8 - 1 - Non-linear Hypotheses (10 min).srt

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In this and in the
在这节课和接下来的课程中
(字幕整理：中国海洋大学 黄海广,haiguang2000@qq.com )

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next set of videos, I'd like
我将给大家介绍

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to tell you about a learning
一种叫“神经网络”(Neural Network)

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algorithm called a Neural Network.
的机器学习算法

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We're going to first talk about
我们将首先讨论

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the representation and then
神经网络的表层结构

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in the next set of videos
在后续课程中

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talk about learning algorithms for it.
再来具体讨论的学习算法

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Neutral networks is actually
神经网络实际上是一个

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a pretty old idea, but had
相对古老的算法

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fallen out of favor for a while.
并且后来沉寂了一段时间

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But today, it is the
不过到了现在

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state of the art technique for
它又成为许多机器学习问题

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many different machine learning problems.
的首选技术

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So why do we need yet another learning algorithm?
不过我们为什么还需要这个学习算法？

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We already have linear regression and
我们已经有线性回归和逻辑回归算法了

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we have logistic regression, so why do we need, you know, neural networks?
为什么还要研究神经网络？

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In order to motivate the discussion
为了阐述研究

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of neural networks, let me
神经网络算法的目的

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start by showing you a few
我们首先来看几个

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examples of machine learning
机器学习问题作为例子

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problems where we need
这几个问题的解决

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to learn complex non-linear hypotheses.
都依赖于研究复杂的非线性分类器

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Consider a supervised learning classification
考虑这个监督学习分类的问题

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problem where you have a training set like this.
我们已经有了对应的训练集

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If you want to apply logistic
如果利用逻辑回归算法

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regression to this problem, one
来解决这个问题

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thing you could do is apply
首先需要构造

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logistic regression with a
一个包含很多非线性项的

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lot of nonlinear features like that.
逻辑回归函数

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So here, g as usual
这里g仍是s型函数 (即f(x)=1/(1+e^-x) )

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is the sigmoid function, and we
我们能让函数

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can include lots of polynomial terms like these.
包含很多像这样的多项式项

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And, if you include enough polynomial
事实上  当多项式项数足够多时

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terms then, you know, maybe
那么可能

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you can get a hypotheses
你能够得到一个

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that separates the positive and negative examples.
分开正样本和负样本的分界线

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This particular method works well
当只有两项时

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when you have only, say, two
比如 x1 x2

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features - x1 and x2
这种方法确实能得到不错的结果

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- because you can then include
因为你可以

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all those polynomial terms of
把x1和x2的所有组合

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x1 and x2.
都包含到多项式中

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But for many interesting machine learning
但是对于许多

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problems would have a
复杂的机器学习问题

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lot more features than just two.
涉及的项往往多于两项

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We've been talking for a while
我们之前已经讨论过

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about housing prediction, and suppose
房价预测的问题

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you have a housing classification
假设现在要处理的是

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problem rather than a
关于住房的分类问题

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regression problem, like maybe
而不是一个回归问题

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if you have different features of
假设你对一栋房子的多方面特点

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a house, and you want
都有所了解

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to predict what are the
你想预测

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odds that your house will
房子在未来半年内

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be sold within the next
能被卖出去的概率

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six months, so that will be a classification problem.
这是一个分类问题

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And as we saw we can
我们可以想出

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come up with quite a
很多特征

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lot of features, maybe a hundred
对于不同的房子有可能

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different features of different houses.
就有上百个特征

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For a problem like this, if
对于这类问题

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you were to include all the
如果要包含

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quadratic terms, all of
所有的二次项

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these, even all of the
即使只包含

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quadratic that is the second
二项式或多项式的计算

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or the polynomial terms, there would be a lot of them.
最终的多项式也可能有很多项

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There would be terms like x1 squared,
比如x1^2

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x1x2, x1x3, you know, x1x4
x1x2   x1x3   x1x4

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up to x1x100 and then
直到x1x100

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you have x2 squared, x2x3
还有x2^2  x2x3

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and so on.
等等很多项

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And if you include just
因此

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the second order terms, that
即使只考虑二阶项

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is, the terms that are
也就是说

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a product of, you know,
两个项的乘积

