如何通过正则表达式在数据框中拆分列？

Question

Example:

ID <- c(1:5)
v1 <- c("abc1", "d2", "eF34", "GHi567", "NoNumber")
df <- data.frame(ID, v1, stringsAsFactors = FALSE)

I want to do something like the following, but in simpler code and possibly one function. str_match_all in stringr would be good, but it requires an atomic vector. I suppose I could write in a row loop, but I'd like something already vectorized. Will something in stringi be useful here?

pattern1 <- "([A-Za-z]+)([0-9]+)"
df$v2 <- sub(pattern = pattern1, replacement = "\\1", x = df$v1)
df$v3 <- sub(pattern = pattern1, replacement = "\\2", x = df$v1)

Also, I'd like to be able to handle the matching issue in Row 5, making df$v3[5] <- NA .

Answer 1

One option would be splitting the string ( strsplit ) by specifying the lookaround as pattern, and convert the 'list' output to 'matrix using stri_list2matrix from stringi . This will pad NA to list elements that have length less than the maximum list element length.

library(stringi)
df[paste0('v', 2:3)] <- stri_list2matrix(strsplit(df$v1,
                  '(?<=[A-Za-z])(?=[0-9])', perl=TRUE), byrow=TRUE)

 df
 #  ID       v1       v2   v3
 #1  1     abc1      abc    1
 #2  2       d2        d    2
 #3  3     eF34       eF   34
 #4  4   GHi567      GHi  567
 #5  5 NoNumber NoNumber <NA>

Or use extract from tidyr . We can paste the strings that don't have a numeric element at the end with NA and use extract . This also have the option to convert the 'class' by specifying convert=TRUE .

library(tidyr)
df$v1 <- with(df, ifelse(grepl('\\d+$', v1),v1, paste0(v1,NA)) )
extract(df, v1, into=c('v2', 'v3'), '([A-Za-z]+)([0-9]+|NA)', 
             remove=FALSE, convert=TRUE)
#   ID         v1       v2  v3
#1  1       abc1      abc   1
#2  2         d2        d   2
#3  3       eF34       eF  34
#4  4     GHi567      GHi 567
#5  5 NoNumberNA NoNumber  NA

Or a base R option would be

df[paste0('v', 2:3)] <- read.table(text=gsub('([A-Za-z]*)([0-9]*)',
    '\\1 \\2', df$v1), header=FALSE, stringsAsFactors=FALSE, fill=TRUE)

Answer 2

Slight mod to pattern and using as.numeric for coercion to proper class:

pattern1 <- "([A-Za-z]+)([0-9]*)"
df$v2 <- sub(pattern = pattern1, replacement = "\\1", x = df$v1)
 df$v3 <- as.numeric(sub(pattern = pattern1, replacement = "\\2", x = df$v1))
 df

  ID       v1       v2  v3
1  1     abc1      abc   1
2  2       d2        d   2
3  3     eF34      eF4  34
4  4   GHi567    GHi67 567
5  5 NoNumber NoNumber  NA

Answer 3

Using development version of data.table, v1.9.5 :

require(data.table) #v1.9.5
setDT(df)[, c("c1", "c2") := tstrsplit(v1, "(?<=[[:alpha:]])(?=[0-9])", perl=TRUE)]
#    ID       v1       c1  c2
# 1:  1     abc1      abc   1
# 2:  2       d2        d   2
# 3:  3     eF34       eF  34
# 4:  4   GHi567      GHi 567
# 5:  5 NoNumber NoNumber  NA

regex borrowed from @akrun. Use type.convert=TRUE if you want c2 to be converted to numeric automatically during tstrsplit() .