What you call "continuous" I call "cyclic" (with period NROW(x) in this case).
An easy to way to deal with this is to repeat the column three times, do the
original analysis, and return the results corresponding to the middle part of
the
repeated x. I think the following does the right thing, but have not tested it
much.
fCyclic <- function (x, threshold = 0.8 * max(x), ...)
{
if (all(x > threshold)) {
index <- which.max(x)
data.frame(Index=index, Value=x[index])
} else {
ret <- f(x = rep(x, 3), threshold, ...)
ret <- ret[ret$Index > length(x) & ret$Index <= 2 * length(x), ]
ret$Index <- ret$Index - length(x)
ret
}
}
It does require a test for all of x being above the threshold, which f didn't
need. You could put the test in f and take it out of fCyclic.
Again, use apply() if you to want to run it over each column.
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
From: eliza botto [mailto:eliza_bo...@hotmail.com]
Sent: Friday, May 24, 2013 4:41 PM
To: ruipbarra...@sapo.pt; William Dunlap
Cc: r-help@r-project.org
Subject: RE: [R] Continuous columns of matrix
Dear Rui,
i just noticed that there is one slight problem in the William' code. If you
see the results of column number 4 which are
[[4]]
Index Value
1 1 2.061959
2 12 2.245474
where as we already dicussed that columns are continuous and 1 and 12 cant be
picked together.
so the results should have been
[[4]]
Index Value
1 12 2.245474
how to get rid of this error??
Thanks in advance...
Elisa
> Date: Fri, 24 May 2013 23:38:51 +0100
> From: ruipbarra...@sapo.pt<mailto:ruipbarra...@sapo.pt>
> To: wdun...@tibco.com<mailto:wdun...@tibco.com>
> CC: eliza_bo...@hotmail.com<mailto:eliza_bo...@hotmail.com>;
> r-help@r-project.org<mailto:r-help@r-project.org>
> Subject: Re: [R] Continuous columns of matrix
>
> Hello,
>
> Just use
>
> apply(mat, 2, f)
>
> Rui Barradas
>
> Em 24-05-2013 23:25, William Dunlap escreveu:
> > Are you trying to identify the highest point in each run of points higher
> > than a threshold,
> > akin to the problem of naming mountains so that each minor bump on a high
> > ridge does not
> > get its own name? The following does that:
> >
> > f <- function (x, threshold = 0.8 * max(x), plot=FALSE)
> > {
> > if (plot) { plot(x, type="l") ; abline(h=threshold) }
> > big <- x > threshold
> > n <- length(big)
> > startRunOfBigs <- which(c(big[1], !big[-n] & big[-1]))
> > if (plot) abline(v=startRunOfBigs, col="green")
> > endRunOfBigs <- which(c(big[-n] & !big[-1], big[n]))
> > if (plot) abline(v=endRunOfBigs, col="red")
> > stopifnot(length(startRunOfBigs) == length(endRunOfBigs),
> > all(startRunOfBigs <= endRunOfBigs))
> > index <- vapply(seq_along(startRunOfBigs),
> > function(i)which.max(x[startRunOfBigs[i]:endRunOfBigs[i]])+startRunOfBigs[i]-1L,
> > 0L)
> > if (plot && length(index)>0) points(cex=1.5, index, x[index])
> > data.frame(Index=index, Value=x[index])
> > }
> >
> > E.g., with your data:
> >> x <- c(0.563879907485297, 0.749077642331436, 0.681023650957497,
> >> 1.30140773002346,
> > 1.46377246795771, 1.20312609775816, 0.651886452442823, 0.853749099839423,
> > 1.041608733313, 0.690719733451964, 1.49415144965002, 1.30559703478921
> > )
> >> f(x, plot=TRUE) # the plot illustrates what it is doing
> > Index Value
> > 1 5 1.463772
> > 2 11 1.494151
> >
> > Bill Dunlap
> > Spotfire, TIBCO Software
> > wdunlap tibco.com
> >
> >
> >> -----Original Message-----
> >> From: r-help-boun...@r-project.org<mailto:r-help-boun...@r-project.org>
> >> [mailto:r-help-boun...@r-project.org]<mailto:[mailto:r-help-boun...@r-project.org]>
> >> On Behalf
> >> Of eliza botto
> >> Sent: Friday, May 24, 2013 2:10 PM
> >> To: ruipbarra...@sapo.pt<mailto:ruipbarra...@sapo.pt>;
> >> r-help@r-project.org<mailto:r-help@r-project.org>
> >> Subject: Re: [R] Continuous columns of matrix
> >>
> >> Dear Rui,
> >> Regarding your last reply there is a small additional question which i
> >> want to ask. For the
> >> following column
> >>> dput(mat[,1])c(0.563879907485297, 0.749077642331436, 0.681023650957497,
> >> 1.30140773002346, 1.46377246795771, 1.20312609775816, 0.651886452442823,
> >> 0.853749099839423, 1.041608733313, 0.690719733451964, 1.49415144965002,
> >> 1.30559703478921)
> >> Your loop gives following results
> >> [,1] [,2] [,3] [,4]
> >> index 4.000000 5.000000 6.000000 11.000000
> >> values 1.301408 1.463772 1.203126 1.494151
> >> which is very accurate. Index 11 had the maximum value and 4,5,6 were the
> >> neibhours. i
> >> want to add an other condition which is that if a column has more
> >> than 1 neibhours, than just like the maximum value, those neibhour can not
> >> be next to
> >> each other as well.