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two of these terms, x1
x1乘以x1

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times x1 and so on, then,
等等类似于此的项

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for the case of n equals
那么  在n=100的情况下

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100, you end up with about five thousand features.
最终也有5000个二次项

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And, asymptotically, the
而且渐渐地

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number of quadratic features grows
随着特征个数n的增加

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roughly as order n
二次项的个数大约以n^2的量级增长

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squared, where n is the
其中

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number of the original features,
n是原始项的个数

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like x1 through x100 that we had.
即我们之前说过的x1到x100这些项

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And its actually closer to n squared over two.
事实上二次项的个数大约是(n^2)/2

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So including all the
因此要包含所有的

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quadratic features doesn't seem
二次项是很困难的

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like it's maybe a good
所以这可能

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idea, because that is a
不是一个好的做法

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lot of features and you
而且由于项数过多

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might up overfitting the training
最后的结果很有可能是过拟合的

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set, and it can
此外

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also be computationally expensive, you know, to
在处理这么多项时

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be working with that many features.
也存在运算量过大的问题

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One thing you could do is
当然  你也可以试试

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include only a subset of
只包含上边这些二次项的子集

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these, so if you include only the
例如 我们只考虑

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features x1 squared, x2 squared,
x1^2  x2^2

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x3 squared, up to
x3^2直到

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maybe x100 squared, then
x100^2 这些项

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the number of features is much smaller.
这样就可以将二次项的数量大幅度减少

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Here you have only 100 such
减少到只有100个二次项

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quadratic features, but this
但是由于

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is not enough features and
忽略了太多相关项

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certainly won't let you fit
在处理类似左上角的数据时

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the data set like that on the upper left.
不可能得到理想的结果

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In fact, if you include
实际上

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only these quadratic features together
如果只考虑x1的平方

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with the original x1, and
到x100的平方

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so on, up to x100 features,
这一百个二次项

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then you can actually fit very
那么你可能会

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interesting hypotheses. So, you
拟合出一些特别的假设

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can fit things like, you know, access a
比如可能拟合出

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line of the ellipses like these, but
一个椭圆状的曲线

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you certainly cannot fit a more
但是肯定不能拟合出

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complex data set like that shown here.
像左上角这个数据集的分界线

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So 5000 features seems like
所以5000个二次项看起来已经很多了

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a lot, if you were
而现在假设

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to include the cubic, or
包括三次项

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third order known of each others,
或者三阶项

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the x1, x2, x3.
例如x1 x2 x3

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You know, x1 squared,
x1^2

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x2, x10 and
x2 x10

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x11, x17 and so on.
x11 x17等等

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You can imagine there are gonna be a lot of these features.
类似的三次项有很多很多

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In fact, they are going to be
事实上

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order and cube such features
三次项的个数是以n^3的量级增加

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and if any is 100
当n=100时

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you can compute that, you
可以计算出来

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end up with on the order
最后能得到

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of about 170,000 such cubic
大概17000个三次项

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features and so including
所以

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these higher auto-polynomial features when
当初始特征个数n增大时

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your original feature set end
这些高阶多项式项数

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is large this really dramatically
将以几何级数递增

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blows up your feature space and
特征空间也随之急剧膨胀

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this doesn't seem like a
当特征个数n很大时

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good way to come up with
如果找出附加项

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additional features with which
来建立一些分类器

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to build none many classifiers when n is large.
这并不是一个好做法

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For many machine learning problems, n will be pretty large.
对于许多实际的机器学习问题 特征个数n是很大的

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Here's an example.
我们看看下边这个例子

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Let's consider the problem of computer vision.
关于计算机视觉中的一个问题

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And suppose you want to
假设你想要

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use machine learning to train
使用机器学习算法

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a classifier to examine an
来训练一个分类器

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image and tell us whether
使它检测一个图像

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or not the image is a car.
来判断图像是否为一辆汽车

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Many people wonder why computer vision could be difficult.
很多人可能会好奇  这对计算机视觉来说有什么难的

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I mean when you and I
当我们自己看这幅图像时

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look at this picture it is so obvious what this is.
里面有什么是一目了然的事情

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You wonder how is it
你肯定会很奇怪

