将字符拆分为数据框中的两个变量-splitcharactersintotwovariablesindataframe

7  

substr is another way to do it.

substr是另一种方法。

> variable <- c(rep("A1", 4), rep("A2", 4), rep("B1", 4))
> data.frame(treatment=substr(variable, 1,1), time=as.numeric(substr(variable,2,2)))
   treatmen time
1         A    1
2         A    1
3         A    1
4         A    1
5         A    2
6         A    2
7         A    2
8         A    2
9         B    1
10        B    1
11        B    1
12        B    1

9  

Update: 24 December 2017

Somewhere along the line, the "stringr" package (which is imported with "reshape2" and which is responsible for the splitting that takes place with colsplit) started to use "stringi" for several of its functions. Some behavior seems to have changed because of that.

沿着这条线的某个地方，“stringr”包（使用“reshape2”导入并负责使用colsplit进行拆分）开始使用“stringi”来实现其几个功能。由于这一点，一些行为似乎已经改变。

Using the current "reshape2" (and current "stringr" package), colsplit works the way you would have expected it to with your code:

使用当前的“reshape2”（以及当前的“stringr”包），colsplit的工作方式与您对代码的预期方式相同：

packageVersion("reshape2")
## [1] ‘1.4.3’
packageVersion("stringr")
## [1] ‘1.2.0’

colsplit(variable, "", names = c("treatment", "time"))
##    treatment time
## 1          A    1
## 2          A    1
## 3          A    1
## 4          A    1
## 5          A    2
## 6          A    2
## 7          A    2
## 8          A    2
## 9          B    1
## 10         B    1
## 11         B    1
## 12         B    1

Original Answer: 24 April 2013

If a pattern can be detected in your "variable" but there is no clean split character that can be used, then add one :)

如果可以在“变量”中检测到模式但是没有可以使用的干净分割字符，那么添加一个:)

library(reshape2)
variable <- c("A1", "A1", "A1", "A1", "A2", "A2", 
              "A2", "A2", "B1", "B1", "B1", "B1")
## Here, we add a "." between upper case letters and numbers
colsplit(gsub("([A-Z])([0-9])", "\\1\\.\\2", variable), 
         "\\.", c("Treatment", "Time"))
#    Treatment Time
# 1          A    1
# 2          A    1
# 3          A    1
# 4          A    1
# 5          A    2
# ::::: snip :::: #
# 11         B    1
# 12         B    1

Additional Options: 23 December 2017

My "splitstackshape" package has a single-purpose non-exported helper function called NoSep that can be used for this:

我的“splitstackshape”包有一个名为NoSep的单用途非导出辅助函数，可用于此：

splitstackshape:::NoSep(variable)
##    .var .time_1
## 1     A       1
## 2     A       1
## 3     A       1
## 4     A       1
## 5     A       2
## ::: snip :::: #
## 11    B       1
## 12    B       1

The "tidyverse" (specifically the "tidyr" package) has a couple of convenient functions for splitting values into different columns: separate and extract. separate has already been demonstrated by jazzuro, but the solution is very specific to this particular problem. Also, it generally works better with a delimiter. extract expects you to specify a regular expression with the groups you want to capture:

“tidyverse”（特别是“tidyr”包）有几个方便的功能，可以将值分成不同的列：单独和提取。 jazzuro已经证明了分离，但解决方案非常特定于这一特定问题。此外，它通常使用分隔符更好。 extract希望您指定包含要捕获的组的正则表达式：

library(tidyverse)
data.frame(variable) %>% 
  extract(variable, into = c("Treatment", "Time"), regex = "([A-Z]+)([0-9]+)")
#    Treatment Time
# 1          A    1
# 2          A    1
# 3          A    1
# 4          A    1
# 5          A    2
# ::::: snip :::: #
# 11         B    1
# 12         B    1

5  

You can use substr to split it:

您可以使用substr来拆分它：

e.g.

例如

df <- data.frame(treatment =   substr(variable, start = 1, stop = 1),
                 time =        substr(variable, start = 2, stop = 2) )

5  

If you create a data frame with the vector, variable, you could use separate() from the tidyr package now.

