还是举一个例子引出高级数据管理的操作
首先创建一个数据框
> Student<-c("John Davis","Angela Williams","Bullwinkle Moose","David Jones","Janice Markhammer","Cheryl Cushing","Reuven Ytzrhak","Greg Knox","Joel Enghland","Marry Rayburn")
> math<-c(502,600,412,358,495,512,410,625,573,522)
> Science<-c(95,99,80,82,75,85,80,95,89,86)
> English<-c(25,22,18,15,20,28,15,30,27,18)
> grade<-data.frame(Student,math,Science,English)
> grade
Student math Science English
1 John Davis 502 95 25
2 Angela Williams 600 99 22
3 Bullwinkle Moose 412 80 18
4 David Jones 358 82 15
5 Janice Markhammer 495 75 20
6 Cheryl Cushing 512 85 28
7 Reuven Ytzrhak 410 80 15
8 Greg Knox 625 95 30
9 Joel Enghland 573 89 27
10 Marry Rayburn 522 86 18
现在要解决的问题如下：
1.给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
2.将成绩前20%的评为A，接下来的20%评定为B，以此类推
3.按学生名字的字母顺序对学生进行排序
在解决问题之前我们有必要了解数值和字符处理函数
数学函数
abs(x) 绝对值
sqrt(x) 平方根
ceiling(x) 不小于x的最小整数
floor(x) 不大于x的最大整数
trunc(x) 取整
round(x,digits=n) 将x舍入为指定位的小数
signif(x,digits=n) 将x舍入为指定的有效数字位数
统计函数
mean(x) 平均数
median(x) 中位数
sd(x) 标准差
var(x) 方差
mad(x) 绝对中位差
quantile(x,probs) 求分位数
range(x) 求值域
sum(x) 求和
diff(x,lag=n) 滞后差分
min(x) 最小值
max(x) 最大值
scale() 将各向量值中心化后标准化
字符处理函数
nchar(x) 统计x中字符数量
sub(pattern,replacement,ignore.case=FALSE,fixed=FALSE) 在x中搜索
pattern，并以文本replacement替换
strsplit(x,split,fixed=FALSE) 在x处分割字符向量x中的元素
接下来我们着手处理一下上面提到的问题
1.将学生的各科成绩组合为单一的成绩衡量指标
> grade
Student math Science English
1 John Davis 502 95 25
2 Angela Williams 600 99 22
3 Bullwinkle Moose 412 80 18
4 David Jones 358 82 15
5 Janice Markhammer 495 75 20
6 Cheryl Cushing 512 85 28
7 Reuven Ytzrhak 410 80 15
8 Greg Knox 625 95 30
9 Joel Enghland 573 89 27
10 Marry Rayburn 522 86 18
> z<-scale(grade[,2:4]) #将各科分数分别中心化后标准化以便于比较
> z
math Science English
[1,] 0.01269128 1.07806562 0.58685145
[2,] 1.14336936 1.59143020 0.03667822
[3,] -1.02568654 -0.84705156 -0.69688609
[4,] -1.64871324 -0.59036927 -1.24705932
[5,] -0.06807144 -1.48875728 -0.33010394
[6,] 0.12806660 -0.20534583 1.13702468
[7,] -1.04876160 -0.84705156 -1.24705932
[8,] 1.43180765 1.07806562 1.50380683
[9,] 0.83185601 0.30801875 0.95363360
[10,] 0.24344191 -0.07700469 -0.69688609
attr(,"scaled:center")
math Science English
500.9 86.6 21.8
attr(,"scaled:scale")
math Science English
86.673654 7.791734 5.452828
> score<-apply(z,1,mean) #分别对z的每行求均值
> score
[1] 0.5592028 0.9238259 -0.8565414 -1.1620473 -0.6289776 0.3532485 -1.0476242
[8] 1.3378934 0.6978361 -0.1768163
> grade<-cbind(grade,score)
> grade #将比较得分与数据框结合
Student math Science English score
1 John Davis 502 95 25 0.5592028
2 Angela Williams 600 99 22 0.9238259
3 Bullwinkle Moose 412 80 18 -0.8565414
4 David Jones 358 82 15 -1.1620473
5 Janice Markhammer 495 75 20 -0.6289776
6 Cheryl Cushing 512 85 28 0.3532485
7 Reuven Ytzrhak 410 80 15 -1.0476242
8 Greg Knox 625 95 30 1.3378934
9 Joel Enghland 573 89 27 0.6978361
10 Marry Rayburn 522 86 18 -0.1768163
> y<-quantile(grade$score,c(0.8,0.6,0.4,0.2)) #用quantile函数计算出各个分位数的临界点
> y
80% 60% 40% 20%
0.7430341 0.4356302 -0.3576808 -0.8947579
#对各个学生得分的百分数排名重编码为一个新的类别型等级变量
> grade$level[grade$score>=y[1]]<-"A"
> grade$level[grade$score<y[1] & grade$score>=y[2]]<-"B"
> grade$level[grade$score<y[2] & grade$score>=y[3]]<-"C"
> grade$level[grade$score<y[3] & grade$score>=y[4]]<-"D"
> grade$level[grade$score<y[4]]<-"F"
> grade
Student math Science English score level
