CODEWITHSUNDEEP
rloopsvariables
Samir
Samir

Reputation: 23

how to use a variable in grepl function within a loop-R

I am new to R and, have some problems with looping and grepl functions I have a data from like: 

str(peptidesFilter)
  'data.frame':   78389 obs. of  130 variables:
 $ Sequence                      : chr  "AAAAAIGGR" "AAAAAIGGRPNYYGNEGGR" "AAAAASSNPGGGPEMVR" "AAAAAVGGR" ...
 $ First.amino.acid              : chr  "A" "A" "A" "A" ...
 $ Protein.group.IDs             : chr  "1" "1;2;4" "2;5 "3" "4;80" ...

I want to filter the data according to $ Protein.group.IDs by using grepl function below 

    peptidesFilter.new <- peptidesFilter[grepl('(^|;)2($|;)',
peptidesFilter$Protein.group.IDs),]

I want to do it with a loop for every individual data ( e.g 1, 2, 3, etc...) and re-write name of data frame containing variable peptidesFilter.i

   i =1
   while( i <= N) { peptidesFilter.[[i]] <- 
   peptidesFilter[grepl('(^|;)i($|;)',
   peptidesFilter$Protein.group.IDs),] 
    i=i+1 }

i have two problems, main one i in the grep1 function does not recognized as a variable and how i can re-name filtered data in a way it will contain variable.

any ideas?

Upvotes: 1

Views: 2452

Answers (1)

agstudy
agstudy

Reputation: 121568

For grepl problem you can use paste0 for example:

paste0('(^|;)',i,'($|;)')

For the loop , you can so something like this : 

ll <- lapply(seq(1:4),function(x)
         peptidesFilter[grepl(paste0('(^|;)',x,'($|;)'),
                           peptidesFilter$Protein.group.IDs),])

then you can transform it to a data.frame:

do.call(rbind,ll)

            Sequence First.amino.acid Protein.group.IDs
1            AAAAAIGGR                A                 1
2  AAAAAIGGRPNYYGNEGGR                A             1;2;4
21 AAAAAIGGRPNYYGNEGGR                A             1;2;4
3    AAAAASSNPGGGPEMVR                A               2;5
4            AAAAAVGGR                A                 3
22 AAAAAIGGRPNYYGNEGGR                A             1;2;4

Upvotes: 2

Related Questions