原理
我们需要的是后验概率分布
%E2%80%8B#card=math&code=P%28Y%7CX%29%E2%80%8B)，有了这个之后，我们根据预测样本的X，就可以预测出Y。 但是这没办法直接学习，所以我们转而通过训练数据集学习联合概率分布
%E2%80%8B#card=math&code=P%28X%2CY%29%E2%80%8B)，如果我们知道联合分布%E2%80%8B#card=math&code=P%28X%2CY%29%E2%80%8B)，只需要计算
%E2%80%8B#card=math&code=P%28X%29%E2%80%8B)，就可以通过条件概率公式，得到%E2%80%8B#card=math&code=P%28Y%7CX%29%E2%80%8B)。 但是直接学习联合概率分布也是没办法的，所以，我们进一步转化为学习
%E2%80%8B#card=math&code=P%28Y%29%E2%80%8B)和
%E2%80%8B#card=math&code=P%28X%7CY%29%E2%80%8B)，然后通过贝叶斯定理，求出%E2%80%8B#card=math&code=P%28X%2CY%29%E2%80%8B)。
所以整体过程表述如下：
#card=math&code=%C2%A0P%28Y%7CX%29)<—— %EF%BC%8CP(X)#card=math&code=P%28X%2CY%29%EF%BC%8CP%28X%29)<—— %EF%BC%8CP(Y)%EF%BC%8CP(X%7CY)#card=math&code=P%28X%29%EF%BC%8CP%28Y%29%EF%BC%8CP%28X%7CY%29)
下面我们来看一下学习过程，已知：
- 输入空间：
为n维向量集合，也就是每个样本有n个特征 - 输出空间：
，表示有k个分类 - X是定义在输入空间上的随机向量，Y是定义在输出空间上的随机变量，%E2%80%8B#card=math&code=P%28X%2CY%29%E2%80%8B)是X，Y的联合概率分布
- 训练数据集
%2C(x_2%2Cy_2)%2C%E2%80%A6(x_N%2Cy_N)%5C%7D%E2%80%8B#card=math&code=T%3D%5C%7B%28x_1%2Cy_1%29%2C%28x_2%2Cy_2%29%2C%E2%80%A6%28x_N%2Cy_N%29%5C%7D%E2%80%8B)由%E2%80%8B#card=math&code=P%28X%2CY%29%E2%80%8B)独立同分布产生
%26%3D%5Cfrac%7BP(X%3Dx%2CY%3Dck)%7D%7BP(X%3Dx)%7D%20%26(1)%20%5C%5C%20%0A%26%3D%20%5Cfrac%7BP(X%3Dx%7CY%3Dc_k)P(Y%3Dc_k)%7D%7B%5Csum%7Bi%7D%20P(X%3Dx%7CY%3Dci)P(Y%3Dc_i)%7D%20%26(2)%5C%5C%20%0A%26%3D%20%5Cfrac%7BP(X%5E%7B(1)%7D%3Dx%5E%7B(1)%7D%2C…%2CX%5E%7B(n)%7D%3Dx%5E%7B(n)%7D%7CY%3Dc_k)P(Y%3Dc_k)%7D%7B%5Csum%7Bi%7D%20P(X%5E%7B(1)%7D%3Dx%5E%7B(1)%7D%2C…%2CX%5E%7B(n)%7D%3Dx%5E%7B(n)%7D%7CY%3Dci)P(Y%3Dc_i)%7D%20%26(3)%20%5C%5C%0A%26%3D%20%5Cfrac%7BP(Y%3Dc_k)%5Cprod_j%20P(X%3Dx%5E%7B(j)%7D%7CY%3Dc_k)%7D%7B%5Csum%7Bi%7D%20P(Y%3Dci)%5Cprod_j%20P(X%3Dx%5E%7B(j)%7D%7CY%3Dc_i)%7D%20