R语言英文回归教案Logistic Regression Assumptions
r语言多元logistic回归代码 -回复
r语言多元logistic回归代码-回复如何在R语言中进行多元logistic回归分析。
多元logistic回归是一种用于分析多个自变量与一个多分类响应变量之间关系的统计方法。
它可以用于预测多个类别中一个特定类别的概率，并且可以帮助我们理解哪些自变量对于预测分类最为重要。
接下来，我将逐步介绍如何在R语言中进行多元logistic回归分析。
步骤1：准备数据首先，我们需要准备需要进行分析的数据集。
数据集应包含一个或多个自变量（可以是定量或分类型数据）和一个多分类的响应变量。
在这里，我们以一个虚拟数据集为例，数据集包含3个自变量（age、gender和income）和一个二分类的响应变量（purchase）。
我们将使用"mlbench"包中的"BreastCancer"数据集来演示。
Rinstall.packages("mlbench")library(mlbench)data("BreastCancer")步骤2：拟合多元logistic回归模型在R语言中，我们可以使用“multinom”函数从“nnet”包中来进行多元logistic回归分析。
Rinstall.packages("nnet")library(nnet)model <- multinom(purchase ~ age + gender + income, data = BreastCancer)在这里，我们使用"multinom"函数来拟合一个多元logistic回归模型。
参数"purchase ~ age + gender + income"表示我们将使用age、gender和income作为自变量来预测purchase变量。
步骤3：模型拟合结果我们可以使用"summary"函数来查看多元logistic回归模型的拟合结果。
r语言逻辑回归_roc曲线_理论说明
r语言逻辑回归roc曲线理论说明1. 引言1.1 概述逻辑回归（logistic regression）是一种广泛应用于分类问题的统计学习方法，其基本原理是通过建立一个线性回归模型来预测概率，并使用sigmoid函数将预测结果转化为一个二分类几率。
R语言作为一种流行且功能强大的数据分析和统计建模工具，在逻辑回归模型的应用上具有很大优势。
ROC曲线（Receiver Operating Characteristic curve）则是评估分类模型性能的重要工具之一。
它以假阳性率（false positive rate）作为横坐标、真阳性率（true positive rate）作为纵坐标，绘制出一条曲线来反映模型在各个阈值下识别正例和负例的表现，从而提供了更全面的性能评估指标。
本文将结合R语言逻辑回归和ROC曲线两个主题，详细说明逻辑回归在分类问题中的理论基础和建立步骤，并介绍如何使用R语言进行逻辑回归模型建立和ROC曲线绘制。
通过一个实际案例的分析，我们将展示如何运用这些知识来解读模型结果并进行讨论。
1.2 文章结构本文将按照以下结构进行展开讨论：- 第2部分将介绍R语言逻辑回归的理论基础，包括相关概念和建模步骤。
- 第3部分将详细阐述ROC曲线的概念、绘制方法以及解读和应用。
- 第4部分将通过一个实例分析，演示如何使用R语言进行逻辑回归模型建立和ROC曲线绘制，并对结果进行解读和讨论。
- 最后，在第5部分中，我们将总结研究成果并指出存在的不足之处，提出改进方向，并展望未来关于逻辑回归和ROC曲线的研究方向。
1.3 目的本文旨在全面介绍R语言逻辑回归和ROC曲线的理论知识，并通过实例演示其应用。
希望读者能够通过阅读本文了解逻辑回归的基本概念、建模步骤以及如何使用R语言进行建模与评估。
同时，通过对ROC曲线的学习，读者能够了解该曲线在分类模型性能评估中的重要性，并学会如何解读和应用。
最后，我们也希望为未来关于逻辑回归和ROC曲线领域的研究提出一些建议和展望。
R语言RobustRegression回归英文教案
R语言RobustRegression回归英文教案Hilary Term2015,Week8Practical SB1bMarch11,2015There are three exercises in this practical but only Exercise3will be assessed and it contributes8.5% to your raw SB1total mark.The deadline for submission is12noon Monday week2,Trinity Term2015at the Statistics Department reception,1South Parks Road.1Wilcoxon Rank TestThe data sitka258is part of the dataset Sitka(R library MASS)which contains repeated measurements on the log-size of79Sitka spruce trees,54of which were grown in ozone-enriched chambers and25were controls.The size was measured ?ve times in1988,at roughly monthly intervals.sitka258contains only the last measurements taken on these trees.#read the data>sitka258<-read.table("http://vollmer.ms/sebastian/teaching/2015/sb1b/sit ka258.txt",header=T)#Size values of the two populations>ozon<-sitka258[which(sitka258$treat=="ozone"),]$size>ctrl<-sitka258[which(sitka258$treat=="control"),]$size#Perform a Wilcoxon’s rank test>wilcox.test(ozon,ctrl,exact=TRUE,conf.int=TRUE)Calculate the value of W using the de?nition of the lectures(R calls W the Mann–Whitney statistic ni=mj=11(0,∞)(Y j?X i).from the lectures).2Robust and Nonlinear RegressionThe dataset stackloss,known as Brownlee’s Stack Loss Plant Data(R