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that a learning algorithm could possibly
为什么学习算法竟可能会不知道

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fail to know what this picture is.
图像是什么

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To understand why computer vision
为了解答这个疑问

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is hard let's zoom
我们取出这幅图片中的一小部分

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into a small part of the
将其放大

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image like that area where the
比如图中

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little red rectangle is.
这个红色方框内的部分

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It turns out that where you
结果表明

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and I see a car, the computer sees that.
当人眼看到一辆汽车时 计算机实际上看到的却是这个

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What it sees is this matrix,
一个数据矩阵

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or this grid, of pixel
或像这种格网

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intensity values that tells
它们表示了像素强度值

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us the brightness of each pixel in the image.
告诉我们图像中每个像素的亮度值

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So the computer vision problem is
因此 对于计算机视觉来说问题就变成了

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to look at this matrix of
根据这个像素点亮度矩阵

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pixel intensity values, and tell
来告诉我们

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us that these numbers represent the door handle of a car.
这些数值代表一个汽车门把手

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Concretely, when we use
具体而言

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machine learning to build a
当用机器学习算法构造

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car detector, what we do
一个汽车识别器时

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is we come up with a
我们要想出

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label training set, with, let's
一个带标签的样本集

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say, a few label examples
其中一些样本

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of cars and a few
是各类汽车

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label examples of things that
另一部分样本

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are not cars, then we
是其他任何东西

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give our training set to
将这个样本集输入给学习算法

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the learning algorithm trained a
以训练出一个分类器

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classifier and then, you
训练完毕后

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know, we may test it and show
我们输入一幅新的图片

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the new image and ask, "What is this new thing?".
让分类器判定  “这是什么东西？”

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And hopefully it will recognize that that is a car.
理想情况下 分类器能识别出这是一辆汽车

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To understand why we
为了理解引入

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need nonlinear hypotheses, let's take
非线性分类器的必要性

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a look at some of the
我们从学习算法的训练样本中

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images of cars and maybe
挑出一些汽车图片

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non-cars that we might feed to our learning algorithm.
和一些非汽车图片

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Let's pick a couple of pixel
让我们从其中

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locations in our images, so
每幅图片中挑出一组像素点

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that's pixel one location and
这是像素点1的位置

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pixel two location, and let's
这是像素点2的位置

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plot this car, you know, at the
在坐标系中标出这幅汽车的位置

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location, at a certain
在某一点上

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point, depending on the intensities
车的位置取决于

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of pixel one and pixel two.
像素点1和像素点2的亮度

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And let's do this with a few other images.
让我们用同样的方法标出其他图片中汽车的位置

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So let's take a different example
然后我们再举一个

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of the car and you know,
关于汽车的不同的例子

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look at the same two pixel locations
观察这两个相同的像素位置

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and that image has a
这幅图片中

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different intensity for pixel one
像素1有一个像素强度

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and a different intensity for pixel two.
像素2也有一个不同的像素强度

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So, it ends up at a different location on the figure.
所以在这幅图中它们两个处于不同的位置

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And then let's plot some negative examples as well.
我们继续画上两个非汽车样本

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That's a non-car, that's a
这个不是汽车

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non-car  .
这个也不是汽车

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And if we do this for
然后我们继续

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more and more examples using
在坐标系中画上更多的新样本

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the pluses to denote cars
用''+"表示汽车图片

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and minuses to denote non-cars,
用“-”表示非汽车图片

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what we'll find is that
我们将发现

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the cars and non-cars end up
汽车样本和非汽车样本

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lying in different regions of
分布在坐标系中的不同区域

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the space, and what we
因此

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need therefore is some sort
我们现在需要一个

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of non-linear hypotheses to try
非线性分类器

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to separate out the two classes.
来尽量分开这两类样本

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What is the dimension of the feature space?
这个分类问题中特征空间的维数是多少？

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Suppose we were to use just 50 by 50 pixel images.
假设我们用50*50像素的图片

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Now that suppose our images were
我们的图片已经很小了

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pretty small ones, just 50 pixels on the side.
长宽只各有50个像素

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Then we would have 2500 pixels,
但这依然是2500个像素点

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and so the dimension of
因此

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our feature size will be N
我们的特征向量的元素数量

