\vec\{y\}=\\left(\\begin\{array\}\{c\} y^\{(1)\} \\\\ y^\{(2)\} \\\\ \\vdots \\\\ y^\{(m)\} \\end\{array\}\\right)~~~~~~~\\theta=\\left(\\begin\{array\}\{c\} \\theta\_\{0\} \\\\ \\theta\_\{1\} \\\\ \\vdots \\\\ \\theta\_\{n\} \\end\{array\}\\right)~~~~~~X=\\left\[\\begin\{array\}\{cccc\} x\_\{0\}^\{(1)\} & x\_\{1\}^\{(1)\} & \\cdots & x\_\{n\}^\{(1)\} \\\\ x\_\{0\}^\{(2)\} & x\_\{1\}^\{(2)\} & \\cdots & x\_\{n\}^\{(2)\} \\\\ \\vdots & & &\\vdots\\\\ x\_\{0\}^\{(m)\} & x\_\{1\}^\{(m)\} & \\cdots & x\_\{n\}^\{(m)\} \\end\{array\}\\right\]\_\{m \\times(n+1)\} y=⎝⎜⎜⎜⎛y(1)y(2)⋮y(m)⎠⎟⎟⎟⎞       θ=⎝⎜⎜⎜⎛θ0θ1⋮θn⎠⎟⎟⎟⎞      X=⎣⎢⎢⎢⎢⎡x0(1)x0(2)⋮x0(m)x1(1)x1(2)x1(m)⋯⋯⋯xn(1)xn(2)⋮xn(m)⎦⎥⎥⎥⎥⎤m×(n\+1)

X θ = \[ θ 0 x 0 ( 1 ) + θ 1 x 1 ( 1 ) + θ 2 x 2 ( 1 ) + ⋯ + θ n x n ( 1 ) θ 0 x 0 ( 2 ) + θ 1 x 2 ( 2 ) + θ 2 x 2 ( 2 ) + ⋯ + θ n x n ( 2 ) ⋯ θ 0 x 0 ( m ) + θ 1 x 1 ( m ) + θ 2 x 2 ( m ) + ⋯ + θ n x n ( m ) \]                     g ( X θ ) = \[ h θ ( x ( 1 ) ) h θ ( x ( 2 ) ) ⋯ h θ ( x ( m ) ) \] X \\theta=\\left\[\\begin\{array\}\{c\} \\theta\_\{0\} x\_\{0\}^\{(1)\}+\\theta\_\{1\} x\_\{1\}^\{(1)\}+\\theta\_\{2\} x\_\{2\}^\{(1)\}+\\cdots+\\theta\_\{n\} x\_\{n\}\{ \}^\{(1)\} \\\\ \\theta\_\{0\} x\_\{0\}^\{(2)\}+\\theta\_\{1\} x\_\{2\}^\{(2)\}+\\theta\_\{2\} x\_\{2\}^\{(2)\}+\\cdots+\\theta\_\{n\} x\_\{n\}^\{(2)\} \\\\ \\cdots \\\\ \\theta\_\{0\} x\_\{0\}^\{(m)\}+\\theta\_\{1\} x\_\{1\}^\{(m)\}+\\theta\_\{2\} x\_\{2\}^\{(m)\}+\\cdots+\\theta\_\{n\} x\_\{n\}^\{(m)\} \\end\{array\}\\right\]~~~~~~~~~~~~~~~~~~~ g(X \\theta)=\\left\[\\begin\{array\}\{c\} h\_\{\\theta\}\\left(x^\{(1)\}\\right) \\\\ h\_\{\\theta\}\\left(x^\{(2)\}\\right) \\\\ \\cdots\\\\ h\_\\theta\\left(x^\{(m)\}\\right) \\end\{array\}\\right\] Xθ=⎣⎢⎢⎢⎡θ0x0(1)\+θ1x1(1)\+θ2x2(1)\+⋯\+θnxn(1)θ0x0(2)\+θ1x2(2)\+θ2x2(2)\+⋯\+θnxn(2)⋯θ0x0(m)\+θ1x1(m)\+θ2x2(m)\+⋯\+θnxn(m)⎦⎥⎥⎥⎤                   g(Xθ)=⎣⎢⎢⎡hθ(x(1))hθ(x(2))⋯hθ(x(m))⎦⎥⎥⎤