manual), contains operational data for a plant for the oxidation of ammonia to nitric acid; there are21observations on4variables.The dependent variable is stack.loss is 10times the percentage of the ingoing ammonia to the plant that escapes from the absorption column unabsorbed,i.e.it is an inverse measure of the e?ciency of the plant.The predictor variable is Water.Temp,the temperature of cooling water circulated through coils in the absorption tower.2.1Robust Regression>#Load Brownlee’s Stack Loss Plant Data>data(stackloss)>head(stackloss)>attach(stackloss)>stack.lm<-lm(stack.loss~Water.T emp)>plot(stack.lm)#consider the diagnostics of OLS>#Fit a least squares regression model>plot(Water.Temp,stack.loss,main="Comparison of regression methods",+xlab="Water temperature",ylab="Stack loss")>abline(stack.lm,lty=2,col=’red’)Now try?tting some robust regression models.If the following packages are notalready installed,go to Packages→Install package(s).Alternatively,if you’re usinga Mac or Linux machine,use the function install.packages.library(MASS)stack.rlm=rlm(stack.loss~Water.Temp)abline(stack.rlm,lty=2,col=’green’)data.frame(item=1:21,resid=stack.rlm$resid,weight=stack.rl m$w)abline(s tack.rlm,lty=2,col=’gree Try least median of squares,check out the help?le of the lqs command and add a cyan coloured line.>Adding a legendlegend(18,40,c("Least squares","Huber","LMS"),lty=c(2,3,4), +col=c(’red’,’green’,’cyan’),bty="n",ncol=1)Calculate the paired bootstrap estimate of standard deviation of slope coe?cient>m<-1000>n<-dim(stackloss)[1]>#Paired bootstrap>pols<-numeric(m)>phuber<-numeric(m)>plms<-numeric(m)>for(i in1:m){+ind<-sample(1:n,n,replace=T)+pols[i]<-coef(lm(stack.loss[ind]~Water.Temp[ind]))[2]+phuber[i]<-coef(rlm(stack.loss[ind]~Water.Temp[ind]))[2] +plms[i]<-coef(lqs(stack.loss[ind]~Water.Temp[ind],method=’lms’))[2] +}>cat(sqrt(var(pols)),sqrt(var(phuber)),sqrt(var(plms)))Perform a residual bootstrap estimate of the standard deviation of slope coe?/doc/3512512918.html,ing the commands likey1=fitted(stack.lm)youcan express the?tted values and in turn the residuals.By sampling indices with replacement ind<-sample(1:n,n,replace=T) the bootstrap data takes the form of y1+residual1[ind].2.2Multivariate linear regressionConsider the multivariate linear of the stack.loss for the predictors Air.Flow,Water.Temp and Acid.Conc.mfit=lm(stack.loss~Air.Flow+Water.T emp+Acid.Conc.)plot(mfit)#does the output suggest outliers?For comparison perform LTSmfitlts=lqs(stack.loss~Air.Flow+Water.Temp+Acid.Conc.,met hod=’lms’)plot(1:21,mfitlts$residuals/mad(mfitlts$residuals))#explain what this plot illustrates2.3Nadaraya-Watson kernel estimatorCompute Nadaraya-Watson kernel estimator with Gaussian kernel>#Perform non-linear regression by kernel smoothing>stack.k1<-ksmooth(Water.Temp,stack.loss,kernel="normal",bandwidth=0.8 5)>stack.k2<-ksmooth(Water.Temp,stack.loss,kernel="normal",bandwidth=1.5) >stack.k3<-ksmooth(Water.Temp,stack.loss,kernel="normal",bandwidth=3) >plot(Water.Temp,stack.loss,main="Comparison