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equals 2500 where our feature
N=2500

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vector x is a list
特征向量X

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of all the pixel testings, you
包含了所有像素点的亮度值

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know, the pixel brightness of pixel
这是像素点1的亮度

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one, the brightness of pixel
这是像素点2的亮度

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two, and so on down
如此类推

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to the pixel brightness of the
直到最后一个

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00:08:01,400 --> 00:08:03,420
last pixel where, you know, in a
像素点的亮度

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typical computer representation, each of
对于典型的计算机图片表示方法

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00:08:05,540 --> 00:08:07,190
these may be values between say
如果存储的是每个像素点的灰度值 (色彩的强烈程度)

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0 to 255 if it gives
那么每个元素的值

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us the grayscale value.
应该在0到255之间

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So we have n equals 2500,
因此 这个问题中n=2500

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and that's if we
但是

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were using grayscale images.
这只是使用灰度图片的情况

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If we were using RGB
如果我们用的是RGB彩色图像

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images with separate red, green
每个像素点包含红、绿、蓝三个子像素

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and blue values, we would have n equals 7500.
那么n=7500

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So, if we were to
因此 如果我们非要

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try to learn a nonlinear
通过包含所有的二次项

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hypothesis by including all
来解决这个非线性问题

250
00:08:32,300 --> 00:08:33,710
the quadratic features, that is
那么

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all the terms of the form, you know,
这就是式子中的所有条件

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Xi times Xj, while with the
xi*xj

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2500 pixels we would end
连同开始的2500像素

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up with a total of three million features.
总共大约有300万个

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00:08:43,050 --> 00:08:44,620
And that's just too large to
这数字大得有点离谱了

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00:08:44,720 --> 00:08:46,430
be reasonable; the computation would
对于每个样本来说

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00:08:46,600 --> 00:08:48,680
be very expensive to find and
要发现并表示

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00:08:48,840 --> 00:08:50,070
to represent all of these
所有这300万个项

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00:08:50,310 --> 00:08:52,250
three million features per training example.
这计算成本太高了

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00:08:55,470 --> 00:08:57,580
So, simple logistic regression together
因此 只是简单的增加

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00:08:58,100 --> 00:08:59,230
with adding in maybe the
二次项或者三次项

262
00:08:59,300 --> 00:09:00,510
quadratic or the cubic features
之类的逻辑回归算法

263
00:09:00,930 --> 00:09:01,910
- that's just not a
并不是一个解决

264
00:09:01,980 --> 00:09:03,950
good way to learn complex
复杂非线性问题的好办法

265
00:09:04,550 --> 00:09:06,090
nonlinear hypotheses when n
因为当n很大时

266
00:09:06,310 --> 00:09:08,410
is large because you just end up with too many features.
将会产生非常多的特征项

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00:09:09,370 --> 00:09:10,620
In the next few videos, I
在接下来的视频课程中

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would like to tell you about Neural
我将为大家讲解神经网络

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00:09:12,080 --> 00:09:13,670
Networks, which turns out
它在解决复杂的非线性分类问题上

270
00:09:13,980 --> 00:09:15,370
to be a much better way to
被证明是

271
00:09:15,650 --> 00:09:17,720
learn complex hypotheses, complex nonlinear
是一种好得多的算法

272
00:09:17,960 --> 00:09:19,780
hypotheses even when your
即使你输入特征空间

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00:09:20,070 --> 00:09:22,080
input feature space, even when n is large.
或输入的特征维数n很大也能轻松搞定

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00:09:22,860 --> 00:09:24,080
And along the way I'll
在后面的课程中

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00:09:24,420 --> 00:09:25,580
also get to show
我将给大家展示

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00:09:25,780 --> 00:09:26,730
you a couple of fun videos
一些有趣的视频

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of historically important applications
视频中讲述了神经网络在历史上的重要应用

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00:09:30,300 --> 00:09:31,290
of Neural networks as well that I
我也希望

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00:09:32,100 --> 00:09:33,480
hope those videos that
这些我们即将看到的视频

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00:09:33,570 --> 00:09:35,460
we'll see later will be fun for you to watch as well.
能给你的学习过程带来一些乐趣 ^ ^