substitute string when there is a dot number ':'-CodePudding

I have strings that look like these:

> ABCD.1:f_HJK
> ABFD.1:f_HTK
> CJD:f_HRK
> QQYP.2:f_HDP

So basically, I have always a string in the first part, I could have a part with . and a number, and after this part I always have ':' and a string. I would like to remove the '. number' when it is included in the string, using R.

I know that maybe regular expressions could be useful but I have not idea about I can apply them in this context. I know that I can substitute the '.' with gsub, but not idea about how I can add the information about number and ':'.

Thank you for your help.

CodePudding user response：

Does this work:

v <- c('ABCD.1:f_HJK','ABFD.1:f_HTK','CJD:f_HRK','QQYP.2:f_HDP')
v
[1] "ABCD.1:f_HJK" "ABFD.1:f_HTK" "CJD:f_HRK"    "QQYP.2:f_HDP"
gsub('([A-Z]{,4})(\\.\\d)?(:.*)','\\1\\3',v)
[1] "ABCD:f_HJK" "ABFD:f_HTK" "CJD:f_HRK"  "QQYP:f_HDP"

CodePudding user response：

You could also use any of the following depending on the structure of your string

If no other period and numbers in the string

 sub("\\.\\d ", "", v)
 [1] "ABCD:f_HJK" "ABFD:f_HTK" "CJD:f_HRK"  "QQYP:f_HDP"

If you are only interested in the first pattern matched.

 sub("^([A-Z] )\\.\\d :", "\\1:", v)
 [1] "ABCD:f_HJK" "ABFD:f_HTK" "CJD:f_HRK"  "QQYP:f_HDP"

Same as above, invoking perl. ie no captured groups

  sub("^[A-Z] \\K\\.\\d ", "", v, perl = TRUE)
  [1] "ABCD:f_HJK" "ABFD:f_HTK" "CJD:f_HRK"  "QQYP:f_HDP"

CodePudding user response：

If I understood your explanation correctly, this should do the trick:

gsub("(\\.\\d )", "", string)