of non-linear regression methods",xlab="Water temperature",ylab="Stack loss")>lines(stack.k1,col=’red’)>lines(stack.k2,col=’blue’)>lines(stack.k3,col=’green’)Perform LOO-CV and calculate the MSE for bandwidth=1.5 >y.hat<-ksmooth(Water.Temp,stack.loss,kernel="normal",bandwidth=1.5,x.points=Water.Temp)$y>h<-1.5#help(ksmooth)explains for the bandwidth#bandwidth:the bandwidth.The kernels are scaled so that their#quartiles(viewed as probability densities)are at+/-#’0.25*bandwidth’.#So we need to adjust in order to obtain the actual smoothing#parameter>h<-abs(0.25*h/qnorm(.25))#Calculate the smoothing matrix under gaussian kernel>S<-matrix(NA,ncol=n,nrow=n)for(i in1:n){for(j in1:n){S[i,j]<-dnorm((Water.Temp[i]-Water.Temp[j])/h)/sum(dnorm((Water.Temp[i]-Water.Temp)/h))}}#The points x are ordered when used in prediction#For this reason we should arrange stack.loss and Sii accordingly>ox<-order(Water.Temp)#First check that we got S right>cbind((S%*%stack.loss)[ox],y.hat)#Then calculate the MSE>MSE<-mean(((stack.loss[ox]-y.hat)/(1-diag(S)[ox]))^2)Compute LOO-CV usingMSE(h)=n?1i(Y i??m(?i)h(x i))2.You might?nd the following R code helpful>x=c(1,5,3) >x[-2] [1]13PlotMSE against h by?lling into the following code.hs=seq(0.8,10,0.1)MSEhats=numeric(length(hs))#Compute the estimate of the MSE for each of the h’splot(hs,MSEhats)3Environmental Behaviour in the OECD Countries(34marks) The data set http://vollmer.ms/sebastian/teaching/2015/sb1b/recycling.csv has been compiled from[2][1][3]. It contains the recycling behaviour of28OECD countries along with their GDP per person and spending on organic /doc/3512512918.html,eread.table("http://vollmer.ms/sebastian/teaching/2015/sb1b /recycling.csv",fill=TRUE,sep=",",stringsAsFactors=FALSE,header=TRUE)in order to obtain the data./doc/3512512918.html,pare the average recycling percentage between European and non-European countries using the Wilcoxonrank test(4marks).The test produces a warning,explain what this means.Research how this problem can be?xed in general(presentation in your own words!)or argue how theproblem can be?xed in the case above (2marks).2.We modelaverage recycling in%=a+bGDP per capita+noise.Estimate a and b as in Exercise2using OLS and identify the most in?uential observation using the cook distance(3marks).Use Huber regression,the LMS and the LTS to estimate a and b(3marks).Additionally, change a parameter of the command rlm in order to use the Tukeys bisquare choice ofψ(2marks).Robust regression methods are used for protection against outliers.Assuming the data at hand is reliable,justify the use of robust methods(1mark).Perform the bootstrap analysis of the estimate for a(4marks).3.Similar to Exercise2compute the Nadaraya-Watson kernel estimator(based on the Gaussian kernel)ofthe average recycling in%based on the GDP per capita and plot the leave-one-out cross validation of the estimated MSE against the smoothing parameter h(4marks).Do the same for the case of local linear regression(4marks).4.Gather data for an additional predictor column(stating the source)and take as dependent variable one of%paper&cardboard,%glass,%plastic,%average or organic food spending per person,analyse it using one of the above methods and give reasons for your choice of the method(5marks).