以前没有接触过R语言，第一天学习R语言，生信技能树团队的齐老师安排作业给我，我觉得有作业才有目标性和动力去搜索资源。通过网络搜索和参考相关书籍以及齐老师的指导，顺利地完成作业：
1.创建一个向量元素为a,b,e,2,5
work <- c(a,b,e)
data <- c(2,5)
2.确定上述向量中每个元素的数据类型
> class(work[1])
[1] "character"
> class(work[2])
[1] "character"
> class(work[3])
[1] "character"
> class(data[1])
[1] "numeric"
> class(data[2])
[1] "numeric"
3.创建一个包含字符和数值的数据框（5行3列）
patientID <- c(1,2,3,4,5)
age <- c(25,32,38,43,60)
breastcancer <- c("Type1","Type1","Type3","Type1","Type2")
patientdata <- data.frame(patientID,age,breastcancer)
patientdata
patientID age breastcancer
1 1 25 Type1
2 2 32 Type1
3 3 38 Type3
4 4 43 Type1
5 5 60 Type2
4.对上述数据框的行名和列名进行更改
rownames(patientdata) <- c("a","b","c","d","e")
colnames(patientdata) <- c("personalities","years","cancer")
patientdata
personalities years cancer
a 1 25 Type1
b 2 32 Type1
c 3 38 Type3
d 4 43 Type1
e 5 60 Type2
5.确定数据框中各列的数据类型
class(patientdata[,1])
class(patientdata[,2])
class(patientdata[,3])
> class(patientdata[,1])
[1] "numeric"
> class(patientdata[,2])
[1] "numeric"
> class(patientdata[,3])
[1] "factor"
or
class(patientdata$personalities)
class(patientdata$years)
class(patientdata$cancer)
> class(patientdata$personalities)
[1] "numeric"
> class(patientdata$years)
[1] "numeric"
> class(patientdata$cancer)
[1] "factor"
6.不同方法检索数据框的第3行第2列的元素
patientdata[3,2]
patientdata['c','years']
> patientdata[3,2]
[1] 38
> patientdata['c','years']
[1] 38
7.创建一个矩阵（4行6列）
Y <- matrix(1:20,nrow=5,ncol=4)
Y
> Y
[,1] [,2] [,3] [,4]
[1,] 1 6 11 16
[2,] 2 7 12 17
[3,] 3 8 13 18
[4,] 4 9 14 19
[5,] 5 10 15 20
8.对上述矩阵的行名和列名进行更改
colnames(Y)<-c("R1","R2","R3","R4")
rownames(Y)<-c("C1","C2","C3","C4","C5")
R1 R2 R3 R4
C1 1 6 11 16
C2 2 7 12 17
C3 3 8 13 18
C4 4 9 14 19
C5 5 10 15 20
9.不同方法检索数据框的第3行第2列的元素
Y[3,2]
Y["C3","R2"]
> Y[3,2]
[1] 8
> Y["C3","R2"]
[1] 8
将作业写至简书邮件jmzeng1314@163.com，抄送juanyuanjing@1163.com
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