Remove all text before colon

Question

I have a file containing a certain number of lines. Each line looks like this:

TF_list_to_test10004/Nus_k0.345_t0.1_e0.1.adj:PKMYT1

I would like to remove all before ":" character in order to retain only PKMYT1 that is a gene name. Since I'm not an expert in regex scripting can anyone help me to do this using Unix (sed or awk) or in R?

Solution 1

Here are two ways of doing it in R:

foo <- "TF_list_to_test10004/Nus_k0.345_t0.1_e0.1.adj:PKMYT1"

# Remove all before and up to ":":
gsub(".*:","",foo)

# Extract everything behind ":":
regmatches(foo,gregexpr("(?<=:).*",foo,perl=TRUE))

Solution 2

A simple regular expression used with gsub():

x <- "TF_list_to_test10004/Nus_k0.345_t0.1_e0.1.adj:PKMYT1"
gsub(".*:", "", x)
"PKMYT1"

See ?regex or ?gsub for more help.

Solution 3

There are certainly more than 2 ways in R. Here's another.

unlist(lapply(strsplit(foo, ':', fixed = TRUE), '[', 2))

If the string has a constant length I imagine substr would be faster than this or regex methods.

Solution 4

Using sed:

sed 's/.*://' < your_input_file > output_file

This will replace anything followed by a colon with nothing, so it'll remove everything up to and including the last colon on each line (because * is greedy by default).

As per Josh O'Brien's comment, if you wanted to only replace up to and including the first colon, do this:

sed "s/[^:]*://"

That will match anything that isn't a colon, followed by one colon, and replace with nothing.

Note that for both of these patterns they'll stop on the first match on each line. If you want to make a replace happen for every match on a line, add the 'g' (global) option to the end of the command.

Also note that on linux (but not on OSX) you can edit a file in-place with -i eg:

sed -i 's/.*://' your_file

Solution 5

You can use awk like this:

awk -F: '{print $2}' /your/file

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