Thanks rui..That was all i wanted...
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.
> >>>>>
> >>>
> >>>
> >
> >
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