> >> So what i should have is the following
> >> [,1] [,2]
> >> index 5.000000 11.000000
> >> values 1.463772 1.494151
> >> Is there anyway of doing it??
> >> Thanks in advance
> >> Elisa
> >>
> >>> Date: Fri, 24 May 2013 17:02:56 +0100
> >>> From: ruipbarra...@sapo.pt<mailto:ruipbarra...@sapo.pt>
> >>> To: eliza_bo...@hotmail.com<mailto:eliza_bo...@hotmail.com>
> >>> CC: b...@xs4all.nl<mailto:b...@xs4all.nl>;
> >>> r-help@r-project.org<mailto:r-help@r-project.org>
> >>> Subject: Re: [R] Continuous columns of matrix
> >>>
> >>> Hello,
> >>>
> >>> No problem. Just change <0 to >= and Inf to -Inf:
> >>>
> >>> fun2 <- function(x){
> >>> n <- length(x)
> >>> imx <- which.max(x)
> >>> if(imx == 1){
> >>> x[2] <- x[n] <- -Inf
> >>> }else if(imx == n){
> >>> x[1] <- x[n - 1] <- -Inf
> >>> }else{
> >>> x[imx - 1] <- -Inf
> >>> x[imx + 1] <- -Inf
> >>> }
> >>> index <- which(x >= 0.8*x[imx])
> >>> values <- x[index]
> >>> list(index = index, values = values)
> >>> }
> >>>
> >>> apply(mat, 2, fun2)
> >>>
> >>>
> >>> Rui Barradas
> >>>
> >>> Em 24-05-2013 16:23, eliza botto escreveu:
> >>>> Dear Rui,
> >>>>
> >>>> I infact wanted to have something like the following..
> >>>> suppose the columns are
> >>>>
> >>>> structure(c(0.706461987893674, 0.998391468394261, 0.72402995269242,
> >> 1.70874688194537, 1.93906363083693, 0.89540353128442, 0.328327645695443,
> >> 0.427434603701202, 0.591932250254601, 0.444627635494183, 1.44407704434405,
> >> 1.79150336746345, 2.06195904001605, 1.41493262330451, 1.35748791897328,
> >> 1.19490680241894, 0.702488756183322, 0.338258418490199, 0.123398398622741,
> >> 0.138548982660226, 0.16170889185798, 0.414543218677095, 1.84629295875002,
> >> 2.24547399004563), .Dim = c(12L, 2L))
> >>>>
> >>>> For col 1
> >>>> [[1]]
> >>>> $Index
> >>>> 5 12
> >>>> $value
> >>>> 1.939 1.79
> >>>> Although value 1.708 of index 4 also has value which is above 80% of the
> >>>> maximum
> >> value but as it is in the neighbor of maxmimum value so we wont consider
> >> it.
> >>>> similarly for the column 2
> >>>> [[1]]
> >>>> $Index
> >>>> 12
> >>>> $value
> >>>> 2.245
> >>>> Although values 1.846 of index 11 and 2.0619 of index 1 also have values
> >>>> which are
> >> above 80% of the maximum value but as they are in the neighbor of maxmimum
> >> value so
> >> we wont consider them.
> >>>> i am sorry if the manner in which i asked my question was not conclusive.
> >>>> i hope you wont mind...
> >>>> Elisa
> >>>>> Date: Fri, 24 May 2013 15:59:50 +0100
> >>>>> From: ruipbarra...@sapo.pt<mailto:ruipbarra...@sapo.pt>
> >>>>> To: eliza_bo...@hotmail.com<mailto:eliza_bo...@hotmail.com>
> >>>>> CC: b...@xs4all.nl<mailto:b...@xs4all.nl>;
> >>>>> r-help@r-project.org<mailto:r-help@r-project.org>
> >>>>> Subject: Re: [R] Continuous columns of matrix
> >>>>>
> >>>>> Hello,
> >>>>>
> >>>>> Something like this?
