如何用几行代码将字符数组转换为 data.frame？

Question

我有以下数组
my_list <- c("Jan-01--Dec-31|00:00--24:00", "Jan-01--Jun-30|12:00--18:00", "Jul-06--Dec-31|09:00--19:00")

导致以下结果的最短代码是什么：

  x1     x2     x3
1 Jan-01 Jan-01 Jul-06
2 Dec-31 Jun-30 Dec-31

和

  x2    x2    x3
1 00:00 12:00 09:00
2 24:00 18:00 19:00

目前我有（不是很好）代码

df <- as.data.frame(strsplit(my_list, split = "|", fixed = T),
                    stringsAsFactors = F)
date_list <- strsplit(as.character(df[1, ]), split = "--", fixed = T)
date_df <- as.data.frame(date_list, col.names = c(1:length(date_list)),
                         stringsAsFactors = F)
time_list <- strsplit(as.character(df[2, ]), split = "--", fixed = T)
time_df <- as.data.frame(time_list, col.names = c(1:length(date_list)),
                         stringsAsFactors = F)

我现在最好的是

date_list <- sapply(strsplit(schedule$schedule, split = "|", fixed = T), "[", 1)
date_df <- t(data.frame(x1=sapply(strsplit(df1, split = "--", fixed = T), "[", 1),
                   x2=sapply(strsplit(df1, split = "--", fixed = T), "[", 2),
                   stringsAsFactors = F))
# and similarly for time_list and time_df.

还有更优雅的吗？

Answer 1

data.table 包中的

tstrsplit 和 stringr 包中的 str_split_fixed 是在拆分字符串向量时获取正确形状数据的非常有用的函数；前者提供 transpose 的拆分字符串，无需使用 apply 函数即可单独提取日期和时间，后者将字符串拆分为指定列的矩阵：

library(data.table); library(stringr)
lapply(tstrsplit(my_list, "\|"), function(s) t(str_split_fixed(s, "--", 2)))

#[[1]]
#     [,1]     [,2]     [,3]    
#[1,] "Jan-01" "Jan-01" "Jul-06"
#[2,] "Dec-31" "Jun-30" "Dec-31"

#[[2]]
#     [,1]    [,2]    [,3]   
#[1,] "00:00" "12:00" "09:00"
#[2,] "24:00" "18:00" "19:00"

Answer 2

my_results <- sapply(strsplit(my_list,"|",fixed=T),function(x) strsplit(x,"--",fixed=T))
my_dates <- t(Reduce("rbind",myresults[1,]))
my_times <- t(Reduce("rbind",myresults[2,]))

Answer 3

使用 stringr 的另一种选择：

library(stringr)
a <- t(str_split_fixed(my_list, "\||--", 4))

#     [,1]     [,2]     [,3]    
#[1,] "Jan-01" "Jan-01" "Jul-06"
#[2,] "Dec-31" "Jun-30" "Dec-31"
#[3,] "00:00"  "12:00"  "09:00" 
#[4,] "24:00"  "18:00"  "19:00"

要获得最终输出，data.frame(a[1:2,]) 和 data.frame(a[3:4,])

更新

my_list <- "Jan-01--Dec-31|00:00--24:00"
a <- t(str_split_fixed(my_list, "\||--", 4))

     [,1]    
[1,] "Jan-01"
[2,] "Dec-31"
[3,] "00:00" 
[4,] "24:00"

data.frame(a[1:2,])

  a.1.2...
1   Jan-01
2   Dec-31

data.frame(a[3:4,])

  a.3.4...
1    00:00
2    24:00

Answer 4

strsplit 接受可以在一次通过中进行拆分的 greppish 模式。然后可以使用 lapply（或 sapply）并以 setNames.

结束

 setNames( data.frame(lapply( strsplit( my_vec, split="\-\-|\|"),  "[", 1:2) ), paste0("x",1:3) )

      x1     x2     x3
1 Jan-01 Jan-01 Jul-06
2 Dec-31 Jun-30 Dec-31

显然可以通过在上面的代码中用 3:4 代替 1:2 来处理时间。

Answer 5

这里有一个base R选项

lst <- strsplit(scan(text=my_list, sep="|", what ="", quiet=TRUE), "--")
do.call(cbind, lst[c(TRUE, FALSE)])
#     [,1]     [,2]     [,3]    
#[1,] "Jan-01" "Jan-01" "Jul-06"
#[2,] "Dec-31" "Jun-30" "Dec-31"

do.call(cbind, lst[c(FALSE, TRUE)])
#     [,1]    [,2]    [,3]   
#[1,] "00:00" "12:00" "09:00"
#[2,] "24:00" "18:00" "19:00"

或单行base R选项

lapply(split(scan(text=my_list, sep="|", what ="", quiet=TRUE), 1:2), 
                      function(x) do.call(cbind, strsplit(x, "--")))
#$`1`
#     [,1]     [,2]     [,3]    
#[1,] "Jan-01" "Jan-01" "Jul-06"
#[2,] "Dec-31" "Jun-30" "Dec-31"

#$`2`
#    [,1]    [,2]    [,3]   
#[1,] "00:00" "12:00" "09:00"
#[2,] "24:00" "18:00" "19:00"

如何用几行代码将字符数组转换为 data.frame？

How can I transform an array of characters with a few lines of code to a data.frame?

r

lapply

sapply