Correction to my code. I created a "doc" variable because I was thinking of
doing something faster, but I never did the change. grep needed to work on
the original source "dat" to be used for counting.
Fixed:
combs = structure(list(V1 = c(65L, 77L, 55L, 23L, 34L), V2 = c(23L, 34L,
34L, 77L, 65L), V3 = c(77L, 65L, 23L, 34L, 55L)), .Names = c("V1",
"V2", "V3"), class = "data.frame", row.names = c(NA, -5L))
dat = list(
c(77,65,34,23,55, 65,23,77, 44),
c(65,23,77,65,55,34, 77, 34,65, 10),
c(77,34,65),
c(55,78,56),
c(98,23,77,65,34, 65, 23, 77, 34))
words = unlist(apply(combs, 1 , function(d) paste(as.character(d),
collapse=" ")))
dat = lapply(dat, function(d) paste( as.character(d), collapse= " "))
#doc = paste(dat, collapse = " ## ") # just some arbitrary separator
character that isn't in your words
counts = sapply(words, function(w) length(grep(w, dat)))
names(counts) = words
counts
cbind(combs, data.frame(N = counts))
On Wed, Jul 27, 2016 at 11:27 AM, sri vathsan <srivib...@gmail.com> wrote:
> Hi,
>
> It is not a just 79 triplets. As I said, there are 79 codes. I am making
> triplets out of that 79 codes and matching the triplets in the list.
>
> Please find the dput of the data below.
>
> > dput(head(newd,10))
> structure(list(uniq_id = c("1", "2", "3", "4", "5", "6", "7",
> "8", "9", "10"), hi = c("11, 22, 84, 85, 108, 111", "18, 84, 85,
> 87, 122, 134",
> "2, 18, 22", "18, 108, 122, 134, 176", "19, 85, 87, 100, 107",
> "79, 85, 111", "11, 88, 108", "19, 88, 96", "19, 85, 96",
> "19, 100, 103")), .Names = c("uniq_id", "hi"), row.names = c(NA,
> -10L), class = c("tbl_df", "tbl", "data.frame"))
> >
>
> I am trying to count the frequency of the triplets in the above data using
> the below code.
>
> # split column into a list
> myList <- strsplit(newd$hi, split=",")
> # get all pairwise combinations
> myCombos <- t(combn(unique(unlist(myList)), 3))
> # count the instances where the pair is present
> myCounts <- sapply(1:nrow(myCombos), FUN=function(i) {
> sum(sapply(myList, function(j) {
> sum(!is.na(match(c(myCombos[i,]), j)))})==3)})
> #final matrix
> final <- cbind(matrix(as.integer(myCombos), nrow(myCombos)), myCounts)
>
> I hope I made my point clear. Please let me know if I miss anything.
>
> Regards,
> Sri
>
>
>
>
> On Wed, Jul 27, 2016 at 11:19 PM, Sarah Goslee <sarah.gos...@gmail.com>
> wrote:
>
> > You said you had 79 triplets and 8000 records.
> >
> > When I compared 100 triplets to 10000 records it took 86 seconds.
> >
> > So obviously there is something you're not telling us about the format
> > of your data.
> >
> > If you use dput() to provide actual examples, you will get better
> > results than if we on Rhelp have to guess. Because we tend to guess in
> > ways that make the most sense after extensive R experience, and that's
> > probably not what you have.
> >
> > Sarah
> >
> > On Wed, Jul 27, 2016 at 1:29 PM, sri vathsan <srivib...@gmail.com>
> wrote:
> > > Hi,
> > >
> > > Thanks for the solution. But I am afraid that after running this code
> > still
> > > it takes more time. It has been an hour and still it is executing. I
> > > understand the delay because each triplet has to compare almost 9000
> > > elements.
> > >
> > > Regards,
> > > Sri
> > >
> > > On Wed, Jul 27, 2016 at 9:02 PM, Sarah Goslee <sarah.gos...@gmail.com>
> > > wrote:
> > >>
> > >> Hi,
> > >>
> > >> It's really a good idea to use dput() or some other reproducible way
> > >> to provide data. I had to guess as to what your data looked like.
> > >>
> > >> It appears that order doesn't matter?
