Split a string vector at whitespace

Question

I have the following vector:

tmp3 <- c("1500 2", "1500 1", "1510 2", "1510 1", "1520 2", "1520 1", "1530 2", 
"1530 1", "1540 2", "1540 1")

I would like to just retain the second number in each of the atoms of this vector, so it would read:

c(2,1,2,1,2,1,2,1,2,1)

Solution 1

There's probably a better way, but here are two approaches with strsplit():

as.numeric(data.frame(strsplit(tmp3, " "))[2,])
as.numeric(lapply(strsplit(tmp3," "), function(x) x[2]))

The as.numeric() may not be necessary if you can use characters...

Solution 2

One could use read.table on textConnection:

X <- read.table(textConnection(tmp3))

then

> str(X)
'data.frame':   10 obs. of  2 variables:
 $ V1: int  1500 1500 1510 1510 1520 1520 1530 1530 1540 1540
 $ V2: int  2 1 2 1 2 1 2 1 2 1

so X$V2 is what you need.

Solution 3

It depends a little bit on how closely your actual data matches the example data you've given. I you're just trying to get everything after the space, you can use gsub:

gsub(".+\\s+", "", tmp3)
[1] "2" "1" "2" "1" "2" "1" "2" "1" "2" "1"

If you're trying to implement a rule more complicated than "take everything after the space", you'll need a more complicated regular expresion.

Solution 4

What I think is the most elegant way to do this

>     res <- sapply(strsplit(tmp3, " "), "[[", 2)

If you need it to be an integer

>     storage.mode(res) <- "integer"

Solution 5

substr(x = tmp3, start = 6, stop = 6)

So long as your strings are always the same length, this should do the trick.

(And, of course, you don't have to specify the argument names - substr(tmp3, 6, 6) works fine, too)

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Comments

[Zak]

This is an elegant solution. Just what I was looking for. Thanks!

[user295691]

Also, res <- as.numeric(sapply(...)) works as well; storage.mode is a little scary

[pedrosaurio]

I tried to use your solution but instead, using the column of a data frame and it didn't work straight away. I would add that for those cases you need to turn it into a list. as.numeric(data.frame(strsplit(as.list(df$columnx), " "))[2,])

[user3067923]

can you explain this…

[Corey Levinson]

the function gsub is for replacing regex matches with something else. In this case, we use the regex .+\\s+ and replace any matches we find with the empty string "". The regex translates to "Match anything in the beginning, but it has to end with a space" (the character space is written as \\s)