%26(4)%20%5C%5C%0A%5Cend%7Baligned%7D%0A#card=math&code=%5Cbegin%7Baligned%7D%0AP%28Y%3Dc_k%7CX%3Dx%29%26%3D%5Cfrac%7BP%28X%3Dx%2CY%3Dc_k%29%7D%7BP%28X%3Dx%29%7D%20%26%281%29%20%5C%5C%20%0A%26%3D%20%5Cfrac%7BP%28X%3Dx%7CY%3Dc_k%29P%28Y%3Dc_k%29%7D%7B%5Csum%7Bi%7D%20P%28X%3Dx%7CY%3Dci%29P%28Y%3Dc_i%29%7D%20%26%282%29%5C%5C%20%0A%26%3D%20%5Cfrac%7BP%28X%5E%7B%281%29%7D%3Dx%5E%7B%281%29%7D%2C…%2CX%5E%7B%28n%29%7D%3Dx%5E%7B%28n%29%7D%7CY%3Dc_k%29P%28Y%3Dc_k%29%7D%7B%5Csum%7Bi%7D%20P%28X%5E%7B%281%29%7D%3Dx%5E%7B%281%29%7D%2C…%2CX%5E%7B%28n%29%7D%3Dx%5E%7B%28n%29%7D%7CY%3Dci%29P%28Y%3Dc_i%29%7D%20%26%283%29%20%5C%5C%0A%26%3D%20%5Cfrac%7BP%28Y%3Dc_k%29%5Cprod_j%20P%28X%3Dx%5E%7B%28j%29%7D%7CY%3Dc_k%29%7D%7B%5Csum%7Bi%7D%20P%28Y%3Dc_i%29%5Cprod_j%20P%28X%3Dx%5E%7B%28j%29%7D%7CY%3Dc_i%29%7D%20%26%284%29%20%5C%5C%0A%5Cend%7Baligned%7D%0A)
上式（3）->（4）步，是假设
%7D%2C%E2%80%A6%2CX%5E%7B(n)%7D#card=math&code=X%5E%7B%281%29%7D%2C%E2%80%A6%2CX%5E%7B%28n%29%7D)彼此独立，即
%7D%2C%E2%80%A6%2CX%5E%7B(n)%7D)%3D%5Cprod%7Bi%3D1%7D%5En%20P(X_i)#card=math&code=%5Cdisplaystyle%20P%28X%5E%7B%281%29%7D%2C%E2%80%A6%2CX%5E%7B%28n%29%7D%29%3D%5Cprod%7Bi%3D1%7D%5En%20P%28Xi%29)，这是个较强的假设，朴素贝叶斯也因此得名。 如果不假设独立的话，那么每个
%7D#card=math&code=x%5E%7B%28i%29%7D)表示样本的第i个特征，假设可能取值有
个，而Y有k个，所以参数个数为，这是非常大的，是不可直接估计的。
所以，问题转化为，求最大的
%E2%80%8B#card=math&code=P%28Y%3Dc_k%7CX%3Dx%29%E2%80%8B)，此时的类别
，就是我们要预测的类别：
%3D%5Carg%5Cmax%7Bc_k%7D%20%5Cfrac%7BP(Y%3Dc_k)%5Cprod_j%20P(X%3Dx%5E%7B(j)%7D%7CY%3Dc_k)%7D%7B%5Csum%7Bi%7D%20P(Y%3Dci)%5Cprod_j%20P(X%3Dx%5E%7B(j)%7D%7CY%3Dc_i)%7D%0A#card=math&code=y%3Df%28x%29%3D%5Carg%5Cmax%7Bck%7D%20%5Cfrac%7BP%28Y%3Dc_k%29%5Cprod_j%20P%28X%3Dx%5E%7B%28j%29%7D%7CY%3Dc_k%29%7D%7B%5Csum%7Bi%7D%20P%28Y%3Dc_i%29%5Cprod_j%20P%28X%3Dx%5E%7B%28j%29%7D%7CY%3Dc_i%29%7D%0A)