# 详细向量化过程 #

∑ i = 1 m ( h θ ( x ( i ) ) − y ( i ) ) x j ( i ) = \[ h θ ( x ( 1 ) ) − y ( 1 ) \] x j ( 1 ) + \[ h θ ( x ( 2 ) ) − y ( 2 ) \] x j ( 2 ) + ⋯ + \[ h θ ( x ( m ) ) − y ( m ) \] x j ( m ) = ( x j ( 1 ) , x j ( 2 ) , ⋯   , x j ( m ) ) ⋅ ( h θ ( x ( 1 ) ) − y ( 1 ) h θ ( x ( 2 ) ) − y ( 2 ) ⋮ h θ ( x ( m ) ) − y ( m ) ) = ( x j ( 1 ) , x j ( 2 ) , ⋯   , x j ( m ) ) ⋅ \[ ( h θ ( x ( 1 ) ) h θ ( x ( 2 ) ) ⋮ h θ ( x ( m ) ) ) − ( y ( 1 ) y ( 2 ) ⋮ y ( m ) ) \] = x j ⋅ \[ g ( X θ ) − y ⃗ \] \\begin\{aligned\} &\\sum\_\{i=1\}^\{m\}\\left(h\_\{\\theta\}\\left(x^\{(i)\}\\right)-y^\{(i)\}\\right) x\_\{j\}^\{(i)\} \\\\\\\\ =&\{\\left\[h\_\{\\theta\}\\left(x^\{(1)\}\\right)-y^\{(1)\}\\right\]x\_\{j\}^\{(1)\}+\\left\[h\_\{\\theta\}\\left(x^\{(2)\}\\right)-y^\{(2)\}\\right\] x\_\{j\}^\{(2)\}\} +\\cdots+\\left\[h\_\{\\theta\}\\left(x^\{(m)\}\\right)-y^\{(m)\}\\right\] x\_\{j\}^\{(m)\} \\\\\\\\ = &\\left(x\_\{j\}^\{(1)\}, x\_\{j\}^\{(2)\}, \\cdots, x\_\{j\}^\{(m)\}\\right) \\cdot\\left(\\begin\{array\}\{c\} h\_\{\\theta\}\\left(x^\{(1)\}\\right)-y^\{(1)\} \\\\ h\_\{\\theta\}\\left(x^\{(2)\}\\right)-y^\{(2)\} \\\\ \\vdots \\\\ h\_\{\\theta\}\\left(x^\{(m)\}\\right)-y^\{(m)\} \\end\{array\}\\right) \\\\\\\\ =& \\left(x\_\{j\}^\{(1)\}, x\_\{j\}^\{(2)\}, \\cdots, x\_\{j\}^\{(m)\}\\right)\\cdot\\left\[\\left(\\begin\{array\}\{c\} h\_\{\\theta\}\\left(x^\{(1)\}\\right) \\\\ h\_\{\\theta\}\\left(x^\{(2)\}\\right) \\\\ \\vdots \\\\ h\_\{\\theta\}\\left(x^\{(m)\}\\right) \\end\{array\}\\right)-\\left(\\begin\{array\}\{c\} y^\{(1)\} \\\\ y^\{(2)\} \\\\ \\vdots \\\\ y^\{(m)\} \\end\{array\}\\right)\\right\] \\\\\\\\ =& x\_\{j\} \\cdot\[g(X \\theta)-\\vec\{y\}\] \\end\{aligned\} ====i=1∑m(hθ(x(i))−y(i))xj(i)\[hθ(x(1))−y(1)\]xj(1)\+\[hθ(x(2))−y(2)\]xj(2)\+⋯\+\[hθ(x(m))−y(m)\]xj(m)(xj(1),xj(2),⋯,xj(m))⋅⎝⎜⎜⎜⎛hθ(x(1))−y(1)hθ(x(2))−y(2)⋮hθ(x(m))−y(m)⎠⎟⎟⎟⎞(xj(1),xj(2),⋯,xj(m))⋅⎣⎢⎢⎢⎡⎝⎜⎜⎜⎛hθ(x(1))hθ(x(2))⋮hθ(x(m))⎠⎟⎟⎟⎞−⎝⎜⎜⎜⎛y(1)y(2)⋮y(m)⎠⎟⎟⎟⎞⎦⎥⎥⎥⎤xj⋅\[g(Xθ)−y\]

则：  
 θ j : = θ j − α m x j \[ g ( X θ ) − y ⃗ \] \\theta\_\{j\}:=\\theta\_\{j\}-\\frac\{\\alpha\}\{m\}x\_\{j\}\[g(X \\theta)-\\vec\{y\}\] θj:=θj−mαxj\[g(Xθ)−y\]

\[ θ 0 θ 1 ⋮ θ n \] : = \[ θ 0 θ 1 ⋮ θ n \] − α m \[ x 0 x 1 ⋮ x n \] \[ g ( X θ ) − y ⃗ \] \\left\[\\begin\{array\}\{c\} \\theta\_\{0\} \\\\ \\theta\_\{1\} \\\\ \\vdots \\\\ \\theta\_\{n\} \\end\{array\}\\right\]:=\\left\[\\begin\{array\}\{c\} \\theta\_\{0\} \\\\ \\theta\_\{1\} \\\\ \\vdots \\\\ \\theta\_\{n\} \\end\{array\}\\right\]-\\frac\{\\alpha\}\{m\}\\left\[\\begin\{array\}\{c\} x\_\{0\} \\\\ x\_\{1\} \\\\ \\vdots \\\\ x\_\{n\} \\end\{array\}\\right\]\\left\[g\\left(X\\theta\\right)-\\vec\{y\}\\right\] ⎣⎢⎢⎢⎡θ0θ1⋮θn⎦⎥⎥⎥⎤:=⎣⎢⎢⎢⎡θ0θ1⋮θn⎦⎥⎥⎥⎤−mα⎣⎢⎢⎢⎡x0x1⋮xn⎦⎥⎥⎥⎤\[g(Xθ)−y\]

最终得：

θ : = θ − α m X T ( g ( X θ ) − y ⃗ ) \\theta:=\\theta-\\frac\{\\alpha\}\{m\} X^\{T\}(g(X \\theta)-\\vec\{y\}) θ:=θ−mαXT(g(Xθ)−y)