如何选择R数据帧中不在向量中的列？
R数据帧可以具有很多列，我们可能希望选择其中几列。在这种情况下，最好通过取消选择不需要的列而不是选择我们需要的列来提取列，因为需要的列数多于不需要的列。借助！可以轻松完成此操作！标志和单个方括号。
示例
请看以下数据帧-
> Age<-sample(20:50,20)
> Gender<-rep(c("Male","Female"),times=10)
> Salary<-sample(20000:40000,20)
> ID<-1:20
> Education<-rep(c("Grad","Post-Grad","PhD","Highschool"),times=5)
> Experience<-sample(1:5,20,replace=TRUE)
> df<-data.frame(ID,Gender,Age,Salary,Experience,Education)
> df输出结果
ID Gender Age Salary Experience Education 1 1 Male 35 38245 4 Grad 2 2 Female 43 21995 5 Post-Grad 3 3 Male 21 38941 4 PhD 4 4 Female 26 36599 2 Highschool 5 5 Male 50 27477 2 Grad 6 6 Female 28 25281 2 Post-Grad 7 7 Male 33 20310 4 PhD 8 8 Female 24 30171 2 Highschool 9 9 Male 38 28779 3 Grad 10 10 Female 46 31213 3 Post-Grad 11 11 Male 36 27697 4 PhD 12 12 Female 41 36929 2 Highschool 13 13 Male 42 35367 2 Grad 14 14 Female 29 28711 1 Post-Grad 15 15 Male 22 29253 3 PhD 16 16 Female 30 28982 5 Highschool 17 17 Male 39 39458 4 Grad 18 18 Female 27 31891 2 Post-Grad 19 19 Male 48 29931 2 PhD 20 20 Female 31 34817 2 Highschool
选择除性别和年龄以外的列-
> df[,!names(df)%in%c("Gender","Age")]输出结果
ID Salary Experience Education 1 1 38245 4 Grad 2 2 21995 5 Post-Grad 3 3 38941 4 PhD 4 4 36599 2 Highschool 5 5 27477 2 Grad 6 6 25281 2 Post-Grad 7 7 20310 4 PhD 8 8 30171 2 Highschool 9 9 28779 3 Grad 10 10 31213 3 Post-Grad 11 11 27697 4 PhD 12 12 36929 2 Highschool 13 13 35367 2 Grad 14 14 28711 1 Post-Grad 15 15 29253 3 PhD 16 16 28982 5 Highschool 17 17 39458 4 Grad 18 18 31891 2 Post-Grad 19 19 29931 2 PhD 20 20 34817 2 Highschool
选择ID和教育以外的列-
> df[,!names(df)%in%c("ID","Education")]输出结果
Gender Age Salary Experience 1 Male 35 38245 4 2 Female 43 21995 5 3 Male 21 38941 4 4 Female 26 36599 2 5 Male 50 27477 2 6 Female 28 25281 2 7 Male 33 20310 4 8 Female 24 30171 2 9 Male 38 28779 3 10 Female 46 31213 3 11 Male 36 27697 4 12 Female 41 36929 2 13 Male 42 35367 2 14 Female 29 28711 1 15 Male 22 29253 3 16 Female 30 28982 5 17 Male 39 39458 4 18 Female 27 31891 2 19 Male 48 29931 2 20 Female 31 34817 2
选择ID，教育和性别以外的列-
> df[,!names(df)%in%c("ID","Education","Gender")]输出结果
Age Salary Experience 1 35 38245 4 2 43 21995 5 3 21 38941 4 4 26 36599 2 5 50 27477 2 6 28 25281 2 7 33 20310 4 8 24 30171 2 9 38 28779 3 10 46 31213 3 11 36 27697 4 12 41 36929 2 13 42 35367 2 14 29 28711 1 15 22 29253 3 16 30 28982 5 17 39 39458 4 18 27 31891 2 19 48 29931 2 20 31 34817 2
选择ID，年龄和性别以外的列-
> df[,!names(df)%in%c("ID","Age","Gender")]Salary Experience Education 1 38245 4 Grad 2 21995 5 Post-Grad 3 38941 4 PhD 4 36599 2 Highschool 5 27477 2 Grad 6 25281 2 Post-Grad 7 20310 4 PhD 8 30171 2 Highschool 9 28779 3 Grad 10 31213 3 Post-Grad 11 27697 4 PhD 12 36929 2 Highschool 13 35367 2 Grad 14 28711 1 Post-Grad 15 29253 3 PhD 16 28982 5 Highschool 17 39458 4 Grad 18 31891 2 Post-Grad 19 29931 2 PhD 20 34817 2 Highschool