对于
这个样本来说，分母
#card=math&code=P%28X%3Dx%29)都是一样的，所以我们只需要比较分子#card=math&code=P%28Y%3Dc_k%7CX%3Dx%29)，当
时的大小，并不需要具体的值，所以我们可以忽略分母，只求：
%3D%5Carg%5Cmax%7Bc_k%7D%20P(Y%3Dc_k)%5Cprod_j%20P(X%3Dx%5E%7B(j)%7D%7CY%3Dc_k)%0A#card=math&code=y%3Df%28x%29%3D%5Carg%5Cmax%7Bc_k%7D%20P%28Y%3Dc_k%29%5Cprod_j%20P%28X%3Dx%5E%7B%28j%29%7D%7CY%3Dc_k%29%0A)
得到的最大的，就是该样本的预测类别。
后验概率最大化的含义
为什么书中说”朴素贝叶斯法将实例分到后验概率最大的类中，这等价于期望风险最小化？”
极大似然法求
自己理解
根据上面的分析，我们需要首先计算：
%E2%80%8B#card=math&code=P%28Y%3Dc_k%29%E2%80%8B)
就是在训练集中计算每个分类的频率
%7D%7CY%3Dc_k)#card=math&code=P%28X%3Dx_j%5E%7B%28i%29%7D%7CY%3Dc_k%29)
联合分布，即算每个特征的每个取值，在每个分类下出现的频率- 利用1，2，就可以进行预测了。
正规的写法
#card=math&code=I%28x%29)叫做指示函数，实际上就是用来计数的，当满足x时，
，例如
#card=math&code=I%28y_i%3Dc_k%29)的意思就是当
取值等于
时，记为1。再结合求和使用，就是计算
的样本数。
- 根据给定的实例预测
%20%5Cdisplaystyle%5Cprod%7Bj%3D1%7D%5En%20P(X%5E%7B(j)%7D%3Dx%5E%7B(j)%7D%7CY%3Dc_k)%2C%20%5Cqquad%20k%3D1%2C2%2C…%2CK%0A#card=math&code=P%28Y%3Dc_k%29%20%5Cdisplaystyle%5Cprod%7Bj%3D1%7D%5En%20P%28X%5E%7B%28j%29%7D%3Dx%5E%7B%28j%29%7D%7CY%3Dc_k%29%2C%20%5Cqquad%20k%3D1%2C2%2C…%2CK%0A)
- 选择最大的概率的类
%20%5Cdisplaystyle%5Cprod%7Bj%3D1%7D%5En%20P(X%5E%7B(j)%7D%3Dx%5E%7B(j)%7D%7CY%3Dc_k)%2C%20%5Cqquad%20k%3D1%2C2%2C…%2CK%0A#card=math&code=y%3D%5Carg%5Cmax%7Bck%7DP%28Y%3Dc_k%29%20%5Cdisplaystyle%5Cprod%7Bj%3D1%7D%5En%20P%28X%5E%7B%28j%29%7D%3Dx%5E%7B%28j%29%7D%7CY%3Dc_k%29%2C%20%5Cqquad%20k%3D1%2C2%2C…%2CK%0A)
示例
给定训练数据如下，%7D%2CX%5E%7B(2)%7D#card=math&code=X%5E%7B%281%29%7D%2CX%5E%7B%282%29%7D)表示特征，Y表示类别标签：
|
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
11 |
12 |
13 |
14 |
15 |
| %7D#card=math&code=X%5E%7B%281%29%7D) |
1 |
1 |
1 |
1 |
1 |
2 |
2 |
