On Apr 15, 2013, at 2:30 PM, arun wrote:
> Hi,
> vec1<- letters
alp <- as.list(letters) would have constructed the vector that was described.
> vec1[!grepl("b|r|x",alp)]
> # [1] "a" "c" "d" "e" "f" "g" "h" "i" "j" "k" "l" "m" "n" "o" "p" "q" "s" "t"
> "u"
> #[20] "v" "w" "y" "z"
> vec1[!vec1%in% c("b","r","x") ]
> # [1] "a" "c" "d" "e" "f" "g" "h" "i" "j" "k" "l" "m" "n" "o" "p" "q" "s" "t"
> "u"
> #[20] "v" "w" "y" "z"
>
> alp<-lapply(seq_along(vec1),function(i) vec1[i])
> res<-alp[!grepl("b|r|x",alp)]
> unlist(res)
> # [1] "a" "c" "d" "e" "f" "g" "h" "i" "j" "k" "l" "m" "n" "o" "p" "q" "s" "t"
> "u"
> #[20] "v" "w" "y" "z"
> unlist(alp[!alp%in%c("b","r","x")])
> #[1] "a" "c" "d" "e" "f" "g" "h" "i" "j" "k" "l" "m" "n" "o" "p" "q" "s" "t"
> "u"
> #[20] "v" "w" "y" "z"
> A.K.
All good. There would be an additional way to do this if the list were first
assigned names. (At the moment the list only has positions for reference.)
alp <- setNames(alp, letters[1:26])
# Something like that initial code would succeed:
alp <- alp[ !names(alp) %in% c("b","r","x")]
The result is still a list.
>
>> If I for example have a list (or vector) that contains all the letters in
>> the alphabet.
>>
>> alp <- list("a","b","c",......................."z") this is of course not
>> the exact code
>>
>> How can I remove multiple elements at one time without knowing their
>> location in the list. Say I want to remove b,r,x?
>> Same question if apl is a vector
>>
>> I have tried
>>
>> alp <- alp[-c("b","r","x")]
You _cannot_use_negative_indexing_with_names (or values). You could have used
logical indexing:
alp [ sapply(alp, function(x) !x %in% c("b","r","x") ) ]
# OR perhaps the most compact solution offered so far; numeric indexing with
the minus unary operator:
alp[ -grep("b|r|x", alp) ]
--
David Winsemius
Alameda, CA, USA
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