CS224n Assignment 1

https://web.stanford.edu/class/archive/cs/cs224n/cs224n.1174/syllabus.html 第一次作業筆記

Softmax

softmax常數不變性

softmax(x_i) = \frac{e^{x_i}}{\sum_j{e^{x_j}}}

\begin{align} (softmax(x + c))_{i}= \frac{e^{x_{i} + c}}{\sum_{j} e^{x_{j} + c}} = \frac{e^{x_{i}} \times e^{c}}{e^{c} \times \sum_{j} e^{x_{j}}} \nonumber \\ = \frac{e^{x_{i}} \times {e^{c}}}{{e^{c}} \times \sum_{j} e^{x_{j}}} = (softmax(x))_{i} \nonumber \end{align}

由于e^{x+y}=e^x * e^y,因此多余的e^c可以上下消除,于是:
這里

softmax(x) = softmax(x + c)

發現了一個Softmax非常好的性質,即使兩個數都很大比如 10001001,其結果與 12的結果相同,即其只關注數字之間的差,而不是差占的比例。

Python實現

之所以介紹Softmax常數不變性,是因為發現給定的測試用例非常大,直接計算e^x次方

import numpy as np


def softmax(x):
    orig_shape = x.shape

    if len(x.shape) > 1:
        # Matrix
        ### YOUR CODE HERE
        x_max = np.max(x, axis=1).reshape(x.shape[0], 1)
        x -= x_max
        exp_sum = np.sum(np.exp(x), axis=1).reshape(x.shape[0], 1)
        x = np.exp(x) / exp_sum 
        ### END YOUR CODE
    else:
        # Vector
        ### YOUR CODE HERE
        x_max = np.max(x)
        x -= x_max
        exp_sum = np.sum(np.exp(x))
        x = np.exp(x) / exp_sum
        ### END YOUR CODE
        #or:  x = (np.exp(x)/sum(np.exp(x)))   

    assert x.shape == orig_shape
    return x

def test_softmax_basic():
    """
    Some simple tests to get you started.
    Warning: these are not exhaustive.
    """
    print("Running basic tests...")
    test1 = softmax(np.array([1,2]))
    print(test1)
    ans1 = np.array([0.26894142,  0.73105858])
    assert np.allclose(test1, ans1, rtol=1e-05, atol=1e-06)

    test2 = softmax(np.array([[1001,1002],[3,4]]))
    print(test2)
    ans2 = np.array([
        [0.26894142, 0.73105858],
        [0.26894142, 0.73105858]])
    assert np.allclose(test2, ans2, rtol=1e-05, atol=1e-06)

    test3 = softmax(np.array([[-1001,-1002]]))
    print(test3)
    ans3 = np.array([0.73105858, 0.26894142])
    assert np.allclose(test3, ans3, rtol=1e-05, atol=1e-06)

    print("You should be able to verify these results by hand!\n")

if __name__ == "__main__":
    test_softmax_basic()

神經網絡基礎

梯度檢查

image

Sigmoid導數

定義\sigma(x)如下,發現\sigma(x) + \sigma(-x) = 1
\sigma(x) &= \frac{1}{1 + e^{-x}} = \frac{e^x}{e^x + 1}\\ \sigma(-x) &= \frac{1}{1 + e^{x}} = \frac{e^x+1}{e^x + 1} - \frac{e^x}{e^x + 1}=1-\sigma(x)

image

即: \sigma' = \sigma(x) \times (1-\sigma(x))

交叉熵定義

當使用交叉熵作為評價指標時,求梯度:

  • 已知: \hat{y} = softmax(\theta)
  • 交叉熵: CE(y,\hat{y}) = - \sum_i{y_i \times log(\hat{y_i})}

其中\boldsymbol{y}是指示變量,如果該類別和樣本的類別相同就是1,否則就是0。因為y一般為one-hot類型。

\hat{y_i} 表示每種類型的概率,概率已經過softmax計算。

對于交叉熵其實有多重定義的方式,但含義相同:

分別為:

二分類定義

\begin{align}J = ?[y\cdot log(p)+(1?y)\cdot log(1?p)]\end{align} \\

  • y——表示樣本的label,正類為1,負類為0
  • p——表示樣本預測為正的概率

多分類定義

\begin{align}J = -\sum_{c=1}^My_{c}\log(p_{c})\end{align} \\

  • y——指示變量(0或1),如果該類別和樣本的類別相同就是1,否則是0;
  • p——對于觀測樣本屬于類別c的預測概率。

但表示的意思都相同,交叉熵用于反映 分類正確時的概率情況

Softmax導數

進入解答:

  • 首先定義S_i和分子分母。

S_i = softmax(\theta) = \frac{f_i}{g_i} = \frac{e^{\theta_i}}{\sum^{k}_{k=1}e^{\theta_k}}

  • S_i\theta_j求導:

    \begin{align} \frac{\partial{S_i}}{\partial{\theta_j}} &= \frac{f_i'g_i - f_ig_i'}{g_i}\\ &=\frac{(e^{\theta_i})'\sum^{k}_{k=1}e^{\theta_k} - e^{\theta_i}(\sum^{k}_{k=1}e^{\theta_k})'}{(\sum^{k}_{k=1}e^{\theta_k})^2}\end{align}

注意: S_i分子是\theta_i ,分母是所有的\theta ,而求偏微的是\theta_j

  • 因此,根據i與j的關系,分為兩種情況:
  • i == j 時:
$f_i' = e^{\theta_i}$,$g_i' = e^{\theta_j}$ 



$\begin{align} \frac{\partial{S_i}}{\partial{\theta_j}} &=\frac{e^{\theta_i}\sum^{k}_{k=1}e^{\theta_k} - e^{\theta_i}e^{\theta_j}}{(\sum^{k}_{k=1}e^{\theta_k})^2} \\ &= \frac{e^{\theta_{i}}}{\sum_{k} e^{\theta_{k}}} \times \frac{\sum_{k} e^{\theta_{k}} – e^{\theta_{j}}}{\sum_{k} e^{\theta_{k}}} \nonumber \\  &= S_{i} \times (1 – S_{i})   \end{align}$
  • i \not= j時:
$f'_{i} = 0 $,$g'_{i} = e^{\theta_{j}}$ 



$\begin{align}  \frac{\partial{S_i}}{\partial{\theta_j}} &= \frac{0 – e^{\theta_{j}} e^{\theta_{i}}}{(\sum_{k} e^{\theta_{k}})^{2}}  \\&= – \frac{e^{\theta_{j}}}{\sum_{k} ^{\theta_{k}}} \times \frac{e^{\theta_{i}}}{\sum_{k} e^{\theta_{k}}} \\ &=-S_j \times S_i\end{align}$

交叉熵梯度

計算\frac{\partial{CE}}{\partial{\theta_i}} ,根據鏈式法則,\frac{\partial{CE}}{\partial{\theta_i}} = \frac{\partial{CE}}{\partial{S_i}}\frac{\partial{S_i}}{\partial{\theta_j}}

  • \hat{y} = softmax(\theta)

  • CE(y,\hat{y}) = - \sum_k{y_k log(\hat{y_k})}

$\begin{align}      \frac{\partial CE}{\partial \theta_{i}} &= – \sum_{k} y_{k} \frac{\partial log S_{k}}{\partial \theta_{i}}  \\&= – \sum_{k} y_{k} \frac{1}{S_{k}} \frac{\partial S_{k}}{\partial \theta_{i}} \\ &= – y_{i} (1 – S_{i}) – \sum_{k \ne i} y_{k} \frac{1}{S_{k}} (-S_{k} \times S_{i}) \\ &= – y_{i} (1 – S_{i}) + \sum_{k \ne i} y_{k} S_{i} \\ &= S_{i}(\sum_{k} y_{k}) – y_{i}\end{align}$

因為\sum_{k} y_{k}=1,所以\frac{\partial CE}{\partial \theta_{i}} = S_i - y_i = \hat{y} - y