2 |
2 |
2 |
3 |
3 |
3 |
3 |
3 |
 %7D#card=math&code=X%5E%7B%282%29%7D) |
S |
M |
M |
S |
S |
S |
M |
M |
L |
L |
L |
M |
M |
L |
L |
| Y |
-1 |
-1 |
1 |
1 |
-1 |
-1 |
-1 |
1 |
1 |
1 |
1 |
1 |
1 |
1 |
-1 |
求
%5ET#card=math&code=x%3D%282%2C%20S%29%5ET)的类别？
求解:
- 计算先验概率和条件概率
%3D%5Cfrac%7B9%7D%7B15%7D%2CP(Y%3D-1)%3D%5Cfrac%7B6%7D%7B15%7D%20%5C%5C%0A%26P(X%5E%7B(1)%7D%3D1%7CY%3D1)%3D%5Cfrac%7B2%7D%7B9%7D%2CP(X%5E%7B(1)%7D%3D2%7CY%3D1)%3D%5Cfrac%7B3%7D%7B9%7D%2CP(X%5E%7B(1)%7D%3D3%7CY%3D1)%3D%5Cfrac%7B4%7D%7B9%7D%20%5C%5C%0A%26P(X%5E%7B(2)%7D%3DS%7CY%3D1)%3D%5Cfrac%7B1%7D%7B9%7D%2CP(X%5E%7B(2)%7D%3DM%7CY%3D1)%3D%5Cfrac%7B4%7D%7B9%7D%2CP(X%5E%7B(2)%7D%3DL%7CY%3D1)%3D%5Cfrac%7B4%7D%7B9%7D%20%5C%5C%0A%26P(X%5E%7B(1)%7D%3D1%7CY%3D-1)%3D%5Cfrac%7B3%7D%7B6%7D%2CP(X%5E%7B(1)%7D%3D2%7CY%3D-1)%3D%5Cfrac%7B2%7D%7B6%7D%2CP(X%5E%7B(1)%7D%3D3%7CY%3D-1)%3D%5Cfrac%7B1%7D%7B6%7D%20%5C%5C%0A%26P(X%5E%7B(2)%7D%3DS%7CY%3D-1)%3D%5Cfrac%7B3%7D%7B6%7D%2CP(X%5E%7B(2)%7D%3DM%7CY%3D-1)%3D%5Cfrac%7B2%7D%7B6%7D%2CP(X%5E%7B(2)%7D%3DL%7CY%3D-1)%3D%5Cfrac%7B1%7D%7B6%7D%20%0A%5Cend%7Baligned%7D%0A#card=math&code=%5Cbegin%7Baligned%7D%0A%26P%28Y%3D1%29%3D%5Cfrac%7B9%7D%7B15%7D%2CP%28Y%3D-1%29%3D%5Cfrac%7B6%7D%7B15%7D%20%5C%5C%0A%26P%28X%5E%7B%281%29%7D%3D1%7CY%3D1%29%3D%5Cfrac%7B2%7D%7B9%7D%2CP%28X%5E%7B%281%29%7D%3D2%7CY%3D1%29%3D%5Cfrac%7B3%7D%7B9%7D%2CP%28X%5E%7B%281%29%7D%3D3%7CY%3D1%29%3D%5Cfrac%7B4%7D%7B9%7D%20%5C%5C%0A%26P%28X%5E%7B%282%29%7D%3DS%7CY%3D1%29%3D%5Cfrac%7B1%7D%7B9%7D%2CP%28X%5E%7B%282%29%7D%3DM%7CY%3D1%29%3D%5Cfrac%7B4%7D%7B9%7D%2CP%28X%5E%7B%282%29%7D%3DL%7CY%3D1%29%3D%5Cfrac%7B4%7D%7B9%7D%20%5C%5C%0A%26P%28X%5E%7B%281%29%7D%3D1%7CY%3D-1%29%3D%5Cfrac%7B3%7D%7B6%7D%2CP%28X%5E%7B%281%29