1 John Davis 502 95 25 0.5592028 B
2 Angela Williams 600 99 22 0.9238259 A
3 Bullwinkle Moose 412 80 18 -0.8565414 D
4 David Jones 358 82 15 -1.1620473 F
5 Janice Markhammer 495 75 20 -0.6289776 D
6 Cheryl Cushing 512 85 28 0.3532485 C
7 Reuven Ytzrhak 410 80 15 -1.0476242 F
8 Greg Knox 625 95 30 1.3378934 A
9 Joel Enghland 573 89 27 0.6978361 B
10 Marry Rayburn 522 86 18 -0.1768163 C
#用strsplit()函数将学生的姓和名拆分
> name<-strsplit(grade$Student," ")
Error in strsplit(grade$Student, " ") : non-character argument
#这里报错了，因为Student这个变量不是字符串变量
> is.character(grade$Student)
[1] FALSE
> class(grade$Student)
[1] "factor"
#是因子变量，我们将它转化为字符串
> grade$Student<-as.character(grade$Student)
> name<-strsplit((grade$Student)," ")
> name
[[1]]
[1] "John" "Davis"
[[2]]
[1] "Angela" "Williams"
[[3]]
[1] "Bullwinkle" "Moose"
[[4]]
[1] "David" "Jones"
[[5]]
[1] "Janice" "Markhammer"
[[6]]
[1] "Cheryl" "Cushing"
[[7]]
[1] "Reuven" "Ytzrhak"
[[8]]
[1] "Greg" "Knox"
[[9]]
[1] "Joel" "Enghland"
[[10]]
[1] "Marry" "Rayburn"
#用sapply()函数提取列表每个成分的第一个元素作为Firstname第二个元素作为Lastname
> Firstname<-sapply(name,"[",1)
> Lastname<-sapply(name,"[",2)
> Firstname
[1] "John" "Angela" "Bullwinkle" "David" "Janice" "Cheryl"
[7] "Reuven" "Greg" "Joel" "Marry"
> Lastname
[1] "Davis" "Williams" "Moose" "Jones" "Markhammer" "Cushing"
[7] "Ytzrhak" "Knox" "Enghland" "Rayburn"
#删除原有的姓名，将拆分后的姓氏和姓名与数据框结合
> grade<-grade[,-1]
> grade
math Science English score level
1 502 95 25 0.5592028 B
2 600 99 22 0.9238259 A
3 412 80 18 -0.8565414 D
4 358 82 15 -1.1620473 F
5 495 75 20 -0.6289776 D
6 512 85 28 0.3532485 C
7 410 80 15 -1.0476242 F
8 625 95 30 1.3378934 A
9 573 89 27 0.6978361 B
10 522 86 18 -0.1768163 C
> grade<-cbind(Firstname,Lastname,grade)
> grade
Firstname Lastname math Science English score level
1 John Davis 502 95 25 0.5592028 B
2 Angela Williams 600 99 22 0.9238259 A
3 Bullwinkle Moose 412 80 18 -0.8565414 D
4 David Jones 358 82 15 -1.1620473 F
5 Janice Markhammer 495 75 20 -0.6289776 D
6 Cheryl Cushing 512 85 28 0.3532485 C
7 Reuven Ytzrhak 410 80 15 -1.0476242 F
8 Greg Knox 625 95 30 1.3378934 A
9 Joel Enghland 573 89 27 0.6978361 B
10 Marry Rayburn 522 86 18 -0.1768163 C
#最后一步，按照姓名和姓氏进行排序
> grade[order(Firstname,Lastname),]
Firstname Lastname math Science English score level
2 Angela Williams 600 99 22 0.9238259 A
3 Bullwinkle Moose 412 80 18 -0.8565414 D
6 Cheryl Cushing 512 85 28 0.3532485 C
4 David Jones 358 82 15 -1.1620473 F
8 Greg Knox 625 95 30 1.3378934 A
5 Janice Markhammer 495 75 20 -0.6289776 D
9 Joel Enghland 573 89 27 0.6978361 B
1 John Davis 502 95 25 0.5592028 B
10 Marry Rayburn 522 86 18 -0.1768163 C
7 Reuven Ytzrhak 410 80 15 -1.0476242 F
#当然现实一点我们也可以按得分高低排序
> grade[order(-score),]
Firstname Lastname math Science English score level
8 Greg Knox 625 95 30 1.3378934 A
2 Angela Williams 600 99 22 0.9238259 A
9 Joel Enghland 573 89 27 0.6978361 B
1 John Davis 502 95 25 0.5592028 B
6 Cheryl Cushing 512 85 28 0.3532485 C
10 Marry Rayburn 522 86 18 -0.1768163 C
5 Janice Markhammer 495 75 20 -0.6289776 D
3 Bullwinkle Moose 412 80 18 -0.8565414 D
7 Reuven Ytzrhak 410 80 15 -1.0476242 F
4 David Jones 358 82 15 -1.1620473 F
#还可以输出为Excel表格
> grade<-grade[order(-score),]
> write.csv(grade,file = "grade.csv")
image.png
任务完成
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