Question:
I have a list for which i can remove element(s) from it based on the presence of a pattern (see this post).lst <- list(a = 1:4, b = 4:8, c = 8:10) pattern <- "a|c" lstnew <- lst[-grep(pattern, names(lst))] [/code]
The above code removes elements a and c from the list. Great. Sometimes though I have no matching pattern in the list. I then want it to return the full list. If i use the above code, it returns an empty named list.pattern <- "d|e" lstnew <- lst[-grep(pattern, names(lst))] lstnew named list() [/code]
It seems like an ifelse() is a logical choice to achieve this i.e. if pattern has a match in list, remove elements from list, otherwise return full list. Any suggestions?Answer:
Usegrepl to get a logical vector for subsetting:lst <- list(a = 1:4, b = 4:8, c = 8:10) pattern <- "a|c" grepl(pattern, names(lst)) #[1] TRUE FALSE TRUE lst[!grepl(pattern, names(lst))] #$b #[1] 4 5 6 7 8 pattern <- "d|e" grepl(pattern, names(lst)) #[1] FALSE FALSE FALSE lst[!grepl(pattern, names(lst))] #$a #[1] 1 2 3 4 # #$b #[1] 4 5 6 7 8 # #$c #[1] 8 9 10 [/code]
If you have better answer, please add a comment about this, thank you!