[R语言]R包的安装&帮助获取


本文主要参考:Paul Teetor《R语言经典实例》一书

在R语言中,包含的包中有各种应用函数;
1.install.packages(‘packagename’)//安装R包
library(package_name)//载入包,对于base包可省略
2.library(help=”package_name”)//显示包的帮助,包括包的版本和包中函数
数据:对于R语言自带数据大多存在于utils包中,可以直接使用;

show(Titanic)
对于其他安装包中的数据载入
data(Cars93,package=”MASS”)
查看安装包的内容
data(“MASS”)
将从MASS包中载入Cars93数据
3.help(name)=?mean//help文档;
尤其的,help(name)可以用来进一步学习数据包/函数包中的属性内容、
help.start()//启动html文档
args(functionname)//返回函数的参数列表

args(lm)
function (formula, data, subset, weights, na.action, method = "qr", 
    model = TRUE, x = FALSE, y = FALSE, qr = TRUE, singular.ok = TRUE, 
    contrasts = NULL, offset, ...) 
NULL

example(functionname)//查看函数的使用实例
help.search(“name”)=??name//搜索本地计算机上已安装的帮助文档
help.search(“name”)//搜索所有包含该函数的R包列表
help(name,package=”packagename”)//在help.search()返回函数所在包中查询帮助文档
help(package=”tseries’)//查看安装的软件包的内容

>help(package="tseries")
 关于程辑包‘tseries’的信息

描述:

Package:            tseries
Version:            0.10-34
Title:              Time Series Analysis and Computational Finance
Authors@R:          c(person("Adrian", "Trapletti", role = "aut", email
                    = "adrian@trapletti.org"), person("Kurt", "Hornik",
                    role = c("aut", "cre"), email =
                    "Kurt.Hornik@R-project.org"), person("Blake",
                    "LeBaron", role = "ctb", comment = "BDS test
                    code"))
Description:        Time series analysis and computational finance.
Depends:            R (>= 2.10.0)
Suggests:           its
Imports:            graphics, stats, utils, quadprog, zoo
License:            GPL-2
Packaged:           2015-02-20 12:43:13 UTC; hornik
Author:             Adrian Trapletti [aut], Kurt Hornik [aut, cre],
                    Blake LeBaron [ctb] (BDS test code)
Maintainer:         Kurt Hornik @R-project.org>

RSiteSearch(“key phrase”)//通过网络搜索信息

RSiteSearch("RSS");//返回对RSS的网络解释

4.回归分析中的英文:

sst:sum of square of Total=TSS:total sum of squares
sse:sum of square for Error=RSS:residual sum of squares
ssr:sum of square for regression=ESS:explained sum of squares
residual:残余的,残差;
Residual Standard Error:标准差;
intercept:截距;
Multiple R-squared:??;adjusted R-squared:??;
df:degree of freedom;anova:analysis of variance;//方差分析;
AIC:Akaike information criterion,最小信息准则;
RSS:'1.residual sum of squares表示一组统计量的残差平方和;'
     '2.root of sum of squares表示一组统计量平方和的开方值;'
**这两种说法有一定的语境**
sum of sq:偏回归平方和,注意平回归平方和表示剔除该变量后回归平方和减少的数值;

5.数据对象:
基本类型:numeric(),character(),complex(),logical()
复合类型/递归类型:list,function,expression
mode()//返回类型;length()//返回长度;attributes(object)//返回对象的
所有非基本属性(基本属性包括mode和length);attr(object,name)//用于选取
指定的属性;attr(z,"dim")<-c(10,10)//将z的dim属性重写为10,10,即将z属性
变为一个10*10矩阵;数组的维数dim(z)的升序是从第一位开始,即z[1,1],z[2,1]
z[3,1],z[1,2]...但对于矩阵而言维数的意义相同,第一个表示行,第二个表示列;
6.向量赋值时,当向量索引超过定义长度时,赋值后的向量被延长,超出的长度未被赋值的元素值为NA;
7.表达:
1:3//1,2,3;
2*1:3//2,4,6;
R语言在定义中设置属性值,如:

x<-array(1:20,dim<-c(4,5))//表示将array的维数设为4*5,数组元素是1:20;
i<-array(c(1:3,3:1),dim<-c(3,2));//输出为
   [,1][,2]
[1,] 1   3
[2,] 2   2
[3,] 3   1
x[i]=0;//