反向傳播計算神經網絡梯度

根據題目給定的定義:

image

已知損失函數J = CEh = sigmoid(xW_1+b_1), \hat{y} = softmax(hW_2+b_2)

\frac{\partial{J}}{\partial{x}},\frac{\partial{J}}{\partial{W_2}},\frac{\partial{J}}{\partial{W1}},\frac{\partial{J}}{\partial{b2}},\frac{\partial{J}}{\partial{b_1}}

解答:

反向傳播,定義z_2 = hW_2 + b_2z_1 = xW_1 + b_1

對于輸出層\hat{y}來說,\hat{y}的輸入為 z_2 = hW_2+b_2,而輸出則為 \hat{y} = softmax(z_2)

上小節計算得到 CE(y,\hat{y}) = - \sum_k{y_k log(\hat{y_k})} 的梯度為 \frac{\partial CE}{\partial \theta_{i}} = \hat{y} - y,

可以使用 z_2 替代 \theta_i ,得到

  • \delta_1 = \frac{\partial{CE}}{\partial{z_2}} = \hat{y} - y

  • \begin{align} \delta_2 = \frac{\partial{CE}}{\partial{h}} = \frac{\partial{CE}}{\partial{z_2}} \frac{\partial{z_2}}{\partial{h}} = \delta_1W_2^T \end{align}

  • \begin{align}\delta_3 = \frac{\partial{CE}}{z_1} = \frac{\partial{CE}}{\partial{h}}\frac{\partial{h}}{\partial{z_1}} = \delta_2 \frac{\partial{h}}{\partial{z_1}}= \delta_2 \circ \sigma'(z_1)\end{align} # 推測這里使用點乘的原因是\delta_2經過計算后,應該是一個標量,而不是向量。

  • 于是得到:\frac{\partial{CE}}{\partial{x}}=\delta_3\frac{\partial{z_1}}{\partial{x}} = \delta_3W_1^T

與計算\frac{\partial{CE}}{\partial{x}}相似,計算

  • \frac{\partial{CE}}{\partial{W_2}} = \frac{\partial{CE}}{\partial{z_2}}\frac{\partial{z_2}}{\partial{W_2}}=\delta_1 \cdot h
  • \frac{\partial{CE}}{\partial{b_2}} = \frac{\partial{CE}}{\partial{z_2}}\frac{\partial{z_2}}{\partial{b_2}}=\delta_1
  • \frac{\partial{CE}}{\partial{W_1}} = \frac{\partial{CE}}{\partial{z_1}}\frac{\partial{z_1}}{\partial{W_1}}=\delta_3 \cdot x
  • \frac{\partial{CE}}{\partial{b_1}} = \frac{\partial{CE}}{\partial{z_1}}\frac{\partial{z_1}}{\partial{b_1}}=\delta_3

如果仍然對反向傳播有疑惑

參數數量

\begin{align} n_{W_{1}} &= D_{x} \times H \\ n_{b_{1}} &= H \\ n_{W_{2}} &= H \times D_{y} \\ n_{b_{2}} &= D_{y} \\ N &= (D_{x} \times H) + H + (H \times D_{y}) + D_{y} \\ &=(D_x+1)\times H+(H+1)\times D_y \end{align}

代碼實現

  • sigmoid和對應的梯度
def sigmoid(x):
    s = 1 / (1 + np.exp(-x))
    return s

def sigmoid_grad(s):
    ds = s * (1-s)
    return ds
  • 梯度檢查
import numpy as np
import random


# First implement a gradient checker by filling in the following functions
def gradcheck_naive(f, x):
    """ Gradient check for a function f.

    Arguments:
    f -- a function that takes a single argument and outputs the
         cost and its gradients
    x -- the point (numpy array) to check the gradient at
    """

    rndstate = random.getstate()
    random.setstate(rndstate)
    fx, grad = f(x) # Evaluate function value at original point
    h = 1e-4        # Do not change this!

