我有一个带有字符列的数据框,其中包含由换行符\n
分隔的多个字符串形式的电子邮件元数据:
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的不同子串分成不同的列,使它看起来像这样:
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>
我目前的方法是使用tidyr包中的separate
:
library(dplyr)
library(tidyr)
res1 <- df %>%
separate(col = myString, into = paste(rep("email", 3), 1:3), sep = "\\n", extra = "drop")
res1[res1 == ""] <- NA
但是使用这种方法,我必须手动指定要提取三列。
我希望用以下任何一个或两个来改进这个过程:
如果有一个很好的解决方案能够以长篇形式返回数据,而不是宽泛的,那也很棒。
示例数据:
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")
答案 0 :(得分:2)
我会从我的“splitstackshape”软件包中建议cSplit
:
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
您也可以使用参数stri_split_fixed
从“stringi”包中尝试simplify = TRUE
(虽然您的示例数据会在末尾添加一个额外的空列)。方法如下:
library(stringi)
data.frame(person = df$person,
stri_split_fixed(df$myString, "\n",
simplify = TRUE))
答案 1 :(得分:1)
看起来很丑陋,但是你走了......
使用strsplit分割char矢量。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$myString, "\n")
max_len <- max(sapply(a, length))
for(i in 1:max_len){
df[,paste0("email", i)] <- sapply(a, "[", i)
}
答案 2 :(得分:1)
以下是长篇文章的有效途径:
a <- strsplit(df$myString, "\n")
lens <- vapply(a, length, integer(1L)) # or lengths(a) in R 3.2
longdf <- df[rep(seq_along(a), lens),]
longdf$string <- unlist(a)
请注意stack()
通常对这些情况有用。
可以使用IRanges Bioconductor软件包进行简化:
longdf <- df[togroup(a),]
longdf$string <- unlist(a)
然后,如果真的有必要,请转到宽屏:
longdf$myString <- NULL
longdf$token <- sequence(lens)
widedf <- reshape(longdf, timevar="token", idvar="person", direction="wide")
答案 3 :(得分:1)
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
然后可以简单地转换为宽格式:
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+)