类似的

matrix(矩阵值,维数)//生成矩阵

8.step()逐步回归法的实例

>      summary(lm1 <- lm(Fertility ~ ., data = swiss))

Call:
lm(formula = Fertility ~ ., data = swiss)

Residuals:
     Min       1Q   Median       3Q      Max 
-15.2743  -5.2617   0.5032   4.1198  15.3213 

Coefficients:
                 Estimate Std. Error t value Pr(>|t|)    
(Intercept)      66.91518   10.70604   6.250 1.91e-07 ***
Agriculture      -0.17211    0.07030  -2.448  0.01873 *  
Examination      -0.25801    0.25388  -1.016  0.31546    
Education        -0.87094    0.18303  -4.758 2.43e-05 ***
Catholic          0.10412    0.03526   2.953  0.00519 ** 
Infant.Mortality  1.07705    0.38172   2.822  0.00734 ** 
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 7.165 on 41 degrees of freedom
Multiple R-squared:  0.7067,    Adjusted R-squared:  0.671 
F-statistic: 19.76 on 5 and 41 DF,  p-value: 5.594e-10

>      slm1 <- step(lm1)
Start:  AIC=190.69
Fertility ~ Agriculture + Examination + Education + Catholic + 
    Infant.Mortality

                   Df Sum of Sq    RSS    AIC
- Examination       1     53.03 2158.1 189.86
                          2105.0 190.69
- Agriculture       1    307.72 2412.8 195.10
- Infant.Mortality  1    408.75 2513.8 197.03
- Catholic          1    447.71 2552.8 197.75
- Education         1   1162.56 3267.6 209.36

Step:  AIC=189.86
Fertility ~ Agriculture + Education + Catholic + Infant.Mortality

                   Df Sum of Sq    RSS    AIC
                          2158.1 189.86
- Agriculture       1    264.18 2422.2 193.29
- Infant.Mortality  1    409.81 2567.9 196.03
- Catholic          1    956.57 3114.6 205.10
- Education         1   2249.97 4408.0 221.43
>      summary(slm1)

Call:
lm(formula = Fertility ~ Agriculture + Education + Catholic + 
    Infant.Mortality, data = swiss)

Residuals:
     Min       1Q   Median       3Q      Max 
-14.6765  -6.0522   0.7514   3.1664  16.1422 

Coefficients:
                 Estimate Std. Error t value Pr(>|t|)    
(Intercept)      62.10131    9.60489   6.466 8.49e-08 ***
Agriculture      -0.15462    0.06819  -2.267  0.02857 *  
Education        -0.98026    0.14814  -6.617 5.14e-08 ***
Catholic          0.12467    0.02889   4.315 9.50e-05 ***
Infant.Mortality  1.07844    0.38187   2.824  0.00722 ** 
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 7.168 on 42 degrees of freedom
Multiple R-squared:  0.6993,    Adjusted R-squared:  0.6707 
F-statistic: 24.42 on 4 and 42 DF,  p-value: 1.717e-10

>      slm1$anova
           Step Df Deviance Resid. Df Resid. Dev      AIC
1               NA       NA        41   2105.043 190.6913
2 - Examination  1 53.02656        42   2158.069 189.8606
> 

数据类型转换

#1.基本(原子)数据类型之间的转换
as.character(x)
as.complex(x)
as.numeric(x)
as.integer(x)
as.logical(x)
#2.结构化数据类型间转换
as.data.frame(x)
as.list(x)
as.matrix(x)
as.vector(x)
#更多具体变换见《R语言经典实例》P156

添加某一列,假设d为list

#注意该方法对data_frame类型的数据有效,因此要先转换为data_frame类型
d0<-as.data.frame(d)
d0$column_name<-column_value
#这样d0中就加入了column_name这列,值为向量column_name

删除某一列/列

#删除第一行
datatest<-datatest[-1]
#删除第二行
datatest<-datatest[-2]
#删除第一列
datatest<-datatest[,-1]
#删除第二列
datatest<-datatest[,-2]

按某一列排序

#按datatest的第五列,升序排列
datatest[order(datatest[,5],decreasing=F),]
#同理可类推其他排序方法

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