    # Iterate over all indexes in x
    it = np.nditer(x, flags=['multi_index'], op_flags=['readwrite'])
    while not it.finished:
        ix = it.multi_index
        print(ix)
        # Try modifying x[ix] with h defined above to compute
        # numerical gradients. Make sure you call random.setstate(rndstate)
        # before calling f(x) each time. This will make it possible
        # to test cost functions with built in randomness later.

        ### YOUR CODE HERE:
        x[ix] += h
        new_f1 = f(x)[0]
        x[ix] -= 2*h
        random.setstate(rndstate)
        new_f2 = f(x)[0]
        x[ix] += h
        numgrad = (new_f1 - new_f2) / (2 * h)
        ### END YOUR CODE

        # Compare gradients
        reldiff = abs(numgrad - grad[ix]) / max(1, abs(numgrad), abs(grad[ix]))
        if reldiff > 1e-5:
            print("Gradient check failed.")
            print("First gradient error found at index %s" % str(ix))
            print("Your gradient: %f \t Numerical gradient: %f" % (
                grad[ix], numgrad))
            return

        it.iternext() # Step to next dimension

    print("Gradient check passed!")

  • 反向傳播
  def forward_backward_prop(data, labels, params, dimensions):
      """
      Forward and backward propagation for a two-layer sigmoidal network
      Compute the forward propagation and for the cross entropy cost,
      and backward propagation for the gradients for all parameters.
      Arguments:
      data -- M x Dx matrix, where each row is a training example.
      labels -- M x Dy matrix, where each row is a one-hot vector.
      params -- Model parameters, these are unpacked for you.
      dimensions -- A tuple of input dimension, number of hidden units
                    and output dimension
      """
      ### Unpack network parameters (do not modify)
      ofs = 0
      Dx, H, Dy = (dimensions[0], dimensions[1], dimensions[2])
      W1 = np.reshape(params[ofs:ofs+ Dx * H], (Dx, H))
      ofs += Dx * H
      b1 = np.reshape(params[ofs:ofs + H], (1, H))
      ofs += H
      W2 = np.reshape(params[ofs:ofs + H * Dy], (H, Dy))
      ofs += H * Dy
      b2 = np.reshape(params[ofs:ofs + Dy], (1, Dy))
      ### YOUR CODE HERE: forward propagation
      h = sigmoid(np.dot(data,W1) + b1)
      yhat = softmax(np.dot(h,W2) + b2)
      ### END YOUR CODE
      ### YOUR CODE HERE: backward propagation
      cost = np.sum(-np.log(yhat[labels==1])) 
      
      d1 = (yhat - labels)
      gradW2 = np.dot(h.T, d1)
      gradb2 = np.sum(d1,0,keepdims=True)
      
      d2 = np.dot(d1,W2.T)
      # h = sigmoid(z_1)
      d3 = sigmoid_grad(h) * d2
      gradW1 = np.dot(data.T,d3)
      gradb1 = np.sum(d3,0)
      
      ### END YOUR CODE
      ### Stack gradients (do not modify)
      grad = np.concatenate((gradW1.flatten(), gradb1.flatten(),
          gradW2.flatten(), gradb2.flatten()))
      return cost, grad

word2vec

關于詞向量的梯度

在以softmax為假設函數的word2vec中

\begin{align} \hat{\boldsymbol{y}}_{o} = p(\boldsymbol{o} \vert \boldsymbol{c}) =\frac{exp(\boldsymbol{u}_{0}^{T} \boldsymbol{v}_{c})}{\sum\limits_{w=1}^{W} exp(\boldsymbol{u}_{w}^{T} \boldsymbol{v}_{c})} \end{align}

\boldsymbol{v}_{c}是中央單詞的詞向量

\boldsymbol{u}_{w} (w = 1,...,W) 是第 w個詞語的詞向量。

假設使用交叉熵作為損失函數, \boldsymbol{o} 為正確單詞 (one-hot向量的第 \boldsymbol{o}維為1),請推導損失函數關于\boldsymbol{v}_c的梯度。

提示:

\begin{align}J_{softmax-CE}(\boldsymbol{o}, \boldsymbol{v}_{c}, \boldsymbol{U}) = CE(\boldsymbol{y}, \hat{\boldsymbol{y}})\end{align}