> > >>
> > >> Given than, here's one approach:
> > >>
> > >> combs <- structure(list(V1 = c(65L, 77L, 55L, 23L, 34L), V2 = c(23L,
> > 34L,
> > >> 34L, 77L, 65L), V3 = c(77L, 65L, 23L, 34L, 55L)), .Names = c("V1",
> > >> "V2", "V3"), class = "data.frame", row.names = c(NA, -5L))
> > >>
> > >> dat <- list(
> > >> c(77,65,34,23,55),
> > >> c(65,23,77,65,55,34),
> > >> c(77,34,65),
> > >> c(55,78,56),
> > >> c(98,23,77,65,34))
> > >>
> > >>
> > >> sapply(seq_len(nrow(combs)), function(i)sum(sapply(dat,
> > >> function(j)all(combs[i,] %in% j))))
> > >>
> > >> On a dataset of comparable time to yours, it takes me under a minute
> > and a
> > >> half.
> > >>
> > >> > combs <- combs[rep(1:nrow(combs), length=100), ]
> > >> > dat <- dat[rep(1:length(dat), length=10000)]
> > >> >
> > >> > dim(combs)
> > >> [1] 100 3
> > >> > length(dat)
> > >> [1] 10000
> > >> >
> > >> > system.time(test <- sapply(seq_len(nrow(combs)),
> > >> > function(i)sum(sapply(dat, function(j)all(combs[i,] %in% j)))))
> > >> user system elapsed
> > >> 86.380 0.006 86.391
> > >>
> > >>
> > >>
> > >>
> > >> On Wed, Jul 27, 2016 at 10:47 AM, sri vathsan <srivib...@gmail.com>
> > wrote:
> > >> > Hi,
> > >> >
> > >> > Apologizes for the less information.
> > >> >
> > >> > Basically, myCombos is a matrix with 3 variables which is a triplet
> > that
> > >> > is
> > >> > a combination of 79 codes. There are around 3lakh combination as
> such
> > >> > and
> > >> > it looks like below.
> > >> >
> > >> > V1 V2 V3
> > >> > 65 23 77
> > >> > 77 34 65
> > >> > 55 34 23
> > >> > 23 77 34
> > >> > 34 65 55
> > >> >
> > >> > Each triplet will compare in a list (mylist) having 8177 elements
> > which
> > >> > will looks like below.
> > >> >
> > >> > 77,65,34,23,55
> > >> > 65,23,77,65,55,34
> > >> > 77,34,65
> > >> > 55,78,56
> > >> > 98,23,77,65,34
> > >> >
> > >> > Now I want to count the no of occurrence of the triplet in the above
> > >> > list.
> > >> > I.e., the triplet 65 23 77 is seen 3 times in the list. So my output
> > >> > looks
> > >> > like below
> > >> >
> > >> > V1 V2 V3 Freq
> > >> > 65 23 77 3
> > >> > 77 34 65 4
> > >> > 55 34 23 2
> > >> >
> > >> > I hope, I made it clear this time.
> > >> >
> > >> >
> > >> > On Wed, Jul 27, 2016 at 7:00 PM, Bert Gunter <
> bgunter.4...@gmail.com>
> > >> > wrote:
> > >> >
> > >> >> Not entirely sure I understand, but match() is already vectorized,
> so
> > >> >> you
> > >> >> should be able to lose the supply(). This would speed things up a
> > lot.
> > >> >> Please re-read ?match *carefully* .
> > >> >>
> > >> >> Bert
> > >> >>
> > >> >> On Jul 27, 2016 6:15 AM, "sri vathsan" <srivib...@gmail.com>
> wrote:
> > >> >>
> > >> >> Hi,
> > >> >>
> > >> >> I created list of 3 combination numbers (mycombos, around 3 lakh
> > >> >> combinations) and counting the occurrence of those combination in
> > >> >> another
> > >> >> list. This comparision list (mylist) is having around 8000
> records.I
> > am
> > >> >> using the following code.
> > >> >>
> > >> >> myCounts <- sapply(1:nrow(myCombos), FUN=function(i) {
> > >> >> sum(sapply(myList, function(j) {
> > >> >> sum(!is.na(match(c(myCombos[i,]), j)))})==3)})
> > >> >>
> > >> >> The above code takes very long time to execute and is there any
> other
> > >> >> effecting method which will reduce the time.
> > >> >> --
> > >> >>
> > >> >> Regards,
> > >> >> Srivathsan.K
> > >> >>
> > >
> > >
> > >
> > >
> >
>
>
>
> --
>
> Regards,
> Srivathsan.K
> Phone : 9600165206
>
> [[alternative HTML version deleted]]
>
> ______________________________________________
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> and provide commented, minimal, self-contained, reproducible code.
>
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