I skipped pre-populating MyDF$C as unnecessary:
> MyDF <- data.frame(A=c(1,2,3,4,5),B=c("aa ab ac","bb bc bd","cc
cf","dd","ee"),
+ stringsAsFactors = FALSE)
## I think this does what you want:
> choices<- sapply(MyDF$B, strsplit, split = " +")
> nm <- names(MyList)
> MyDF$C <- nm[sapply(choices, function(x)match(TRUE,
sapply(MyList,function(tbl)any(x %in% tbl))))]
> MyDF$C
[1] "Y" "z" "X" NA NA
You could of course make this even more opaque by making it a one-liner. ;-)
Cheers,
Bert
On Sat, Apr 6, 2019 at 10:45 AM Ek Esawi <esaw...@gmail.com> wrote:
> Thank you. Sorry i forgot to turn off the html
>
> Below is a sample of my data. My original data frame has over 10,000 rows.
> I want to check each element on my data frame column B
> (MyDF$B) to see if it contains any element(s) of MYList. if os, change
> the value of MyDF$C to the name of the vector of the list that has
> match(s).
>
> I solved this via loops and if statements, using &in& but I am hoping for
> a more compact solution using the apply family functions. I tried
> something like
> this but did not work.
>
> lapply(strsplit(MyDF$B," "),function(x) lapply(MyList,function(y) if(sum(y
> %in% x)>0,x$Code==y[[1]]))
>
> Thanks in advance--EK
>
> Sample data
> MyList <-
> list(X=c("a","ba","cc"),Y=c("abs","aa","BA","BB"),z=c("ab","bb","xy","zy","gh"))
> MyDF <- data.frame(A=c(1,2,3,4,5),B=c("aa ab ac","bb bc bd","cc
> cf","dd","ee"), C= c(0,0,0,0,0), stringsAsFactors = FALSE)
>
> > MyDF
>
> A B C
> 1 1 aa ab ac 0
> 2 2 bb bc bd 0
> 3 3 cc cf 0
> 4 4 dd 0
> 5 5 ee 0
>
> > MyList
>
> $X
> [1] "a" "ba" "cc"
>
> $Y
> [1] "abs" "aa" "BA" "BB"
>
> $z
> [1] "ab" "bb" "xy" "zy" "gh"
>
>
> Desired results.
>
> > MyDF
>
> A B C
> 1 1 aa ab ac Y
> 2 2 bb bc bd Y
> 3 3 cc cf X
> 4 4 dd 0
> 5 5 ee 0
>
> ______________________________________________
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> PLEASE do read the posting guide
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> and provide commented, minimal, self-contained, reproducible code.
>
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