%7D%3D2%7CY%3D-1%29%3D%5Cfrac%7B2%7D%7B6%7D%2CP%28X%5E%7B%281%29%7D%3D3%7CY%3D-1%29%3D%5Cfrac%7B1%7D%7B6%7D%20%5C%5C%0A%26P%28X%5E%7B%282%29%7D%3DS%7CY%3D-1%29%3D%5Cfrac%7B3%7D%7B6%7D%2CP%28X%5E%7B%282%29%7D%3DM%7CY%3D-1%29%3D%5Cfrac%7B2%7D%7B6%7D%2CP%28X%5E%7B%282%29%7D%3DL%7CY%3D-1%29%3D%5Cfrac%7B1%7D%7B6%7D%20%0A%5Cend%7Baligned%7D%0A)
- 计算预测结果
%3DP(Y%3D1)P(X%5E%7B(1)%7D%3D2%7CY%3D1)P(X%5E%7B(2)%7D%3DS%7CY%3D1)%3D%5Cdfrac%7B9%7D%7B15%7D%20%5Ccdot%20%5Cdfrac%7B3%7D%7B9%7D%20%5Ccdot%20%5Cdfrac%7B1%7D%7B9%7D%20%3D%20%5Cdfrac%7B1%7D%7B45%7D%5C%5C%0A%26P(X%3Dx%7CY%3D-1)%3DP(Y%3D-1)P(X%5E%7B(1)%7D%3D2%7CY%3D-1)P(X%5E%7B(2)%7D%3DS%7CY%3D-1)%3D%5Cdfrac%7B6%7D%7B15%7D%20%5Ccdot%20%5Cdfrac%7B2%7D%7B6%7D%20%5Ccdot%20%5Cdfrac%7B3%7D%7B6%7D%20%3D%20%5Cdfrac%7B1%7D%7B15%7D%0A%5Cend%7Baligned%7D%0A#card=math&code=%5Cbegin%7Baligned%7D%0A%26P%28X%3Dx%7CY%3D1%29%3DP%28Y%3D1%29P%28X%5E%7B%281%29%7D%3D2%7CY%3D1%29P%28X%5E%7B%282%29%7D%3DS%7CY%3D1%29%3D%5Cdfrac%7B9%7D%7B15%7D%20%5Ccdot%20%5Cdfrac%7B3%7D%7B9%7D%20%5Ccdot%20%5Cdfrac%7B1%7D%7B9%7D%20%3D%20%5Cdfrac%7B1%7D%7B45%7D%5C%5C%0A%26P%28X%3Dx%7CY%3D-1%29%3DP%28Y%3D-1%29P%28X%5E%7B%281%29%7D%3D2%7CY%3D-1%29P%28X%5E%7B%282%29%7D%3DS%7CY%3D-1%29%3D%5Cdfrac%7B6%7D%7B15%7D%20%5Ccdot%20%5Cdfrac%7B2%7D%7B6%7D%20%5Ccdot%20%5Cdfrac%7B3%7D%7B6%7D%20%3D%20%5Cdfrac%7B1%7D%7B15%7D%0A%5Cend%7Baligned%7D%0A)
- 所以，取最大的概率
最终的结果为
。
贝叶斯估计
上面的极大似然估计可能会出现概率为0的情况，于是采用贝叶斯估计修正如下：
%20%3D%20%5Cfrac%7B%5Cdisplaystyle%5Csum%7Bi%3D1%7D%5E%7BN%7DI(y_i%3Dc_k)%2B%5Clambda%7D%7BN%2BK%5Clambda%7D%20%5C%5C%0A%26P%7B%5Clambda%7D(X%5E%7B(j)%7D%7CY%3Dck)%3D%5Cfrac%7B%5Cdisplaystyle%5Csum%7Bi%3D1%7D%5EN%20I(xi%5E%7B(j)%7D%3Da%7Bjm%7D%2Cyi%3Dc_k)%20%2B%5Clambda%20%7D%7B%5Cdisplaystyle%5Csum%7Bi%3D1%7D%5EN%20I(yi%3Dc_k)%2BS_j%5Clambda%7D%20%5C%5C%0A%26(k%3D1%2C2%2C…%2CK%EF%BC%9Bm%3D1%2C2%2C…%2CS_i%EF%BC%9Bj%3D1%2C2%2C…%2Cn)%0A%0A%5Cend%7Baligned%7D%0A#card=math&code=%5Cbegin%7Baligned%7D%0A%26P%7B%5Clambda%7D%28Y%3Dck%29%20%3D%20%5Cfrac%7B%5Cdisplaystyle%5Csum%7Bi%3D1%7D%5E%7BN%7DI%28yi%3Dc_k%29%2B%5Clambda%7D%7BN%2BK%5Clambda%7D%20%5C%5C%0A%26P%7B%5Clambda%7D%28X%5E%7B%28j%29%7D%7CY%3Dck%29%3D%5Cfrac%7B%5Cdisplaystyle%5Csum%7Bi%3D1%7D%5EN%20I%28xi%5E%7B%28j%29%7D%3Da%7Bjm%7D%2Cyi%3Dc_k%29%20%2B%5Clambda%20%7D%7B%5Cdisplaystyle%5Csum%7Bi%3D1%7D%5EN%20I%28y_i%3Dc_k%29%2BS_j%5Clambda%7D%20%5C%5C%0A%26%28k%3D1%2C2%2C…%2CK%EF%BC%9Bm%3D1%2C2%2C…%2CS_i%EF%BC%9Bj%3D1%2C2%2C…%2Cn%29%0A%0A%5Cend%7Baligned%7D%0A)
当
时，称为拉普拉斯平滑。
示例
以拉普拉斯平滑，我们还是用上面的例子演示一下计算过程：
求解:
- 计算先验概率和条件概率
%3D%5Cfrac%7B10%7D%7B17%7D%2CP(Y%3D-1)%3D%5Cfrac%7B7%7D%7B17%7D%20%5C%5C%0A%26P(X%5E%7B(1)%7D%3D1%7CY%3D1)%3D%5Cfrac%7B3%7D%7B12%7D%2CP(X%5E%7B(1)%7D%3D2%7CY%3D1)%3D%5Cfrac%7B4%7D%7B12%7D%2CP(X%5E%7B(1)%7D%3D3%7CY%3D1)%3D%5Cfrac%7B5%7D%7B12%7D%20%5C%5C%0A%26P(X%5E%7B(2)%7D%3DS%7CY%3D1)%3D%5Cfrac%7B2%7D%7B12%7D%2CP(X%5E%7B(2)%7D%3DM%7CY%3D1)%3D%5Cfrac%7B5%7D%7B12%7D%2CP(X%5E%7B(2)%7D%3DL%7CY%3D1)%3D%5Cfrac%7B5%7D%7B12%7D%20%5C%5C%0A%26P(X%5E%7B(1)%7D%3D1%7CY%3D-1)%3D%5Cfrac%7B4%7D%7B9%7D%2CP(X%5E%7B(1)%7D%3D2%7CY%3D-1)%3D%5Cfrac%7B3%7D%7B9%7D%2CP(X%5E%7B(1)%7D%3D3%7CY%3D-1)%3D%5Cfrac%7B2%7D%7B9%7D%20%5C%5C%0A%26P(X%5E%7B(2)%7D%3DS%7CY%3D-1)%3D%5Cfrac%7B4%7D%7B9%7D%2CP(X%5E%7B(2)%7D%3DM%7CY%3D-1)%3D%5Cfrac%7B3%7D%7B9%7D%2CP(X%5E%7B(2)%7D%3DL%7CY%3D-1)%3D%5Cfrac%7B2%7D%7B9%7D%20%0A%5Cend%7Baligned%7D%0A#card=math&code=%5Cbegin%7Baligned%7D%0A%26P%28Y%3D1%29%3D%5Cfrac%7B10%