其中\boldsymbol{U} = [\boldsymbol{u}_{1},\boldsymbol{u}_{1},\dots, \boldsymbol{u}_{W}]是所有詞向量構成的矩陣。

解答:

首先明確本題給定的模型是skip-gram ,通過給定中心詞,來發現周圍詞的。

定義z=U^T \cdot v_cU 表示所有詞向量組成的矩陣,而v_c 也表示的是一個詞向量。

hint: 如果兩個向量相似性越高,則乘積也就越大。想象一下余弦夾角,應該比較好明白。

因為U中所有的詞向量,都和v_c乘一下獲得z

z是干嘛用的呢? z內就有W個值,每個值表示和v_c 相似程度,通過這個相似度softmax選出最大值,然后與實際對比,進行交叉熵的計算。

已知: \frac{\partial z}{\partial v_c} = U\frac{\partial J}{\partial \boldsymbol{z}} = (\hat{\boldsymbol{y}} -\boldsymbol{y})

因此:\frac{\partial J}{\partial{v_c}} =\frac{\partial J}{\partial \boldsymbol{z}} \frac{\partial z}{\partial v_c} = U(\hat{\boldsymbol{y}} -\boldsymbol{y})


除了上述表示之外,還有另一種計算方法

image
image
image

[圖片上傳失敗...(image-53cc75-1557025564256)]

于是: \frac{\partial J}{\partial{v_c}} = -u_i + \sum^{W}_{w=1}\hat{y_w}u_w

仔細觀察這兩種寫法,會發現其實是一回事,都是 觀察與期望的差(\hat{y} - y)。

推導lookup-table梯度

與詞向量相似

\frac{\partial J}{\partial{U}} =\frac{\partial J}{\partial \boldsymbol{z}} \frac{\partial z}{\partial U} = v_c(\hat{\boldsymbol{y}} -\boldsymbol{y})^{T}

負采樣時的梯度推導

假設進行負采樣,樣本數為\boldsymbol{K},正確答案為\boldsymbol{o},那么有o \notin \{1,...,K\}。負采樣損失函數定義如下:

\begin{align}J_{neg-sample}(\boldsymbol{o}, \boldsymbol{v}_{c}, \boldsymbol{U}) = ? log(\sigma(\boldsymbol{u}_{o}^{T} \boldsymbol{v}_{c})) - \sum\limits_{k=1}^{K} log(\sigma(-\boldsymbol{u}_{k}^{T} \boldsymbol{v}_{c})) \end{align}

其中:

\begin{align}\sigma(x) = \frac{1}{1 + e^{-x}} \nonumber \\= \frac{e^{x}}{1 + e^{x}} \nonumber \end{align}

\frac{\partial}{\partial x} \sigma(x) = \sigma(x) \times (1 - \sigma(x))

解答:

首先說明一下,J_{neg-sample}從哪里來的,參考note1 第11頁,會有一個非常詳細的解釋。

\begin{align} \frac{\partial J}{\partial v_c}&=\left(\sigma(u_o^Tv_c)-1\right)u_o-\sum_{k=1}^K\left(\sigma(-u_k^Tv_c)-1\right)u_k\\ \frac{\partial J}{\partial u_o}&=\left(\sigma(u_o^Tv_c)-1\right)v_c\\ \frac{\partial J}{\partial u_k}&=-\left(\sigma(-u_k^Tv_c)-1\right)v_c\\ \end{align}

全部梯度

推導窗口半徑m的上下文[word_{c?m} ,...,word_{c?1} ,word_{c} ,word_{c+1} ,...,word_{c+m} ]時,skip-gram 和 CBOW的損失函數F(\boldsymbol{o}, \boldsymbol{v}_{c}) (\boldsymbol{o} 是正確答案的詞向量)或說J_{softmax-CE}(\boldsymbol{o},\boldsymbol{v}_{c},\dots)J_{neg-sample}(\boldsymbol{o},\boldsymbol{v}_{c},\dots) 關于每個詞向量的梯度。