(2marks are awarded for clarity)References[1]Oecd productivity database[online].URL:/doc/3512512918.html,/I ndex.aspx?DataSetCode=PDB_LV[cited11.03.15].[2]David McCandless.Knowledge Is Beautiful.William Collins,2014.URL:http://bit.ly/KIB_Recyling.[3]Helga Willer and Julia Lernou.The World of Organic Agriculture.Research Institute of Organic Agriculture, 2015.URL:https:///doc/3512512918.html,/fil eadmin/documents/shop/1663-organic-world-2015.pdf.。
logistic逻辑回归公式推导及R语言实现
logistic逻辑回归公式推导及R语⾔实现Logistic逻辑回归Logistic逻辑回归模型线性回归模型简单，对于⼀些线性可分的场景还是简单易⽤的。
Logistic逻辑回归也可以看成线性回归的变种，虽然名字带回归⼆字但实际上他主要⽤来⼆分类，区别于线性回归直接拟合⽬标值，Logistic逻辑回归拟合的是正类和负类的对数⼏率。
假设有⼀个⼆分类问题，输出为y∈{0,1}定义sigmoid函数:⽤sigmoid函数的输出是0，1之间，⽤来拟合y=1的概率，其函数R语⾔画图如下：x = seq(-5, 5, 0.1)y = 1 / (1 + exp(-1*x))plot(x, y, type="line")logistic逻辑回归可以拟合因变量为1的概率,最终分类的时候，我们可以⼀个阈值，⽐如0.5，⼤于阈值的都分为正类，向量化公式如下：还可以换⼀种⽅式理解logistic逻辑回归，他是⽤多元线性函数去拟合因变量为正例与反例的⽐值的⾃然对数，推导如下：Logistic逻辑回归算法假设⾃变量维度为NW为⾃变量的系数，下标0 - NX为⾃变量向量或矩阵，X维度为N,为了能和W0对应，X需要在第⼀⾏插⼊⼀个全是1的列。
Y为因变量W为未知数待求解最⼤似然估计法梯度下降法迭代公式R语⾔实现使⽤iris数据集> head(iris)Sepal.Length Sepal.Width Petal.Length Petal.Width Species1 5.1 3.5 1.4 0.2 setosa2 4.9 3.0 1.4 0.2 setosa3 4.7 3.2 1.3 0.2 setosa4 4.6 3.1 1.5 0.2 setosa5 5.0 3.6 1.4 0.2 setosa6 5.4 3.9 1.7 0.4 setosa将数据分为训练数据和测试数据R语⾔使⽤批量梯度下降法迭代求解iris2 = rbind(subset(iris, Species=='setosa'), subset(iris, Species=='versicolor'))X <- cbind(rep(1, nrow(iris2)), iris2$Sepal.Length, iris2$Sepal.Width, iris2$Petal.Length, iris2$Petal.Width)Y <- as.numeric(iris2$Species) - 1maxIterNum <- 2000;step <- 0.05;W <- rep(0, ncol(X))m = nrow(X)sigmoid <- function(z) { 1 / (1 + exp(-z))}for (i in 1:maxIterNum){grad <- t(X) %*% (sigmoid(X %*% W)-Y);if (sqrt(as.numeric(t(grad) %*% grad)) < 1e-8){print(sprintf('iter times=%d', i));break;}W <- W - grad * step;}print(W);hfunc <- function(a) {if (a > 0.5) return(1) else return (0);}myY = apply(sigmoid(X %*% W), 1, hfunc)print(cbind(Y, myY))输出后，可以看到拟合完全正确，因为本⽂只是为了推导⼀下逻辑回归的算法，所以直接⽤全部数据拟合，没有再抽出⼀部分做测试数据。
R语言-医药研究 logistic regression
科学技术献祖国5医药研究 logistic regression介绍R 语言的应用实际上都是在介绍算法，算法的本质属于脑筋急转弯，所以算法是不能申请组专利保护的。
然而算法一旦固化到电子产品当中，产生了可以摸得到的商品,这种商品就可以注册版权，受知识产权的保护。
同样的新发现的药物也是可以申请专利受法律保护，这就是现在药物发掘的工作如火如荼的原因。
那些制药公司都希望尽可能多的申请药物专利从中谋利。
在这里我们将简单介绍新的药品是如何发现的。
药品实验都是在生物活体上进行的，使用的实验对象可以是：小白鼠，狗，或者是其它的单细胞，多细胞生物。
关键的一点是使用的实验对象必须是完全相同的，实验结果才能有可比性。
举一个例子，双胞胎个体就是完全相同的生物个体，因为它们是从单一的受精卵分裂出来的。
但实验中我们有时需要用到上百个完全相同的实验个体，通过使用双胞胎不可能得到足够的实验对象。
另外一种办法，可以通过选种的方法来生成众多的完全相同的实验个体。
例如说我们可以有一百个小白鼠，它们每一个都有着完全相同的基因组成。
因为这些小白鼠是通过选种培育出来的，它们都有着共同的单一祖先。
在下面的实验中，我们使用200个这样具有相同基因组成的小白鼠。
介绍疼痛实验，一般是使用一束强光照射到小白鼠的尾巴上。
如果小白鼠的尾巴突然间抽动就说明小白鼠有了疼痛的感觉，这种实验是用来研究哺乳类动物对疼痛承受能力的方法。
吗啡是一种止疼药物，使用吗啡后哺乳类动物都会迟钝对疼痛的感觉。