> >>>>>
> >>>>>
> >>>>> fun2 <- function(x){
> >>>>> n <- length(x)
> >>>>> imx <- which.max(x)
> >>>>> if(imx == 1){
> >>>>> x[2] <- x[n] <- Inf
> >>>>> }else if(imx == n){
> >>>>> x[1] <- x[n - 1] <- Inf
> >>>>> }else{
> >>>>> x[imx - 1] <- Inf
> >>>>> x[imx + 1] <- Inf
> >>>>> }
> >>>>> index <- which(x <= 0.8*x[imx])
> >>>>> values <- x[index]
> >>>>> list(index = index, values = values)
> >>>>> }
> >>>>>
> >>>>> apply(mat, 2, fun2)
> >>>>>
> >>>>>
> >>>>> Rui Barradas
> >>>>>
> >>>>> Em 24-05-2013 13:40, eliza botto escreveu:
> >>>>>> Dear Rui,Thankyou very much for your help. just for my own knowledge
> >>>>>> what if
> >> want the values and index, which are less than or equal to 80% of the
> >> maximum value
> >> other than those in the neighbors?? like if maximum is in row number 5 of
> >> any column
> >> then the second maximum can be in any row other than 4 and 6. similarly if
> >> maximum is
> >> in row number 12 than the second maximum can be in any row other than 1 and
> >> 11...thankyou very much for your help
> >>>>>> elisa
> >>>>>>
> >>>>>>> Date: Fri, 24 May 2013 12:37:37 +0100
> >>>>>>> From: ruipbarra...@sapo.pt<mailto:ruipbarra...@sapo.pt>
> >>>>>>> To: eliza_bo...@hotmail.com<mailto:eliza_bo...@hotmail.com>
> >>>>>>> CC: b...@xs4all.nl<mailto:b...@xs4all.nl>;
> >>>>>>> r-help@r-project.org<mailto:r-help@r-project.org>
> >>>>>>> Subject: Re: [R] Continuous columns of matrix
> >>>>>>>
> >>>>>>> Hello,
> >>>>>>>
> >>>>>>> Berend is right, it's at least confusing. To get just the index of the
> >>>>>>> maximum value in each column,
> >>>>>>>
> >>>>>>> apply(mat, 2, which.max)
> >>>>>>>
> >>>>>>>
> >>>>>>> To get that index and the two neighbours (before and after, wraping
> >>>>>>> around) if they are greater than or equal to 80% of the maximum, try
> >>>>>>>
> >>>>>>> fun <- function(x){
> >>>>>>> n <- length(x)
> >>>>>>> imx <- which.max(x)
> >>>>>>> sec <- numeric(2)
> >>>>>>> if(imx == 1){
> >>>>>>> if(x[n] >= 0.8*x[imx]) sec[1] <- n
> >>>>>>> if(x[2] >= 0.8*x[imx]) sec[2] <- 2
> >>>>>>> }else if(imx == n){
> >>>>>>> if(x[n - 1] >= 0.8*x[imx]) sec[1] <- n - 1
> >>>>>>> if(x[1] >= 0.8*x[imx]) sec[2] <- 1
> >>>>>>> }else{
> >>>>>>> if(x[imx - 1] >= 0.8*x[imx]) sec[1] <- imx - 1
> >>>>>>> if(x[imx + 1] >= 0.8*x[imx]) sec[2] <- imx + 1
> >>>>>>> }
> >>>>>>> sec <- sec[sec != 0]
> >>>>>>> c(imx, sec)
> >>>>>>> }
> >>>>>>>
> >>>>>>> apply(mat, 2, fun)
> >>>>>>>
> >>>>>>>
> >>>>>>> Note that the result comes with the maximum first and the others
> >>>>>>> follow.
> >>>>>>>
> >>>>>>> Hope this helps,
> >>>>>>>
> >>>>>>> Rui Barradas
> >>>>>>>
> >>>>>>>
> >>>>>>> Em 24-05-2013 11:41, eliza botto escreveu:
> >>>>>>>> There you go!!!