如果使用vector，variable创建数据框，则可以立即使用tidyr包中的separate（）。

mydf <- data.frame(variable = c(rep("A1", 4), rep("A2", 4), rep("B1", 4)),
                   stringsAsFactors = FALSE)

separate(mydf, variable, c("treatement", "time"), sep = 1)

#   treatement time
#1           A    1
#2           A    1
#3           A    1
#4           A    1
#5           A    2
#6           A    2
#7           A    2
#8           A    2
#9           B    1
#10          B    1
#11          B    1
#12          B    1

4  

Another solution using regular expression

另一种使用正则表达式

require(stringr)
variable <- c(paste0("A", c(rep(1, 4), rep(2, 3))),
              paste0("B", rep(1, 4))
              )

data.frame(
    treatment = str_extract(variable, "[[:alpha:]]"),
    time = as.numeric(str_extract(variable, "[[:digit:]]"))
    )

##    treatment time
## 1          A    1
## 2          A    1
## 3          A    1
## 4          A    1
## 5          A    2
## 6          A    2
## 7          A    2
## 8          B    1
## 9          B    1
## 10         B    1
## 11         B    1

4  

A new function tstrsplit() was introduced in data.table v1.9.5. The t stands for transpose. It's the result of splitting a character vector with strsplit() and then transposing it.

在data.table v1.9.5中引入了一个新函数tstrsplit（）。 t代表转置。这是用strsplit（）分割字符向量然后转置它的结果。

# dummy data
library(data.table)
dt <- data.table(var = c(rep("A1", 4), rep("A2", 4), rep("B1", 4)))

Using tstrsplit():

使用tstrsplit（）：

dt[, tstrsplit(var, "")]

    V1 V2
 1:  A  1
 2:  A  1
 3:  A  1
 4:  A  1
 5:  A  2
 6:  A  2
 7:  A  2
 8:  A  2
 9:  B  1
10:  B  1
11:  B  1
12:  B  1

Yes, it's that easy. :-)

是的，就这么简单。 :-)

3  

You can use substring() to create vectors then join them using the data.frame function.

您可以使用substring（）创建向量，然后使用data.frame函数将它们连接起来。

yyy<-c("A1", "A1", "A1", "A1", "A2", "A2", "A2", "A2", "B1", "B1", "B1", "B1")

treatment<-substring(yyy, 1,1)

time<-as.numeric(substring(yyy,2,2))

data.frame(treatment,time)

2  

You could just use strsplit

你可以使用strsplit

df <- t(data.frame(strsplit(variable, "")))
rownames(df) <- NULL
colnames(df) <- c("treatment" , "time" )
df
      treatment time
 [1,] "A"       "1" 
 [2,] "A"       "1" 
 [3,] "A"       "1" 
 [4,] "A"       "1" 
 [5,] "A"       "2" 
 [6,] "A"       "2" 
 [7,] "A"       "2" 
 [8,] "A"       "2" 
 [9,] "B"       "1" 
[10,] "B"       "1" 
[11,] "B"       "1" 
[12,] "B"       "1" 

Instead of using t you can use rbind and then coerce to data.frame as follows:

您可以使用rbind而不是使用t，然后强制使用data.frame，如下所示：

setNames(as.data.frame(do.call(rbind, strsplit(variable, ""))), 
         c("Treatment", "Time"))
#    Treatment Time
# 1          A    1
# 2          A    1
# 3          A    1
# 4          A    1
# 5          A    2
# 6          A    2
# 7          A    2
# 8          B    1
# 9          B    1
# 10         B    1
# 11         B    1

1  

Based on the comment of @Justin I suggest this (using v <- c("A1", "B2")):

基于@Justin的评论我建议这个（使用v <- c（“A1”，“B2”））：

> t(sapply(strsplit(v, ''), '[', c(1, 2)))
     [,1] [,2]
[1,] "A"  "1" 
[2,] "B"  "2" 

The vector after `'[' selects the items from the split vector. So I split only once, keeping both items. Maybe this is even easier if you want to keep every item:

''['选择分裂向量中的项后的向量。所以我只拆分一次，保留两个项目。如果你想保留每件物品，这可能更容易：

t(sapply(strsplit(v, ''), identity))