R 以这种格式拆分一列

Question

I need to split out this column into 2 columns

• 5/5/2020Tom Tesla

desired outcome is

• Col1 Col2
• 5/5/2020 Tom Tesla

I have tried strAny but need help as Col 1 is not a fixed with as the date field length varies due to 1 or 2 characters for the day of the month. Any suggestions how to do this?

Answer 1

We can use separate with a regex lookaround to split between a digit and a lower case letter

library(tidyr)
separate(df1, 'col1', into = c('date', 'other'), sep="(?<=[0-9])(?=[A-Za-z])")
#     date             other
#1  1/1/2000            yogurt
#2  1/1/2000      toilet paper
#3  2/1/2000              soda
#4 11/1/2000            bagels
#5 12/1/2000            fruits
#6 13/1/2000 laundry detergent

---

Or using base R with strsplit

do.call(rbind, strsplit(as.character(df1$col1),
      "(?<=[0-9])(?=[A-Za-z])", perl = TRUE))

data

df1 <- structure(list(col1 = c("1/1/2000yogurt", "1/1/2000toilet paper", 
"2/1/2000soda", "11/1/2000bagels", "12/1/2000fruits", "13/1/2000laundry detergent"
)), class = "data.frame", row.names = c(NA, -6L))

Answer 2

Here are couple of ways:

Using extract from tidyr :

tidyr::extract(df, col1, c('col1', 'col2'), regex = '(.*\\d)(.*)')

---

Or with dplyr and stringr :

library(dplyr)
library(stringr)

df %>%
  mutate(col2 = str_extract(col1, '\\d+/\\d+/\\d+'), 
         col3 = str_remove(col1, col2))