> >>>>>>>>
> >>>>>>>> structure(c(0.706461987893674, 0.998391468394261, 0.72402995269242,
> >> 1.70874688194537, 1.93906363083693, 0.89540353128442, 0.328327645695443,
> >> 0.427434603701202, 0.591932250254601, 0.444627635494183, 1.44407704434405,
> >> 1.79150336746345, 0.94525563730664, 1.1025988539757, 0.944726401770203,
> >> 0.941068515436361, 1.50874009152312, 0.590015480056925, 0.311905493999476,
> >> 0.596771673581893, 1.01502499067153, 0.803273181849135, 1.6704085033648,
> >> 1.57021117646422, 0.492096635764151, 0.433332688044914, 0.521585941816778,
> >> 1.66472272302545, 2.61878329527404, 2.19154489521664, 0.493876245329722,
> >> 0.4915787202584, 0.889477365620806, 0.609135860199222, 0.739201878930367,
> >> 0.854663750519518, 0.948228727226247, 1.38569091844218, 0.910510759802679,
> >> 1.25991218521949, 0.993123416952421, 0.553640392997634, 0.357487763503204,
> >> 0.368328033777003, 0.344255688489322, 0.423679560916755, 1.32093576037521,
> >> 3.13420679229785, 2.06195904001605, 1.41493262330451, 1.35748791897328,
> >> 1.19490680241894, 0.702488!
> >> 75618332!
> >>>>>>>> 2, 0.338258418490199, 0.123398398622741, 0.138548982660226,
> >> 0.16170889185798, 0.414543218677095, 1.84629295875002, 2.24547399004563,
> >> 0.0849732189580101, 0.070591276171845, 0.0926010253161898,
> >> 0.362209761457517,
> >> 1.45769283057202, 3.16165004659667, 2.74903557756267, 1.94633472878995,
> >> 1.19319875840883, 0.533232612926756, 0.225531074123974, 0.122949089115578),
> >> .Dim = c(12L, 6L))
> >>>>>>>>
> >>>>>>>> Thanks once again..
> >>>>>>>> Elisa
> >>>>>>>>
> >>>>>>>>
> >>>>>>>>> Subject: Re: [R] Continuous columns of matrix
> >>>>>>>>> From: b...@xs4all.nl<mailto:b...@xs4all.nl>
> >>>>>>>>> Date: Fri, 24 May 2013 12:36:47 +0200
> >>>>>>>>> CC: r-help@r-project.org<mailto:r-help@r-project.org>
> >>>>>>>>> To: eliza_bo...@hotmail.com<mailto:eliza_bo...@hotmail.com>
> >>>>>>>>>
> >>>>>>>>>
> >>>>>>>>> On 24-05-2013, at 12:24, eliza botto
> >>>>>>>>> <eliza_bo...@hotmail.com<mailto:eliza_bo...@hotmail.com>> wrote:
> >>>>>>>>>
> >>>>>>>>>> Dear useRs,If i have a matrix, say, 12 rows and 6 columns. The
> >>>>>>>>>> columns are
> >> continuous. I want to find the index of maximum values and the actual
> >> maximum values.
> >> The maximum values in each column are the highest values and the values
> >> greater than
> >> or equal to 80% of the maximum value. Moreover, if a column has more than
> >> one
> >> maximum values than these values should come immediately next to each
> >> other. For
> >> example, if you column 1 has a highest value in 6th row then the second
> >> maximum values
> >> cant be in row 5 or 7. And as the columns are continuous therefore, if
> >> maximum value is
> >> in row 12th, then the second maximum cant be in row 11 and 1.Thankyou very
> >> much
> >> indeed in advance
> >>>>>>>>>
> >>>>>>>>>
> >>>>>>>>> Incomprehensible.
> >>>>>>>>> What is a continuous column?
> >>>>>>>>>
> >>>>>>>>> Please give an example input matrix and and the result you want.
> >>>>>>>>>
> >>>>>>>>> Berend
> >>>>>>>>>
> >>>>>>>>>> Elisa
> >>>>>>>>>> [[alternative HTML version deleted]]
> >>>>>>>>>>
> >>>>>>>>>
> >>>>>>>>> Please post in plain text.
> >>>>>>>>>
> >>>>>>>>
> >>>>>>>> [[alternative HTML version deleted]]
> >>>>>>>>
> >>>>>>>> ______________________________________________
> >>>>>>>> R-help@r-project.org<mailto:R-help@r-project.org> mailing list
> >>>>>>>> https://stat.ethz.ch/mailman/listinfo/r-help
> >>>>>>>> PLEASE do read the posting guide
> >>>>>>>> http://www.R-project.org/posting-guide.html
> >>>>>>>> and provide commented, minimal, self-contained, reproducible code.
> >>>>>>>>
> >>>>>>
> >>>>>>
> >>>>
> >>>>
> >>
> >> [[alternative HTML version deleted]]
> >>
> >> ______________________________________________
> >> R-help@r-project.org<mailto:R-help@r-project.org> mailing list
> >> https://stat.ethz.ch/mailman/listinfo/r-help
> >> PLEASE do read the posting guide
> >> http://www.R-project.org/posting-guide.html
> >> and provide commented, minimal, self-contained, reproducible code.
[[alternative HTML version deleted]]
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