R中字符处理

1. x1=c("1. aa","2. bb","15. cc","101. dd")
   
2. x2=c("(Submitter supplied) a","(Submitter supplied) cd","(Submitter supplied) a1","(Submitter supplied) abc")
   
3. x3=c("Organism:a","Organism:ab","Organism:ac","Organism:ef")
   
4. aaa<-data.frame(x1,x2,x3)
   
5. aaa

复制代码

希望得到的数据框
x1 x2 x3
aa a a
bb cd ab
cc a1 ac
dd abc ef
要求x1去掉字母前的数字 . 和空格。
      x2去掉(Submitter supplied)和空格，最好用识别")和空格"的函数实现
      x3去掉Organism:，最好用识别:的函数实现
谢谢！

1. library(tidyr)
   
2. aaa1<-separate(aaa,col=x1, into = c("num","x1"), sep=" ")
   
3. aaa2<- separate(aaa1,col=x2,into= c("ss","x2"),sep="\\) ")
   
4. aaa3<-separate(aaa2,col=x3,into=c("org","x3"),sep= ":")
   
5. aaa3
   
6. aaa4 <- aaa3[,c(2,4,6)]
   
7. aaa4

复制代码

不过这样做确实有些麻烦哈，应该还有比这更好的办法吧。

飞天玄舞6 发表于 2016-11-8 09:06
不过这样做确实有些麻烦哈，应该还有比这更好的办法吧。

期待更好地办法

.......

1. >  require(stringr)
   
2. > x1=c("1. aa","2. bb","15. cc","101. dd")
   
3. > x2=c("(Submitter supplied) a","(Submitter supplied) cd","(Submitter supplied) a1","(Submitter supplied) abc")
   
4. > x3=c("Organism:a","Organism:ab","Organism:ac","Organism:ef")
   
5. > aaa<-data.frame(x1,x2,x3)
   
6. >  
   
7. > apply(aaa,2,function(x) str_trim(str_sub(x,regexpr("\\. |\\) |:",x)+1),side = "both"))
   
8.      x1   x2    x3  
   
9. [1,] "aa" "a"   "a"
   
10. [2,] "bb" "cd"  "ab"
    
11. [3,] "cc" "a1"  "ac"
    
12. [4,] "dd" "abc" "ef"

复制代码

不懂不装懂 发表于 2016-11-8 17:30

如何将x3中提取的Organism作为x3的列名
colnames(aaa$x3)<-"Organism"
红色部分如何用提取的函数实现？

gsub("[0-9]+\\.+\\ ","",x1)->x1,可以了

> require(stringr)
> x3n<-str_trim(str_sub(x3,1,regexpr(":",x3)-1),side = "both")
> x3n
[1] "Organism" "Organism" "Organism" "Organism"
如果仅限于x3,也可用楼上的方法
> gsub(":[a-z]+","",x3)
[1] "Organism" "Organism" "Organism" "Organism"

hanson.don 发表于 2016-11-9 07:24
gsub("[0-9]+\\.+\\ ","",x1)->x1,可以了

x2 x3用gsub如何实现？谢谢！

dxystata 发表于 2016-11-17 20:34
x2 x3用gsub如何实现？谢谢！

gsub("\\(Submitter supplied\\) ","",x2)->X2
gsub("Organism:","",x3)->X3