In nonlinear least squares you suppose that you know the form of the relationship between the response and predictor. Suppose instead that you do not know that relationship, and also that you are unwilling to assume that the relationship can be well approximated by a linear model. You need a more nonparametric type of regression fitting approach. One such approach is based on 'trees.' 非线性最小二乘法假设自变量和因变量间关系已知. 如果你不知道这种关系, 你也不愿假设该关系可以用线性模型估计. 你需要一种非参数的回归拟合方法. 一种这样的方法就是基于'树' A regression tree is a sequence of questions that can be answered as yes or no, plus a set of fitted response values. Each question asks whether a predictor satisfies a given condition. Predictors can be continuous or discrete. Depending on the answers to one question, you either proceed to another question or arrive at a fitted response value. 回归树是一系列的可以用是否回答的问题,加上一组拟合的因变量值. 每个问题都会问自变量值是否满足某一条件. 自变量值可以是连续或离散的. 根据问题答案, 可以继续下一个问题或得出拟合因变量值. This example fits a regression tree to variables from the carsmall data set. The example uses the same variables as in the Analysis of Covariance example (see The aoctool Demo), so there is one continuous predictor (car weight) and one discrete predictor (model year). 下例将carsmall数据拟合到一棵回归树上. 这是在共方差分析里用过的例子(见aoctool展示). 有一个连续自变量(车重)和一个离散自变量(年份) The object of the example is to model mileage (MPG) as a function of car weight and model year. First load the data and create a matrix x of predictor values and a vector y of response variables. Then fit a regression tree, specifying the model year column as a categorical variable. In this data set there are cars from the three different model years 1970, 1976, and 1982. 下例目标将汽车耗油模拟为车重和年份的一个函数. 首先载入数据,创造自变量x矩阵和因变量y向量. 再拟合回归树, 将年份定义为分类变量(categorical). 在数据里年份可以取1970, 1976和1982三个值之一. load carsmall x = [Weight,Model_Year]; y = MPG; t = treefit(x,y,"catidx",2); treedisp(t,"name",{"Wt" "Yr"}); Now you want to use this model to determine the predicted mileage for a car weighing 3000 pounds from model year 1982. Start at the top node. The weight is less than the cutoff value of 3085.5, so you take the left branch. The model year is not 1970 or 1976, so you take the right branch. Continue moving down the tree until you arrive at a terminal node that gives the predicted value. In this case, the predicted value is 38 miles per gallon. You can use the treeval function to find the fitted value for any set of predictor values. 现在想要求1982年的重3000磅的车的耗油, 从顶部节点开始, 重量少于阙值3085.5, 下到左枝, 模型年份不是1970或1976, 下到右枝. 一直下去直到到达给出预测值的节点. 这个例子中最后是每加仑油开38英里. 可以使用treeval函数求任何自变量租的拟合值. treeval(t,[3000 82]) ans = 38 With a tree like this one, having many branches, there is a danger that it fits the current data set well but would not do a good job at predicting new values. Some of its lower branches might be strongly affected by outliers and other artifacts of the current data set. If possible you would prefer to find a simpler tree that avoids this problem of overfitting. 这样的树, 有很多分枝, 就有风险它能很好地拟合现有数据, 但在预测新数据时就不准确了. 一些底层分枝可能会受到离群值的强烈影响. 