similar to: select from data frame

If I understand correctly, no looping (ave(), for()) or type casting (as.character()) is needed -- indexing and matching suffice: > with(df, ID[!ID %in% unique(ID[samples %in% c("B","C") ])]) [1] 3 3 Cheers, Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into it." -- Opus (aka Berkeley Breathed in

> On Jul 15, 2017, at 4:01 AM, Andras Farkas via R-help <r-help at r-project.org> wrote: > > Dear All, > > wonder if you could please assist with the following > >

... and here is a slightly cleaner and more transparent way of doing the same thing (setdiff() does the matching) > with(df, setdiff(ID,ID[samples %in% c("B","C") ])) [1] 3 -- Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into it." -- Opus (aka Berkeley Breathed in his "Bloom County"

Dear All, would you please provide your thoughts on the following: let us say I have: a <-c(1,5,8,15,32,69) b <-c(8.5,33) and I would like to extract from "a" the two values that are closest to the values in "b", where the length of this vectors may change but b will allways be shorter than "a". So at the end based on this example I should have the result

Dear all,   could you please give me some pointers on how I could make R screen for a value if it falls within a certain range? I looked at the subset function, but is not doing it, perhaps because I only have 1 value to screen?   aptreciate the input   ex: a <-16.5 I would like to screen to see if a is within the range of 15 to 20, (which it is:-)), and I would like the code to return a value

Dear All   I have the following data (somewhat simplyfied):   TINF <-1 a <-c(500,750,1000,1250,1500,1750,2000) b <-c(8,12,18,24,36,48,60,72,96)   following function:   infcprodessa <-function (D, tin, tau, ts)   (D * (1 - exp(-0.048 * tin))/(tin * (0.048*79) * (1 - exp(-0.048 * tau)))) * exp(-0.048 * (ts - tin)) z <-sapply(1:1, function(n) infcprodessa(1000,TINF,12,12-TINF))   is

Dear All,   this is probably an easy one but I can not get a handle on it:   x <-c(1,2,3,4,5) y <-c(6,7,8,9,10) z <-15 w <-ifelse(z>14,x,y)   this will give me a value of 1 for w. What I would like to get is the whole string of x, so that w would become a numeric object of 5 characters exactly the same as x.   Apreciate the help,   Sincerely,   Andras [[alternative HTML version

Dear All,   I have a matrix with 33 columns and 5000 rows. I would like to find 2 specific columns in the set: the one that holds the highest values and the one that holds the lowest values. In this case the column's mean would be apropriate to use to try to find those specific columns because each columns mean is different and they all change together based on the same "change of rate

Dear All,   as a follow up to my previous e-mail (I think I am getting closer...):   I am trying to apply the trapezoidal functions to a matric column by column. I have the following code:   a <-matrix(c(1:100),ncol=10) b <-matrix(c(2,4,6,8,10,12,14,16,18,20))   apply(a,2,function(b,a) sum(diff(b)*(a[-1]+a[-length(a)]))/2)   for some reason i get an error message: Error in FUN(newX[[, i],

Dear All,   I often have to work with certain models in which I try to "reproduce" a distribution the best I can with very little known information avaible. Is there a package or function in R that could best reproduce a probability distribution using only the mean, median and SD values availble without knowing the actual distribution type to begin with and/or the covariance matrix (for

Dear All,   I am working with a Sweave document to be converted into PDF using Rstudio. It seems to me that my R code will also show up after conversion, which I would like not to happen. Is there a way to specify a command that I could place on the beggining and at the end of the R code that would allow R to recognize the area where my R code is and would ignor it when the data is converted?  

Dear All,   I was wondering if you could help me with the following: I have the code:   tin <-0.5 tau <-24 output0 <-10 TIMELOW <-tin TIMEHIGH <-1*tau TIME1 <-c(seq(TIMELOW,TIMEHIGH, by = sign(TIMEHIGH-TIMELOW)*(tau-tin)/3))   then I would like to calculate:   cp1 <-output0*exp(-0.3*TIME1[1]) cp2 <-output0*exp(-0.3*TIME1[2]) cp3 <-output0*exp(-0.3*TIME1[3]) cp4

Dear All,   sorry, got stuck again on the following: let us say we have:   a <-c(1:5) b <-c(6:10) d <-cbind(a,b)     from d I would like to remove total number of rows based on the length of f. So if:   f <-c(1)   my result is working great with the following solution:   d[-length(f),]   so I get: a b [1,] 2 7 [2,] 3 8 [3,] 4 9 [4,] 5 10 but if I do: f <-c(1,2) then I get: 

Dear All,   another quick question, this one is on skipping part of my code, so let us say:   a <-5 b <-2 e <-0   d <-a+b f <-a-b   what I would like to do is to have R NOT to calculate the value for d in case the value of e equals to zero (essentially skip that "chunk"), but instead move on to calculate te value for f. In the code I am working with the value of e changes,

Dear All,   wonder if you have a thought on the following: I am using the round(x,digits=3) command, but some of my values come out as: 0.07099999999999999 AND 0.06900000000000001. Any thoughts on why this maty be happening or how to eliminate the problem?   apreciate the help,   Andras [[alternative HTML version deleted]]

Dear All,   is there a way to set low and high limits to a simulation with rlnorm()?   as an example:  a <-rlnorm(500,0.7,1)     I get the summary of   Min. 1st Qu. Median Mean 3rd Qu. Max. 0.1175 1.0590 2.1270 3.4870 4.0260 45.3800 I would like to set limits so that the simulated values minimum would be greater then 0.5 and maximum of less than 30. If during simulation a

Dear All   I would like to convert matrix rows to columns. I am thinking the t() function should help, but am having a hard time converting the matrix into the dimensions I would like them to. Example:   a <-matrix(c(1:30),ncol=3) gives me:[,1] [,2] [,3] [1,] 1 11 21 [2,] 2 12 22 [3,] 3 13 23 [4,] 4 14 24 [5,] 5 15 25 [6,] 6 16 26 [7,] 7 17

David, thanks for the response. In your response the quantile function (if I see correctly) runs on the columns versus I need to run it on the rows, which is an easy fix, but that is not exactly what I had in mind... essentially we can remove t() from my original code to make "res" look like this: res<-apply(z, 1, quantile, probs=c(0.3)) but after all maybe I did not explain

Dear All,   I have the following code set up:   x <-2000 y <-8 z <-3   I would need to use these numbers to show up in my plot title "mixed" with text. The x,y,z numbers would need to change, the text would not. So my title should look like this   "x txt1 y txt2 z txt3"   so if: txt1=hours txt2=minutes txt3=seconds   then my title of the plot should read:   2000 hours

Dear List,   I have the following code:   x <-c(0, 13.8333333333333, 38.1666666666667, 62.1666666666667, 85.9166666666667, 108.916666666667) y <-c(1.77, 2.39, 3, 2.65, 2.62, 1.8) Interpolated <- approx(x, y,xout=0:tail(x, n=1),method="linear") plot(Interpolated)   in this code x is time in hours (cumulative), and y is a biological variable. I am using linear interpolation