研究人员新近发现了另外一种药物被称作代尔9号也可以用作麻醉剂。
研究人员想用吗啡与代尔9号相比较来明白哪种药物更有效。
研究人员也想知道当这两种药物同时被使用的时候，它们的药性可不可以相互促进。
发现这种相互促进的药性是十分重要的，因为单一使用某种药物可能会毒性非常的大。
如果有另外一种药物能使它们之间互相促进，就可以在两种药物都使用非常小的剂量的状况下达到比较好的医治效果。
这种药物互相作用在研究杀虫剂的课题中也经常被用到。
logistic regression逻辑回归算法 -回复
logistic regression逻辑回归算法-回复Logistic regression, also known as the logit model, is a popular algorithm used in statistics and machine learning for binary classification tasks. It is widely used in various fields such as finance, healthcare, and marketing. In this article, we will explain the concept of logistic regression, its assumptions, the mathematical function behind it, and the steps involved in implementing it.1. Introduction to Logistic Regression:Logistic regression is a statistical model used to predict the likelihood of a binary outcome based on a set of independent variables. It is particularly useful when the dependent variable is categorical, taking values such as 0 or 1, true or false, or yes or no. The algorithm estimates the probability of the outcome using the logistic function, also known as the sigmoid function.2. Assumptions of Logistic Regression:Before applying logistic regression, several assumptions need to be met. These include:- The dependent variable is binary or ordinal.- Independence of observations.- Linearity between the logit of the outcome and the independentvariables.- Absence of multicollinearity.- Sufficient sample size.- Little to no outliers.3. Logistic Regression Function:The logistic regression function, also known as the logit function, calculates the log odds of the outcome variable being 1. It transforms the linear regression output into a range between 0 and 1, representing the probability of the outcome. The function is defined as:p = 1 / (1 + exp(-z))Where p represents the probability, and z is the linear combination of the independent variables:z = β0 + β1*X1 + β2*X2 + ... + βn*XnThe coefficients (β) represent the contribution of each independent variable to the log-odds of the outcome. These coefficients are estimated using the maximum likelihood estimation (MLE) method.4. Steps in Implementing Logistic Regression:a. Data Collection and Preparation:The first step in logistic regression is collecting and preparing the data. This involves deciding on the dependent variable, selecting relevant independent variables, ensuring data quality, handling missing values, and encoding categorical variables if necessary.b. Splitting the Data:To properly evaluate the performance of the logistic regression model, the dataset is divided into training and testing sets. The training set is used to fit the model, while the testing set is used to assess its performance.c. Model Training:The next step is fitting the logistic regression model on the training data. This involves estimating the coefficients (β) using the MLE method, which maximizes the likelihood of observing the given data given the model.d. Model Evaluation:After the model is trained, its