R help - Apr 2019 - Alternative to lops

Hi All--

Sorry i sent the one inadvertently

Her is a sample of my data. A data frame (MyDF) and a list (MyList).  My
own data frame has over 10,000 rows. I want to find out which elements of
MyDF$B contain any element(s) of MYList; then change MyDF$C to the name of
the vector of the list that has match.

I solved this via loops and if statements, using &in&  but I am hoping
for
a better 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

My Sample data
> 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

	[[alternative HTML version deleted]]

Comments inline, but first:

Please review the posting guide and follow the instructions there, especially:

1) "No HTML posting..."

2) "When providing examples, it is best to give an R command that
constructs the data,..."
> On Apr 4, 2019, at 9:41 AM, Ek Esawi <esawiek at gmail.com> wrote:
> 
> Hi All--
> 
> Sorry i sent the one inadvertently
> 
> Her is a sample of my data. A data frame (MyDF) and a list (MyList).  My
> own data frame has over 10,000 rows. I want to find out which elements of
> MyDF$B contain any element(s) of MYList; then change MyDF$C to the name of
> the vector of the list that has match.
> 
> I solved this via loops and if statements, using &in&  but I am
hoping for
> a better 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
> 
> My Sample data
> 
>> 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

Note: You did not tell us if myDF$B is a factor, in which case strsplit needs to
accommodate multiple blanks:

 levels(MyDF$B)
[1] "      dd" "      ee" "   cc cf" "aa ab
ac" "bb bc bd"
> 
> 
>> 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
'aa' matches Y, 'ab' matches z, 'cc' does not match
> 2 2 bb bc bd Y
Huh? 'bb' matches z, 'bc' and 'bd' do not match, 
> 3 3    cc cf    X
'cc' matches X, 'cf' does not match
> 4 4       dd     0
> 5 5       ee     0
> 
Neither match.


You need to clarify what it is you seek. The example is hard to penetrate.

Maybe this helps you:
> queries <- strsplit(as.character(MyDF$B), "[ ]+")
> matches <- match( unlist(queries), unlist(MyList), 0)
> hits <- findInterval( matches, 1+cumsum(c(0,lengths(MyList))))
> hitList <- relist(hits, queries)
> hitList
[[1]]
[1] 2 3 0

[[2]]
[1] 3 0 0

[[3]]
[1] 0 1 0

[[4]]
[1] 0 0

[[5]]
[1] 0 0

You can now process hitList to get the desired vector.

HTH,

Chuck