Question

我有一个如下所示的数据框：

for (N in seq(100,500,50)) {
  P <- rep(NA,1000)
  for (i in 1:1000) {
    P[i] <- function()
  }
  result <- data.frame(P) # this is simplified 
}

我想将特定行的值（基于条件选择或类似的东西）复制到新列。我想定位第二个单词的最后一个字母（RowID ID1 ID2 ID3 Word1 Word2 Word3 Letter 1 1 2 3 the nice apple t 2 1 2 3 the nice apple h 3 1 2 3 the nice apple e 4 1 2 3 the nice apple n 5 1 2 3 the nice apple i 6 1 2 3 the nice apple c 7 1 2 3 the nice apple e 8 1 2 3 the nice apple a 9 1 2 3 the nice apple p 10 1 2 3 the nice apple p 11 1 2 3 the nice apple l 12 1 2 3 the nice apple e 13 6 7 8 yes did you y 14 6 7 8 yes did you e 15 6 7 8 yes did you s 16 6 7 8 yes did you d 17 6 7 8 yes did you i 18 6 7 8 yes did you d 19 6 7 8 yes did you y 20 6 7 8 yes did you o 21 6 7 8 yes did you u中的值与Letter中的值相同），在这些示例的情况下是这些行：

Word2

然后向数据框添加新列7 1 2 3 the nice apple e 18 6 7 8 yes did you d，如下所示：

LastLetterWord2

是否可以在R中执行此操作，如果是，如何执行此操作？（我还不是R的专家）。一个可能的问题可能是单词不是唯一的，可能会多次出现并处于不同的位置（如Word1，Word2或Word3）。

Answer 1

使用substr和nchar，您可以轻松创建一个新列，其最后一个字母为Word2列：

DF$LastLetterWord2 <- substr(DF$Word2,nchar(DF$Word2),nchar(DF$Word2))

> DF
   RowID ID1 ID2 ID3 Word1 Word2 Word3 Letter LastLetterWord2
1      1   1   2   3   the  nice apple      t               e
2      2   1   2   3   the  nice apple      h               e
3      3   1   2   3   the  nice apple      e               e
4      4   1   2   3   the  nice apple      n               e
5      5   1   2   3   the  nice apple      i               e
6      6   1   2   3   the  nice apple      c               e
7      7   1   2   3   the  nice apple      e               e
8      8   1   2   3   the  nice apple      a               e
9      9   1   2   3   the  nice apple      p               e
10    10   1   2   3   the  nice apple      p               e
11    11   1   2   3   the  nice apple      l               e
12    12   1   2   3   the  nice apple      e               e
13    13   6   7   8   yes   did   you      y               d
14    14   6   7   8   yes   did   you      e               d
15    15   6   7   8   yes   did   you      s               d
16    16   6   7   8   yes   did   you      d               d
17    17   6   7   8   yes   did   you      i               d
18    18   6   7   8   yes   did   you      d               d
19    19   6   7   8   yes   did   you      y               d
20    20   6   7   8   yes   did   you      o               d
21    21   6   7   8   yes   did   you      u               d

Answer 2

使用stringr库中的str_sub也会这样做：

df$LastLetterWord2 =stringr::str_sub(df$Word2, start = -1, end = -1) # using minus to indicate from the end

Answer 3

如果您的表名是DF。

数据

RowID <- 1:21 ID1 <- c(rep(1,12),rep(6,9)) ID2 <- c(rep(2,12),rep(7,9)) ID3 <- c(rep(3,12),rep(8,9)) Word1 <- c(rep("the",12),rep("yes",9)) Word2 <- c(rep("nice",12),rep("did",9)) Word3 <- c(rep("apple",12),rep("you",9)) Letter <- c("t","h","e","n","i","c","e","a","p","p","l","e","y","e","s","d","i","d","y","o","u") DF <- data.frame(RowID,ID1,ID2,ID3,Word1,Word2,Word3,Letter)

<强>命令

nchar返回字符向量的大小。 substring从字符向量中提取字母。

numberofcharacter <- data.frame(apply(DF,2,nchar)) DF$LastLetterWord2 <- substring(DF$Word2,numberofcharacter$Word2,numberofcharacter$Word2)

<强> RESULT

RowID ID1 ID2 ID3 Word1 Word2 Word3 Letter LastLetterWord2 1 1 1 2 3 the nice apple t e 2 2 1 2 3 the nice apple h e 3 3 1 2 3 the nice apple e e 4 4 1 2 3 the nice apple n e 5 5 1 2 3 the nice apple i e 6 6 1 2 3 the nice apple c e 7 7 1 2 3 the nice apple e e 8 8 1 2 3 the nice apple a e 9 9 1 2 3 the nice apple p e 10 10 1 2 3 the nice apple p e 11 11 1 2 3 the nice apple l e 12 12 1 2 3 the nice apple e e 13 13 6 7 8 yes did you y d 14 14 6 7 8 yes did you e d 15 15 6 7 8 yes did you s d 16 16 6 7 8 yes did you d d 17 17 6 7 8 yes did you i d 18 18 6 7 8 yes did you d d 19 19 6 7 8 yes did you y d 20 20 6 7 8 yes did you o d 21 21 6 7 8 yes did you u d

R：将特定行（即组的一部分）中的值复制到列

3 个答案: