在R中的数据框的一列中查找每个字符串的长度

Question

我希望计算name 列的每个字符串的字符数。我的数据框sample 如下所示：

date        name           expenditure      type
23MAR2013   KOSH ENTRP     4000             COMPANY
23MAR2013   JOHN DOE       800              INDIVIDUAL
24MAR2013   S KHAN         300              INDIVIDUAL
24MAR2013   JASINT PVT LTD 8000             COMPANY
25MAR2013   KOSH ENTRPRISE 2000             COMPANY
25MAR2013   JOHN S DOE     220              INDIVIDUAL
25MAR2013   S KHAN         300              INDIVIDUAL
26MAR2013   S KHAN         300              INDIVIDUAL

为什么nchar 给了我一个随机数列表？ str_length() 来自 stringr 包也是如此

Length <- aggregate(nchar(sample$name), by=list(sample$name), FUN=nchar)

输出

         Group.1       x
1 JASINT PVT LTD       2
2       JOHN DOE       1
3     JOHN S DOE       2
4     KOSH ENTRP       2
5 KOSH ENTRPRISE       2
6         S KHAN 1, 1, 1

期望的输出：

     Group.1       x
1 JASINT PVT LTD       14
2       JOHN DOE       8
3     JOHN S DOE       10
4     KOSH ENTRP       10
5 KOSH ENTRPRISE       14
6         S KHAN       6

---

上表的csv：

"Date","name","expenditure","type"
"23MAR2013","KOSH ENTRP",4000,"COMPANY"
"23MAR2013 ","JOHN DOE",800,"INDIVIDUAL"
"24MAR2013","S KHAN",300,"INDIVIDUAL"
"24MAR2013","JASINT PVT LTD",8000,"COMPANY"
"25MAR2013","KOSH ENTRPRISE",2000,"COMPANY"
"25MAR2013","JOHN S DOE",220,"INDIVIDUAL"
"25MAR2013","S KHAN",300,"INDIVIDUAL"
"26MAR2013","S KHAN",300,"INDIVIDUAL"

Answer 1

你也可以applynchar到你的dataframe并从对应的列中获取结果：

data.frame(names=temp$name,chr=apply(temp,2,nchar)[,2])
      names chr
1     KOSH ENTRP  10
2       JOHN DOE   8
3         S KHAN   6
4 JASINT PVT LTD  14
5 KOSH ENTRPRISE  14
6     JOHN S DOE  10
7         S KHAN   6
8         S KHAN   6

Answer 2

如果“Desired Output”中的最后一行是错字，

 aggregate(name~name1, transform(sample, name1=name),
                         FUN=function(x) nchar(unique(x)))
 #            name1 name
 #1 JASINT PVT LTD   14
 #2       JOHN DOE    8
 #3     JOHN S DOE   10
 #4     KOSH ENTRP   10
 #5 KOSH ENTRPRISE   14
 #6         S KHAN    6

或者

 Un1 <- unique(sample$name)
 data.frame(Group=Un1, x=nchar(Un1))

Answer 3

或者，使用data.table

dtx[,PepSeqLen := nchar(PepSeq)]