使用 R 中的字符串值将列转换为多列

Question

我在清理 R 中的数据集时遇到问题。我有一个包含三个变量（名称、日期、数据）的数据集。第三个变量实际上包含我的所有数据，但需要对其进行解析。我需要根据列中的值将此列拆分为多列。例如，在以下数据库中：

x <- data.frame("name" = c("John","John","John","John","John","Sarah","Sarah","Sarah"), "Day" = c(1,1,1,1,1,2,2,2), "Data" = c("Map 28", 2,3,"Transfer","Time","Map 18",2,3))

看起来像：

 name Day     Data
1  John   1   Map 28
2  John   1        2
3  John   1        3
4  John   1 Transfer
5  John   1     Time
6 Sarah   2   Map 18
7 Sarah   2        2
8 Sarah   2        3

我需要查看“数据”列并找到使用“地图”一词的任何时间，然后将其下的所有数据转换为另一列。像这样：

  name Day   Data Val1 Val2     Val3 Val4
1  John   1 Map 28    2    3 Transfer Time
2 Sarah   2 Map 18    2    3     <NA> <NA>

对此的任何帮助将不胜感激！

[编辑]

对不起，我认为我的例子过于简单了......问题是每个人每天都会有多个需要定位的“地图”值。所以看起来更像下面这样。

x <- data.frame("name" = c("John","John","John","John","John","John","John","John","John","John","John","John","Sarah","Sarah","Sarah","Sarah","Sarah","Sarah"), 
                "Day" = c(1,1,1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2), 
                "Data" = c("Map 28", 2,3,"Transfer","Time","Map 15",2,3,"Text","Map3",2,4,"Map 18",2,3,"Map 22",2,3))

    name Day     Data
1   John   1   Map 28
2   John   1        2
3   John   1        3
4   John   1 Transfer
5   John   1     Time
6   John   1   Map 15
7   John   1        2
8   John   1        3
9   John   1     Text
10  John   1     Map3
11  John   1        2
12  John   1        4
13 Sarah   2   Map 18
14 Sarah   2        2
15 Sarah   2        3
16 Sarah   2   Map 22
17 Sarah   2        2
18 Sarah   2        3

然后最终的输出将是......

y <- data.frame("name" = c("John","John","John","Sarah", "Sarah"), 
                "Day" =c(1,1,1,2,2), 
                "Data"= c("Map 28","Map 15","Map 3","Map 18","Map 22"), 
                "Val1" =c(2,2,2,2,2), 
                "Val2"=c(3,3,4,3,3), 
                "Val3"=c("Transfer","Text",NA,NA,NA), 
                "Val4"=c("Time",NA,NA,NA,NA))

name Day   Data Val1 Val2     Val3 Val4
1  John   1 Map 28    2    3 Transfer Time
2  John   1 Map 15    2    3     Text <NA>
3  John   1 Map 3    2    4     <NA> <NA>
4 Sarah   2 Map 18    2    3     <NA> <NA>
5 Sarah   2 Map 22    2    3     <NA> <NA>

Answer 1

在使用ave 添加time 变量后，您可以使用基础R 中的reshape

reshape(transform(x,time=ave(Day,Day,FUN=seq_along)),v.names="Data",dir="wide",idvar = "name")

   name Day Data.1 Data.2 Data.3   Data.4 Data.5
1  John   1 Map 28      2      3 Transfer   Time
6 Sarah   2 Map 18      2      3     <NA>   <NA>

使用您的新编辑：使用基础 R 您可以：

d = transform(x,ID=cumsum(grepl('Map',Data))->a,time=ave(a,a,FUN=seq_along))

reshape(d,v.names="Data",idvar = "ID",dir="wide")
    name Day ID Data.1 Data.2 Data.3   Data.4 Data.5
1   John   1  1 Map 28      2      3 Transfer   Time
6   John   1  2 Map 15      2      3     Text   <NA>
10  John   1  3   Map3      2      4     <NA>   <NA>
13 Sarah   2  4 Map 18      2      3     <NA>   <NA>
16 Sarah   2  5 Map 22      2      3     <NA>   <NA>

使用 tidyverse 你可以做到：

 library(tidyverse)
 x%>%
   group_by(ID = cumsum(str_detect(Data,"Map")))%>%
   mutate(time=1:n())%>%
   spread(time,Data)
# A tibble: 5 x 8
# Groups:   ID [5]
  name    Day    ID `1`    `2`   `3`   `4`      `5`  
  <fct> <dbl> <int> <fct>  <fct> <fct> <fct>    <fct>
1 John      1     1 Map 28 2     3     Transfer Time 
2 John      1     2 Map 15 2     3     Text     NA   
3 John      1     3 Map3   2     4     NA       NA   
4 Sarah     2     4 Map 18 2     3     NA       NA   
5 Sarah     2     5 Map 22 2     3     NA       NA