7D%7B17%7D%2CP%28Y%3D-1%29%3D%5Cfrac%7B7%7D%7B17%7D%20%5C%5C%0A%26P%28X%5E%7B%281%29%7D%3D1%7CY%3D1%29%3D%5Cfrac%7B3%7D%7B12%7D%2CP%28X%5E%7B%281%29%7D%3D2%7CY%3D1%29%3D%5Cfrac%7B4%7D%7B12%7D%2CP%28X%5E%7B%281%29%7D%3D3%7CY%3D1%29%3D%5Cfrac%7B5%7D%7B12%7D%20%5C%5C%0A%26P%28X%5E%7B%282%29%7D%3DS%7CY%3D1%29%3D%5Cfrac%7B2%7D%7B12%7D%2CP%28X%5E%7B%282%29%7D%3DM%7CY%3D1%29%3D%5Cfrac%7B5%7D%7B12%7D%2CP%28X%5E%7B%282%29%7D%3DL%7CY%3D1%29%3D%5Cfrac%7B5%7D%7B12%7D%20%5C%5C%0A%26P%28X%5E%7B%281%29%7D%3D1%7CY%3D-1%29%3D%5Cfrac%7B4%7D%7B9%7D%2CP%28X%5E%7B%281%29%7D%3D2%7CY%3D-1%29%3D%5Cfrac%7B3%7D%7B9%7D%2CP%28X%5E%7B%281%29%7D%3D3%7CY%3D-1%29%3D%5Cfrac%7B2%7D%7B9%7D%20%5C%5C%0A%26P%28X%5E%7B%282%29%7D%3DS%7CY%3D-1%29%3D%5Cfrac%7B4%7D%7B9%7D%2CP%28X%5E%7B%282%29%7D%3DM%7CY%3D-1%29%3D%5Cfrac%7B3%7D%7B9%7D%2CP%28X%5E%7B%282%29%7D%3DL%7CY%3D-1%29%3D%5Cfrac%7B2%7D%7B9%7D%20%0A%5Cend%7Baligned%7D%0A)
所以，以Y为例，因为总共有2个取值，所以
，然后因为是拉普拉斯平滑，所以，所以，分子多了1，分母多了
，从15->17。其他同理。
- 计算预测结果
%3DP(Y%3D1)P(X%5E%7B(1)%7D%3D2%7CY%3D1)P(X%5E%7B(2)%7D%3DS%7CY%3D1)%3D%5Cdfrac%7B10%7D%7B17%7D%20%5Ccdot%20%5Cdfrac%7B4%7D%7B12%7D%20%5Ccdot%20%5Cdfrac%7B2%7D%7B12%7D%20%3D%20%5Cdfrac%7B5%7D%7B153%7D%3D0.0327%5C%5C%0A%26P(X%3Dx%7CY%3D-1)%3DP(Y%3D-1)P(X%5E%7B(1)%7D%3D2%7CY%3D-1)P(X%5E%7B(2)%7D%3DS%7CY%3D-1)%3D%5Cdfrac%7B7%7D%7B17%7D%20%5Ccdot%20%5Cdfrac%7B3%7D%7B9%7D%20%5Ccdot%20%5Cdfrac%7B4%7D%7B9%7D%20%3D%20%5Cdfrac%7B28%7D%7B459%7D%20%3D0.0610%0A%5Cend%7Baligned%7D%0A#card=math&code=%5Cbegin%7Baligned%7D%0A%26P%28X%3Dx%7CY%3D1%29%3DP%28Y%3D1%29P%28X%5E%7B%281%29%7D%3D2%7CY%3D1%29P%28X%5E%7B%282%29%7D%3DS%7CY%3D1%29%3D%5Cdfrac%7B10%7D%7B17%7D%20%5Ccdot%20%5Cdfrac%7B4%7D%7B12%7D%20%5Ccdot%20%5Cdfrac%7B2%7D%7B12%7D%20%3D%20%5Cdfrac%7B5%7D%7B153%7D%3D0.0327%5C%5C%0A%26P%28X%3Dx%7CY%3D-1%29%3DP%28Y%3D-1%29P%28X%5E%7B%281%29%7D%3D2%7CY%3D-1%29P%28X%5E%7B%282%29%7D%3DS%7CY%3D-1%29%3D%5Cdfrac%7B7%7D%7B17%7D%20%5Ccdot%20%5Cdfrac%7B3%7D%7B9%7D%20%5Ccdot%20%5Cdfrac%7B4%7D%7B9%7D%20%3D%20%5Cdfrac%7B28%7D%7B459%7D%20%3D0.0610%0A%5Cend%7Baligned%7D%0A)
- 所以，取最大的概率
最终的结果为。
贝叶斯分类适用场景
从上面的过程来看：
- 输入数据必须是离散的，不然没法计算概率
- 输出可以是多分类的
- 对数据量要求不高
%2C%20(x2%2C%20y_2)%2C%E2%80%A6%2C(x_N%2C%20y_N)%20%5C%7D#card=math&code=T%3D%5C%7B%28x_1%2C%20y_1%29%2C%20%28x_2%2C%20y_2%29%2C%E2%80%A6%2C%28x_N%2C%20y_N%29%20%5C%7D)，其中
%7D%2C%20x_i%5E%7B(2)%7D%2C%E2%80%A6%2Cx_i%5E%7B(n)%7D)%5ET#card=math&code=x_i%3D%28x_i%5E%7B%281%29%7D%2C%20x_i%5E%7B%282%29%7D%2C%E2%80%A6%2Cx_i%5E%7B%28n%29%7D%29%5ET)，
%7D#card=math&code=x_i%5E%7B%28j%29%7D)是第
个样本的第
个特征
；
，表示总共
的分类
%20%3D%20%5Cfrac%7B%5Cdisplaystyle%5Csum%7Bi%3D1%7D%5E%7BN%7DI(y_i%3Dc_k)%7D%7BN%7D%20%5C%5C%0A%26P(X%5E%7B(j)%7D%7CY%3Dc_k)%3D%5Cfrac%7B%5Cdisplaystyle%5Csum%7Bi%3D1%7D%5EN%20I(xi%5E%7B(j)%7D%3Da%7Bjm%7D%2Cyi%3Dc_k)%7D%7B%5Cdisplaystyle%5Csum%7Bi%3D1%7D%5EN%20I(yi%3Dc_k)%7D%20%5C%5C%0A%26(k%3D1%2C2%2C…%2CK%EF%BC%9Bm%3D1%2C2%2C…%2CS_i%EF%BC%9Bj%3D1%2C2%2C…%2Cn)%0A%0A%5Cend%7Baligned%7D%0A#card=math&code=%5Cbegin%7Baligned%7D%0A%26P%28Y%3Dc_k%29%20%3D%20%5Cfrac%7B%5Cdisplaystyle%5Csum%7Bi%3D1%7D%5E%7BN%7DI%28yi%3Dc_k%29%7D%7BN%7D%20%5C%5C%0A%26P%28X%5E%7B%28j%29%7D%7CY%3Dc_k%29%3D%5Cfrac%7B%5Cdisplaystyle%5Csum%7Bi%3D1%7D%5EN%20I%28xi%5E%7B%28j%29%7D%3Da%7Bjm%7D%2Cyi%3Dc_k%29%7D%7B%5Cdisplaystyle%5Csum%7Bi%3D1%7D%5EN%20I%28y_i%3Dc_k%29%7D%20%5C%5C%0A%26%28k%3D1%2C2%2C…%2CK%EF%BC%9Bm%3D1%2C2%2C…%2CS_i%EF%BC%9Bj%3D1%2C2%2C…%2Cn%29%0A%0A%5Cend%7Baligned%7D%0A)
