R Split a column into multiple column by pattern [closed] - r

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I want to separate the digits and character in a column of a dataframe d.df:
col1
ab 12 14 56
xb 23 234 2342 2
ad 23 45
Expected output:
col1 col2
ab 12 14 56
xb 23 234 2342 2
ad 23 45
I recognize it will be something similar to this, but I'm not sure about the separators
t <- as.data.frame(str_match(d$col1,"^(.*)"))
I tried many methods and the output was:
col1 col2
a b 12 14 56
x b 23 234 2342 2
a d 23 45

Answer 1

You can use separate from tidyr.
library(tidyr)
d.df %>% separate(col1, c("col1", "col2"), sep="(?<=[a-z]{2} )")
# col1 col2
# 1 ab 12 14 56
# 2 xb 23 234 2342 2
# 3 ad 23 45
The regex, "(?<=[a-z]{2} )", is a look-behind, meaning "split at the position in the string after two lower case characters followed by a space". tidyr seems to have a limit on the length of look-behinds, so {2} is used to specify the number of letters.

Answer 2

Here is an option with data.table.
library(data.table)#v1.9.5+
setnames(setDT(df1)[, tstrsplit(col1,
'(?<=[^0-9]) (?=[0-9])', perl=TRUE)], paste0('col', 1:2))[]
# col1 col2
#1: ab 12 14 56
#2: xb 23 234 2342 2
#3: ad 23 45
We convert the 'data.frame' to 'data.table' (setDT(df1)). Using tstrsplit from the devel version of 'data.table', split at the space in 'col1' by matching the space after a letter and before a numeric part. We use regex lookarounds ((?<=[^0-9]) and ((?=[0-9])) for matching.

Answer 3

The approach here will vary significantly depending on whether this is actually how your strings look like or just an example. If they are always two letters and numbers, you can substring:
> df <- data.frame(col1 = c("ab 12 14 56", "xb 23 234 2342 2", "ad 23 45"))
>
> df$col1.1 <- sapply(df$col1, substring, 0, 2)
>
> df$col1.2 <- sapply(df$col1, substring, 3)
>
> df
col1 col1.1 col1.2
1 ab 12 14 56 ab 12 14 56
2 xb 23 234 2342 2 xb 23 234 2342 2
3 ad 23 45 ad 23 45
If the length and positions of the strings change, regex might be better suited. Using a base R approach, you can extract only the numbers or letters (keeping the white spaces):
> df <- data.frame(col1 = c("ab 12 14 56", "xb 23 234 2342 2", "ad 23 45"))
> df$col1.1 <- sapply(regmatches(df$col1, gregexpr("[a-zA-Z]", df$col1)), paste, collapse = "")
> df$col1.2 <- sapply(regmatches(df$col1, gregexpr("[0-9]\\s*", df$col1)), paste, collapse = "")
> df
col1 col1.1 col1.2
1 ab 12 14 56 ab 12 14 56
2 xb 23 234 2342 2 xb 23 234 2342 2
3 ad 23 45 ad 23 45