R语言-入门小练习

R语言实战上的题
一组学生参加了数学、科学和英语考试。为了给所有的学生确定一个单一的成绩衡量指标，需要将这些科目的成绩组合起来。另外，还想将前20%的学生评定为A，接下来20%的学生评定为B，以此类推。最后，希望按字母顺序对学生排序。

1、数据录入

>StuId <- c(1:10)

>StuName <- c('John Davis','Angela Williams','Bull >Jones','Cheryl Cushing','Reuven Ytzrhak','Joel Knox','Mary Rayburn','Greg England','Brad Tmac','Tracy Mcgrady')

>Math <- c(502,465,621,575,454,634,576,421,599,666)

>Science <- c(95,67,78,66,96,89,78,56,68,10)

>English <- c(25,12,22,18,15,30,37,12,22,38)

>mydata <- data.frame(StuId,StuName,Math,Science,English)

>mydata

    Stuld         StuName Math Science English

1      1      John Davis  502      95      25

2      2 Angela Williams  465      67      12

3      3      Bull Jones  621      78      22

4      4  Cheryl Cushing  575      66      18

5      5  Reuven Ytzrhak  454      96      15

6      6       Joel Knox  634      89      30

7      7    Mary Rayburn  576      78      37

8      8    Greg England  421      56      12

9      9       Brad Tmac  599      68      22

10    10   Tracy Mcgrady  666      10      38

2、数据预处理-数据中心化、标准化 

scale(data,center=T,scale=T) 
center=T 数据中心化 
scale=T 数据标准化 
数据的中心化：指数据集中的各项数据减去数据集的均值 
数据的标准化：值中心化后的数据在除以数据集的标准差 
数据中心化和标准化的意义是一样的，为了消除量纲对数据结构的影响

>z <- scale(mydata[3:5])

>z

            Math     Science    English

 [1,] -0.5817963  0.98895866  0.2006619

 [2,] -1.0184385 -0.13212808 -1.1722882

 [3,]  0.8225395  0.30829885 -0.1161727

 [4,]  0.2796870 -0.17216689 -0.5386189

 [5,] -1.1482510  1.02899747 -0.8554535

 [6,]  0.9759544  0.74872578  0.7287197

 [7,]  0.2914882  0.30829885  1.4680005

 [8,] -1.5376887 -0.57255501 -1.1722882

 [9,]  0.5629144 -0.09208927 -0.1161727

[10,]  1.3535909 -2.41434036  1.5736120

attr(,"scaled:center")

   Math Science English 

  551.3    70.3    23.1 

attr(,"scaled:scale")

     Math   Science   English 

84.737569 24.975766  9.468662 

3、计算每个学生成绩的均值 

apply(X,MARGIN,FUN,…) 
X:阵列包括矩阵 
MARGIN:1表示矩阵行，2表示矩阵列，也可以是c(1,2) 
FUN:表示具体的运算方式

>mydata$score <- apply(z,1,mean)

>mydata

   Stuld         StuName Math Science English      score

1      1      John Davis  502      95      25  0.2026081

2      2 Angela Williams  465      67      12 -0.7742849

3      3      Bull Jones  621      78      22  0.3382219

4      4  Cheryl Cushing  575      66      18 -0.1436996

5      5  Reuven Ytzrhak  454      96      15 -0.3249024

6      6       Joel Knox  634      89      30  0.8177999

7      7    Mary Rayburn  576      78      37  0.6892625

8      8    Greg England  421      56      12 -1.0941773

9      9       Brad Tmac  599      68      22  0.1182175

10    10   Tracy Mcgrady  666      10      38  0.1709542

4、算出四分位数 

quantile(x,probs) 求分位数。 
x：待求分位数的数值型向量， 
pobs：一个由[0,1]之间的概率值组成的数值向量

>q<-quantile(mydata$score,c(.8,.6,.4,.2)) 

>q

        80%         60%         40%         20% 

 0.40843002  0.18361575  0.01345066 -0.41477887 

5、根据四分位数给出分数等级

>mydata$level[score>=q[1]] <- "A"

>mydata$level[score>=q[2] & score<q[1]] <- "B"

>mydata$level[score>=q[3] & score<q[2]] <- "C"

>mydata$level[score>=q[4] & score<q[3]] <- "D"