對于skip-gram來講,c的上下文對應的損失函數是:

\begin{align} J_{skip-gram}(word_{c-m \dots c+m})= \sum\limits_{-m \leq j \leq m, j \ne 0} F(\boldsymbol{w}_{c+j}, \boldsymbol{v}_{c})\end{align}

這里?\boldsymbol{w}_{c+j} 是離中心詞距離j的那個單詞。

而CBOW稍有不同,不使用中心詞\boldsymbol{v}_{c}而使用上下文詞向量的和\hat{\boldsymbol{v}}作為輸入去預測中心詞:

\begin{align} \hat{\boldsymbol{v}} = \sum\limits_{-m \leq j \leq m, j \ne 0} \boldsymbol{v}_{c+j}\end{align}

然后CBOW的損失函數是:

\begin{align} J_{CBOW}(word_{c-m \dots c+m})= F(\boldsymbol{w}_{c}, \hat{\boldsymbol{v}})\end{align}

解答:

根據前面的推導,知道如何得到梯度\begin{align} \frac{\partial J}{\partial \boldsymbol{v_c}} = \boldsymbol{U}^{T} (\hat{\boldsymbol{y}} - \boldsymbol{y}) \nonumber \end{align}\begin{align} \frac{\partial J}{\partial \boldsymbol{U}} = \boldsymbol{v}_{c} (\hat{\boldsymbol{y}} - \boldsymbol{y})^{T} \nonumber \end{align}。那么所求的梯度可以寫作:

skip-gram

\begin{align} \frac{J_{skip-gram}(word_{c-m \dots c+m})}{\partial \boldsymbol{U}} &= \sum\limits_{-m \leq j \leq m, j \ne 0} \frac{\partial F(\boldsymbol{w}_{c+j}, \boldsymbol{v}_{c})}{\partial \boldsymbol{U}} \nonumber \\ \frac{J_{skip-gram}(word_{c-m \dots c+m})}{\partial \boldsymbol{v}_{c}} &= \sum\limits_{-m \leq j \leq m, j \ne 0} \frac{\partial F(\boldsymbol{w}_{c+j}, \boldsymbol{v}_{c})}{\partial \boldsymbol{v}_{c}} \nonumber \\ \frac{J_{skip-gram}(word_{c-m \dots c+m})}{\partial \boldsymbol{v}_{j}} &= 0, \forall j\ne c \nonumber\end{align}

CBOW

\begin{align} \frac{J_{CBOW}(word_{c-m \dots c+m})}{\partial \boldsymbol{U}}& = \frac{\partial F(\boldsymbol{w}_{c}, \hat{\boldsymbol{v}})}{\partial \boldsymbol{U}} \nonumber \\ \frac{J_{CBOW}(word_{c-m \dots c+m})}{\partial \boldsymbol{v}_{j}} &= \frac{\partial F(\boldsymbol{w}_{c}, \hat{\boldsymbol{v}})}{\partial \hat{\boldsymbol{v}}}, \forall (j \ne c) \in \{c-m \dots c+m\} \nonumber \\ \frac{J_{CBOW}(word_{c-m \dots c+m})}{\partial \boldsymbol{v}_{j}} &= 0, \forall (j \ne c) \notin \{c-m \dots c+m\} \nonumber\end{align}

補充部分:

  • 矩陣的每個行向量的長度歸一化

     x = x/np.linalg.norm(x,axis=1,keepdims=True)
    
  • 在斯坦福情感樹庫上訓練詞向量

    直接運行q3_run即可

    image

情感分析

特征向量

最簡單的特征選擇方法就是取所有詞向量的平均

    sentence_index = [tokens[i] for i in sentence]
    for index in sentence_index:
        sentVector += wordVectors[index, :]

    sentVector /= len(sentence)

正則化

values = np.logspace(-4, 2, num=100, base=10)

調參

bestResult = max(results, key= lambda x: x['dev'])

懲罰因子對效果的影響

image

confusion matrix

關聯性排序的一個東西,對角線上的元素越多,預測越準確。

image
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