Regex 将字符串拆分为未知数量的新数据帧列_Regex_R_String

Regex 将字符串拆分为未知数量的新数据帧列

regex r string

Regex 将字符串拆分为未知数量的新数据帧列,regex,r,string,Regex,R,String,我有一个带有字符列的数据框，其中包含多个字符串形式的电子邮件元数据，这些字符串由换行符分隔\n： person myString 1 John

我有一个带有字符列的数据框，其中包含多个字符串形式的电子邮件元数据，这些字符串由换行符分隔

\n

：

  person                                                                                                                                                 myString
1   John                                                                                                            To name5@email.com by sender6 on 01-12-2014\n
2   Jane To name@email.com,name4@email.com by sender1 on 01-22-2014\nTo name3@email.com by sender2 on 02-03-2014\nTo email5@domain.com by sender1 on 06-21-2014\n
3    Tim                                                                To name2@email.com by sender2 on 05-11-2014\nTo name@email.com by sender2 on 06-03-2015\n

我想将myString的不同子字符串拆分为不同的列，使其如下所示：

  person                                                     email1                                      email2                                        email3
1   John                To name5@email.com by sender6 on 01-12-2014                                        <NA>                                          <NA>
2   Jane To name@email.com,name4@email.com by sender1 on 01-22-2014 To name3@email.com by sender2 on 02-03-2014 To email5@domain.com by sender1 on 06-21-2014
3    Tim                To name2@email.com by sender2 on 05-11-2014  To name@email.com by sender2 on 06-03-2015                                          <NA>

但是使用这种方法，我必须手动指定有三列要提取

我希望通过以下其中一项或两项来改进此过程：

自动计算定界字符最大出现次数的方法（即，需要多少新变量）

拆分为未知列数的其他方法

如果有一个很好的解决方案可以以长格式而不是宽格式返回数据，那也太好了

样本数据：

df <- structure(list(person = c("John", "Jane", "Tim"), myString = c("To name5@email.com by sender6 on 01-12-2014\n", 
    "To name@email.com,name4@email.com by sender1 on 01-22-2014\nTo name3@email.com by sender2 on 02-03-2014\nTo email5@domain.com by sender1 on 06-21-2014\n", 
    "To name2@email.com by sender2 on 05-11-2014\nTo name@email.com by sender2 on 06-03-2015\n"
    )), .Names = c("person", "myString"), row.names = c(NA, -3L), class = "data.frame")

df似乎有点老套，但你看
使用strsplit分割字符向量。获取最大长度，将其用于列
df <- data.frame(
  person = c("John", "Jane", "Tim"),
  myString = c("To name5@email.com by sender6 on 01-12-2014\n",
               "To name@email.com,name4@email.com by sender1 on 01-22-2014\nTo name3@email.com by sender2 on 02-03-2014\nTo email5@domain.com by sender1 on 06-21-2014\n",
               "To name2@email.com by sender2 on 05-11-2014\nTo name@email.com by sender2 on 06-03-2015\n"
  ), stringsAsFactors=FALSE
)

a <- strsplit(df$myString, "\n")
max_len <- max(sapply(a, length))
for(i in 1:max_len){
  df[,paste0("email", i)] <- sapply(a, "[", i)
}

df这是一条通往长型的有效途径：
a <- strsplit(df$myString, "\n")
lens <- vapply(a, length, integer(1L)) # or lengths(a) in R 3.2
longdf <- df[rep(seq_along(a), lens),]
longdf$string <- unlist(a)

然后，如果确实有必要，请转到广泛形式：
longdf$myString <- NULL
longdf$token <- sequence(lens)
widedf <- reshape(longdf, timevar="token", idvar="person", direction="wide")

longdf$myString这可能就足够了：
library(data.table)
dt = as.data.table(df) # or setDT to convert in place

dt[, strsplit(myString, split = "\n"), by = person]
#   person                                                         V1
#1:   John                To name5@email.com by sender6 on 01-12-2014
#2:   Jane To name@email.com,name4@email.com by sender1 on 01-22-2014
#3:   Jane                To name3@email.com by sender2 on 02-03-2014
#4:   Jane              To email5@domain.com by sender1 on 06-21-2014
#5:    Tim                To name2@email.com by sender2 on 05-11-2014
#6:    Tim                 To name@email.com by sender2 on 06-03-2015

然后可以轻松地转换为宽格式：
dcast(dt[, strsplit(myString, split = "\n"), by = person][, idx := 1:.N, by = person],
      person ~ idx, value.var = 'V1')
#   person                                                          1                                           2                                             3
#1:   Jane To name@email.com,name4@email.com by sender1 on 01-22-2014 To name3@email.com by sender2 on 02-03-2014 To email5@domain.com by sender1 on 06-21-2014
#2:   John                To name5@email.com by sender6 on 01-12-2014                                          NA                                            NA
#3:    Tim                To name2@email.com by sender2 on 05-11-2014  To name@email.com by sender2 on 06-03-2015                                            NA

# (load reshape2 and use dcast.data.table instead of dcast if not using 1.9.5+)

我建议从我的“splitstackshape”套餐中选择cSplit
：
您还可以尝试使用参数simplify=TRUE
从“stringi”包中执行stri\u split\u fixed
（尽管对于示例数据，这会在末尾添加一个额外的空列）。该方法类似于：
library(stringi)
data.frame(person = df$person, 
           stri_split_fixed(df$myString, "\n", 
                            simplify = TRUE))

@SamFirke您不需要使用最新版本的data.table重塑2
（最后编辑注释以使其更清晰）感谢您的澄清-我有data.table的最新CRAN版本，但我看到它是1.9.4，1.9.5是GitHub上的当前开发版本。这对我来说很好，这是我见过的解决这个问题的最简单的函数。看起来这个包裹里还有其他好东西。谢谢@桑菲克，谢谢。我希望您注意到，cSplit
还有一个“direction”参数，如果您想要长表单，可以将其设置为“long”。我只是想顺便向您表示感谢。cSplit是一件美丽的东西！
dcast(dt[, strsplit(myString, split = "\n"), by = person][, idx := 1:.N, by = person],
      person ~ idx, value.var = 'V1')
#   person                                                          1                                           2                                             3
#1:   Jane To name@email.com,name4@email.com by sender1 on 01-22-2014 To name3@email.com by sender2 on 02-03-2014 To email5@domain.com by sender1 on 06-21-2014
#2:   John                To name5@email.com by sender6 on 01-12-2014                                          NA                                            NA
#3:    Tim                To name2@email.com by sender2 on 05-11-2014  To name@email.com by sender2 on 06-03-2015                                            NA

# (load reshape2 and use dcast.data.table instead of dcast if not using 1.9.5+)

library(splitstackshape)
cSplit(df, "myString", "\n")
#    person                                                 myString_1
# 1:   John                To name5@email.com by sender6 on 01-12-2014
# 2:   Jane To name@email.com,name4@email.com by sender1 on 01-22-2014
# 3:    Tim                To name2@email.com by sender2 on 05-11-2014
#                                     myString_2
# 1:                                          NA
# 2: To name3@email.com by sender2 on 02-03-2014
# 3:  To name@email.com by sender2 on 06-03-2015
#                                       myString_3
# 1:                                            NA
# 2: To email5@domain.com by sender1 on 06-21-2014
# 3:                                            NA

library(stringi)
data.frame(person = df$person, 
           stri_split_fixed(df$myString, "\n", 
                            simplify = TRUE))