Thanks rui.
you really are an R genius...
it might have taken me 1000 years to figure it out..
thanks
Elisa
> Date: Fri, 24 May 2013 23:38:51 +0100
> From: ruipbarra...@sapo.pt
> To: wdun...@tibco.com
> CC: eliza_bo...@hotmail.com; 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]
> >> On Behalf
> >> Of eliza botto
> >> Sent: Friday, May 24, 2013 2:10 PM
> >> To: ruipbarra...@sapo.pt; 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
> >>> To: eliza_bo...@hotmail.com
> >>> CC: b...@xs4all.nl; 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
> >>>>> To: eliza_bo...@hotmail.com
> >>>>> CC: b...@xs4all.nl; 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
> >>>>>>> To: eliza_bo...@hotmail.com
> >>>>>>> CC: b...@xs4all.nl; 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
> >>>>>>>>> Date: Fri, 24 May 2013 12:36:47 +0200
> >>>>>>>>> CC: r-help@r-project.org
> >>>>>>>>> To: eliza_bo...@hotmail.com
> >>>>>>>>>
> >>>>>>>>>
> >>>>>>>>> On 24-05-2013, at 12:24, eliza botto <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 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 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.
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