如果可能应该选择尽可能简单的树来防止过度拟合问题. You can estimate the best tree size by cross validation. First, compute a resubstitution estimate of the error variance for this tree and a sequence of simpler trees and plot it as the lower (blue) line in the figure. This estimate probably under-estimates the true error variance. Then compute a cross-validation estimate of the same quantity and plot it as the upper (red) line. The cross-validation procedure also provides an estimate of the pruning level, best, needed to achieve the best tree size. 你可以用交叉验证来估计最好的树. 首先计算这棵树以及一系列简化的树的误差离散的再代入估计值, 在下图用蓝线画出. 该估计值有可能低估了实际误差离散. 再画同一值的交叉验证估计,为红线. 交叉验证也提供达到最好的树大小需要的剪枝程度, best. [c,s,ntn] = treetest(t,"resub"); [c2,s2,n2,best] = treetest(t,"cross",x,y); plot(ntn,c,"b-", n2,c2,"r-", n2(best+1),c2(best+1),"mo"); xlabel("Number of terminal nodes") ylabel("Residual variance") legend("Resubstitution error","Cross-validation error","Estimated best tree size") best best = 10 The best tree is the one that has a residual variance that is no more than one standard error above the minimum value along the cross-validation line. In this case the variance is just over 14. The output best takes on values starting with 0 (representing no pruning), so you need to add 1 to use it as an index into the other output arguments. 最好的树其剩余离散不超过交叉验证线上最小值以上一个标准差. 在本例是14多一点. 输出的best值取值是从0开始(代表无剪枝), 所以用作其它语句的索引自变量时要加1. c2(best+1) ans = 14.3440 Use the output best to create a smaller tree that is pruned to the estimated best size. 利用输出best来对树剪枝建立较小的树. t0 = treeprune(t,"level",best); treedisp(t0,"name",{"Wt" "Yr"}) Now plot the original data and overlay the fitted values that you get using this tree. Notice that this tree does not distinguish between cars from 1970 or 1976, so create a vector yold containing fitted values for 1976 and another ynew for year 1982. Cars from 1970 have the same fitted values as those from 1976. 现在再画出原先的数据和用这棵树得出的拟合值. 注意这棵树不能区分1970和1976的车, 建立1976年拟合值的向量yold和1982年的向量ynew. 1970年的车拟合值和1976年的完全一样. xx = (1500:20:5000)"; ynew = treeval(t0,[xx 82*ones(size(xx))]); yold = treeval(t0,[xx 76*ones(size(xx))]); gscatter(Weight,MPG,Model_Year,"rgb","osx"); hold on; plot(xx,yold,"b:", xx,ynew,"r--"); hold off The tree functions (treedisp, treefit, treeprune, treetest, and treeval) can also accept a categorical response variable. In that case, the fitted value from the tree is the category with the highest predicted probability for the range of predictor values falling in a given node. The demo Classification, in the Multivariate Analysis section of the Statistics Toolbox demos, shows how to use decision trees for classification. 树函数(treedisp, treefit, treeprune, treetest, treeval)也可以接受分类因变量. 这种情况下, 树给出的拟合值就是自变量落在某一节点上时概率最高的因变量. 在statistics toolbox展示里的多元分析一节展示了怎么用决策树来分类
从配置上来讲，朗逸除了最低配的车型以外在同档次上配置比新宝来要高，不过价格也要贵个几千元。 新宝来在没正式亮相前叫MODEL X 朗逸叫MODEL Y，都是大众在中国为了适应中国国情在PO34平台上做出来的车，都是中国化的大众车，只不过新宝来德方参与设计的更多，朗逸中方参与设计的更多，所以朗逸看起来更不像大众的车，无论外观还是内饰都跟以往的大众车风格有很大差异。 安全性上两车都是CNCAP五星碰撞的产品，不过新宝来并没有采用激光焊接技术，朗逸采用了。 操控两者基本没区别，如果硬要分出区别那只能说新宝来沿袭宝来的更多一些，操控稍稍好一点点。 新宝来的发动机与速腾1.6的一样都是8气门发动机，只是比原来的捷达发动机调教了一下，基本还是“成熟”产品。 朗逸的发动机跟POLO 还有 明锐1.6是一样的16气门发动机，功率和扭矩比新宝来的药稍高一些，同样也是低转速高扭矩的特点。而且该发动机使用的是正时链条传动，比新宝来的正时皮带要先进，而且基本是免维护的。 油耗两者基本一样，差别很细微。 变速箱两者也是完全相同的，手动变速箱是 MQ200 自动的是 手自一体。
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