accuracy, precision, recall, and otherperformance metrics are evaluated using the testing data. This step helps understand how well the model performs on unseen data.e. Model Improvement:Based on the evaluation results, adjustments can be made to improve the model's performance. This may include variable selection techniques, transforming variables, addressing outliers, or tuning hyperparameters.5. Interpretation of Results:Once the logistic regression model is trained and evaluated, the coefficients can be interpreted. Positive coefficients indicate a positive relationship with the outcome, while negative coefficients indicate a negative relationship. The magnitude of the coefficients represents the strength of the relationship.In conclusion, logistic regression is a powerful algorithm for binary classification tasks. It estimates the probability of a binary outcome using the logistic function and utilizes the MLE method to estimate coefficients. By following the steps mentioned above, one can implement logistic regression and interpret the results to makeinformed decisions based on the analysis.。
Logistics Regression(逻辑回归)
• 如果有一个超大的肿瘤在我们的例子中,阈值就很难设定使用线性的函数来拟合规律后取阈值的办法是行 不通的,离群值(也叫异常值)对结果的影响过大。
逻辑回归模型
逻辑函数的损失函数
若想让预测出的结果全部正确的概率最大,根据最大似然估计(多元线性回归推理中有讲过,
此处不再赘述),就是所有样本预测正确的概率相乘得到的P(总体正确)最大,此时我们
让
,数学表达Leabharlann 式如下:经过变形，取对数，转换，得到逻辑回归的损失函数——交叉熵损失函数。
逻辑回归(logistics regression)
输入变量与输出变量均为连续变量的预测问题是回归问题,输出变量为有限个离散变 量的预测问题成为分类问题.
逻辑回归也被称为对数几率回归，算法名虽然叫做逻辑回归，但是该算法是分类算法， 利用了和回归类似的方法解决问题。
• eg. 根据人的饮食,作息,工作和生存环境等条件预测一个人“有”或者“没有”得恶性肿瘤。通过线性回归加设 定阈值的办法,就可以完成一个简单的二分类任务。
r语言 逻辑斯的回归 阈值
r语言逻辑斯的回归阈值逻辑回归（Logistic Regression）是一种常用的分类算法，它通过将线性回归模型的输出值通过一个逻辑函数（Logistic函数）转换为概率值，从而实现对样本进行分类。
在逻辑回归中，阈值（Threshold）是决定样本分类的重要参数，本文将围绕阈值展开讨论。
让我们回顾一下逻辑回归的基本原理。
逻辑回归的目标是建立一个能够将输入变量映射到输出变量的函数，常用的逻辑函数是Sigmoid函数，其数学表达式为：$$f(z) = \frac{1}{1+e^{-z}}$$其中，$z$是线性回归模型的输出值。
Sigmoid函数具有将$z$映射到0到1之间的特性，这可以看作是样本属于某一类别的概率。
在进行分类时，我们需要设定一个阈值，将Sigmoid函数的输出结果映射为二分类的预测结果。
通常情况下，阈值被设定为0.5，即当Sigmoid函数的输出大于0.5时，将样本预测为正类；当Sigmoid 函数的输出小于0.5时，将样本预测为负类。
然而，在实际应用中，阈值的选择并不是固定的，它会根据具体的问题和需求而变化。
下面我们将从两个方面来探讨阈值的选择对逻辑回归模型的影响。
1. 准确率和召回率的平衡在二分类问题中，准确率（Precision）和召回率（Recall）是衡量模型性能的重要指标。
准确率指的是模型预测为正类的样本中真实为正类的比例，而召回率指的是真实为正类的样本中被模型预测为正类的比例。
当我们将阈值设定为0.5时，可能会导致准确率和召回率的不平衡。
如果将阈值调低（如0.3），将更多的样本预测为正类，可能会提高召回率，但同时也会降低准确率；反之，如果将阈值调高（如0.7），将更少的样本预测为正类，可能会提高准确率，但同时会降低召回率。
因此，在实际应用中，我们需要根据具体问题的需求来选择阈值。
如果在某个场景中更注重准确预测正类样本，可以选择较高的阈值；如果更注重捕捉到尽可能多的正类样本，可以选择较低的阈值。
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Logistic Regression Assumptions
1. The model is correctly specified, i.e.,
The true conditional probabilities are a logistic
function of the independent variables;
No important variables are omitted;
No extraneous variables are included; and
The independent variables are measured
without error.
2. The cases are independent.
3. The independent variables are not linear
combinations of each other.
Perfect multicollinearity makes estimation
impossible,
While strong multicollinearity makes estimates