>mydata$level[score<q[4]] <- "E"

6、拆分StuName列为FirName和LasName两列 

拆分这里有两个方法，方法一是我用的笨方法；方法二使用sapply()函数 
strsplit(x,split=”“) 
x 是向量 
split是以什么分隔 
方法一

>name<-strsplit(mydata$StuName,split=" ")--这句语句错误 x应为向量

--Error in strsplit(mydata$StuName, split = " ") : non-character argument

>name<-strsplit(StuName,split=" ")--以空格分隔,

>name --把strsplit()应用到一个字符串组成的向量上会返回一个列表

[[1]]

[1] "John"  "Davis"



[[2]]

[1] "Angela"   "Williams"



[[3]]

[1] "Bull"  "Jones"



[[4]]

[1] "Cheryl"  "Cushing"

...

>name<-unlist(strsplit(StuName,split=" "))--将list形式转化成向量格式

> name

 [1] "John"     "Davis"    "Angela"   "Williams" "Bull"     "Jones"    "Cheryl"   "Cushing"  "Reuven"  

[10] "Ytzrhak"  "Joel"     "Knox"     "Mary"     "Rayburn"  "Greg"     "England"  "Brad"     "Tmac"    

[19] "Tracy"    "Mcgrady" 

>name<-matrix(name,ncol=2,byrow=T)

>name

      [,1]     [,2]      

 [1,] "John"   "Davis"   

 [2,] "Angela" "Williams"

 [3,] "Bull"   "Jones"   

 [4,] "Cheryl" "Cushing" 

 [5,] "Reuven" "Ytzrhak" 

 [6,] "Joel"   "Knox"    

 [7,] "Mary"   "Rayburn" 

 [8,] "Greg"   "England" 

 [9,] "Brad"   "Tmac"    

[10,] "Tracy"  "Mcgrady"

>mydata<-cbind(mydata,FirName,LasName)

>FirName<-name[,1]

>LasName<-name[,2]

方法二： 
sapply() 提取列表中每个成分的第一个元素， 
“[”是一个可以提取某个对象一部分的函数

>name<-strsplit((StuName)," ")

>lastname<-sapply(name,"[",2)

>lastname

 [1] "Davis"    "Williams" "Jones"    "Cushing"  "Ytzrhak"  "Knox"     "Rayburn"  "England"  "Tmac"    

[10] "Mcgrady" 

>firstname<-sapply(name,"[",1)

>firstname

 [1] "John"   "Angela" "Bull"   "Cheryl" "Reuven" "Joel"   "Mary"   "Greg"   "Brad"   "Tracy" 

7、合并&排序

>mydata<-cbind(mydata,FirName,LasName)

>mydata

   Stuld Math Science English      score FirName  LasName level

4      4  575      66      18 -0.1436996  Cheryl  Cushing     D

1      1  502      95      25  0.2026081    John    Davis     B

8      8  421      56      12 -1.0941773    Greg  England     E

3      3  621      78      22  0.3382219    Bull    Jones     B

6      6  634      89      30  0.8177999    Joel     Knox     A

10    10  666      10      38  0.1709542   Tracy  Mcgrady     c

7      7  576      78      37  0.6892625    Mary  Rayburn     A

9      9  599      68      22  0.1182175    Brad     Tmac     c

2      2  465      67      12 -0.7742849  Angela Williams     E

5      5  454      96      15 -0.3249024  Reuven  Ytzrhak     D

> options(digits=2)--保留小数点后两位

> mydata<-mydata[order(mydata$LasName,mydata$FirName),]--排序

> mydata

   Stuld Math Science English score FirName  LasName level

4      4  575      66      18 -0.14  Cheryl  Cushing     D

1      1  502      95      25  0.20    John    Davis     B

8      8  421      56      12 -1.09    Greg  England     E

3      3  621      78      22  0.34    Bull    Jones     B

6      6  634      89      30  0.82    Joel     Knox     A

10    10  666      10      38  0.17   Tracy  Mcgrady     c

7      7  576      78      37  0.69    Mary  Rayburn     A

9      9  599      68      22  0.12    Brad     Tmac     c

2      2  465      67      12 -0.77  Angela Williams     E

5      5  454      96      15 -0.32  Reuven  Ytzrhak     D