Remove part of a string

Question

How do I remove part of a string? For example in ATGAS_1121 I want to remove everything before _.

Parthenon

Answer 1

Use regular expressions. In this case, you can use gsub:

gsub("^.*?_","_","ATGAS_1121")
[1] "_1121"

This regular expression matches the beginning of the string (^), any character (.) repeated zero or more times (*), and underscore (_). The ? makes the match "lazy" so that it only matches are far as the first underscore. That match is replaced with just an underscore. See ?regex for more details and references

Vachil

Previous regex would match to the last underscore in the case of, e.g., gsub("^.*_","_","ATGAS_1121_xxx"). Now fixed. – Chapbook

@Joshua I find it really useful that you explained the role of the regular expressions. – Cimex

This also works with a vector of strings as the last argument. R is awesome like that. – Integrand

Answer 2

You can use a built-in for this, strsplit:

> s = "TGAS_1121"
> s1 = unlist(strsplit(s, split='_', fixed=TRUE))[2]
> s1    
 [1] "1121"

strsplit returns both pieces of the string parsed on the split parameter as a list. That's probably not what you want, so wrap the call in unlist, then index that array so that only the second of the two elements in the vector are returned.

Finally, the fixed parameter should be set to TRUE to indicate that the split parameter is not a regular expression, but a literal matching character.

Cadman

Answer 3

If you're a Tidyverse kind of person, here's the stringr solution:

R> library(stringr)
R> strings = c("TGAS_1121", "MGAS_1432", "ATGAS_1121") 
R> strings %>% str_replace(".*_", "_")
[1] "_1121" "_1432" "_1121"
# Or:
R> strings %>% str_replace("^[A-Z]*", "")
[1] "_1121" "_1432" "_1121"

Integrand

Answer 4

Here's the strsplit solution if s is a vector:

> s <- c("TGAS_1121", "MGAS_1432")
> s1 <- sapply(strsplit(s, split='_', fixed=TRUE), function(x) (x[2]))
> s1
[1] "1121" "1432"

Richard

Very helpful, thanks! FYI to get the first part of the string (i.e. before the '_'), replace the [2] on the end with a [1]. – Agminate

@Richard do you know how to modify this to keep the first 2 elements of the string – Weatherspoon

@Weatherspoon I believe @Agminate answered that above. Applying his solution gives you s1 <- sapply(strsplit(s, split='_', fixed=TRUE), function(x) (x[1])) – Richard

Answer 5

Maybe the most intuitive solution is probably to use the stringr function str_remove which is even easier than str_replace as it has only 1 argument instead of 2.

The only tricky part in your example is that you want to keep the underscore but its possible: You must match the regular expression until it finds the specified string pattern (?=pattern).

See example:

strings = c("TGAS_1121", "MGAS_1432", "ATGAS_1121")
strings %>% stringr::str_remove(".+?(?=_)")

[1] "_1121" "_1432" "_1121"

Dreadful

Answer 6

Here the strsplit solution for a dataframe using dplyr package

col1 = c("TGAS_1121", "MGAS_1432", "ATGAS_1121") 
col2 = c("T", "M", "A") 
df = data.frame(col1, col2)
df
        col1 col2
1  TGAS_1121    T
2  MGAS_1432    M
3 ATGAS_1121    A

df<-mutate(df,col1=as.character(col1))
df2<-mutate(df,col1=sapply(strsplit(df$col1, split='_', fixed=TRUE),function(x) (x[2])))
df2

  col1 col2
1 1121    T
2 1432    M
3 1121    A

Quay