similar to: problem selecting rows meeting a criterion

Hello. I have the following distance matrix between 8 points: [1,] 0.000000 3.162278 7.280110 8.544004 7.071068 9.899495 6.403124 8.062258 [2,] 3.162278 0.000000 5.000000 6.403124 4.472136 8.944272 6.082763 8.062258 [3,] 7.280110 5.000000 0.000000 1.414214 1.000000 5.000000 4.242641 5.830952 [4,] 8.544004 6.403124 1.414214 0.000000 2.236068 4.123106 4.472136 5.656854 [5,] 7.071068 4.472136

Is there anyway to change any y[i] value (i=2,...6) to make following NLS workable? x <- c(0,5,10,15,20,25,30) y <- c(1.00000,0.82000,0.68000,0.64000,0.66667,0.68667,0.64000) lm(1/y ~~ x) nls(1/y ~~ a+b*x^c, start=list(a=1.16122,b=0.01565,c=1), trace=TRUE) #0.0920573 : 1.16122 0.01565 1.00000 #Error in numericDeriv(form[[3]], names(ind), env) : # Missing value or

This seemingly should be quite simple but I can't solve it: I have a long character vector of geographic data (data frame column named "XY") whose elements vary in length (from 11 to 14 chars). Each element is structured as a set of digits, then an underscore, then more digits, e.g: > data.frame(head(as.character(XY))) head.as.character.XY.. 1 -448623_854854 2

I'm trying to identify and remove rows in a data frame that are duplicated only on particular columns within it (i.e. not on all columns). The "unique" function looks for uniqueness across all columns of a data frame. Identifying unique rows based only on specific columns of interest returns only those columns, not all of the columns in the original frame. I tried this, and then

I'm trying to run this simple random sample procedure and keep getting the error message shown. I don't understand this; I've designated x as a numeric vector, so what is going on here? Thanks. > x = as.vector(c(1:12));x [1] 1 2 3 4 5 6 7 8 9 10 11 12 > mode(x) [1] "numeric" > sample(x, 3) Error in sample(x, 3) : .Random.seed is not an integer vector

dear all, I am stuck on the following problem. Give a string like ss1<- "z:f(5, a=3, b=4, c='1:4', d=2)" or ss2<- "f(5, a=3, b=4, c=\"1:4\", d=2)*z" I would like to remove all entries within parentheses.. Namely, I aim to obtain respectively "z:f()" or "f()*z" I played with sub() and gsub() but without success.. Thank you very

I've read up some more on the subject, and it seems the proper way to do this with GCC is g++ and target attributes. I've refactored the patch that way, and it indeed uses SSSE3 automatically on supporting CPUs, regardless of the build host, so this should be ideal both for home builders and distros. Getting the code to build right in c++ mode (checksum_sse2.cpp only) was a bit of an

I'm trying to obtain the mean of the middle 95% of the values from each row of a matrix (that is, the highest and lowest 2.5% of values in each row are removed before calculating the mean). I am having all sorts of problems with this; for example the command: apply(matrix1,1,function(x) quantile(c(.05,.90),na.rm=T)) returns the exact same quantile values for each row, which is clearly

Dan Nordlund wrote: "It would be necessary to see the code for your 'brief test' before anyone could meaningfully comment on your results. But your results for a single test could have been a valid "random" result." I've re-created what I did below. The problem appears to be with the weighting process: the unweighted sample came out much closer to the actual

OK, what is the trick to extracting the overall p value from an lm object? It shows up in the summary(lm(model)) output but I can't seem to extract it: > test2 = apply(aa, 1, function(x) summary(lm(x[,1] ~ 0 + x[,3] + x[,6]))) > test2[[1]] Call: lm(formula = x[, 1] ~ 0 + x[, 3] + x[, 6]) [omitted summary output] F-statistic: 40.94 on 2 and 7 DF, p-value: 0.0001371 It does not seem

This drop-in patch increases the performance of the get_checksum1() function on x86-64. On the target slow CPU performance of the function increased by nearly 50% in the x86-64 default SSE2 mode, and by nearly 100% if the compiler was told to enable SSSE3 support. The increase was over 200% on the fastest CPU tested in SSSE3 mode. Transfer time improvement with large files existing on both ends

This has got to be incredibly simple but I nevertheless can't figure it out as I am apparently brain dead. I just want to convert the elements of a character vector to variable names, so as to then assign formulas to them, e.g: z = c("model1","model2"); I want to assign formulas, such as lm(y~x[,1]) and lm(y~x[,2]), to the variables "model1" and

Within a very large matrix composed of a mix of values and NAs, e.g, matrix A: [,1] [,2] [,3] [1,] 1 NA NA [2,] 3 NA NA [3,] 3 10 17 [4,] 4 12 18 [5,] 6 16 19 [6,] 6 22 20 [7,] 5 11 NA I need to be able to consecutively number, in new columns, the non-NA values within each column (i.e. A[1,1] A[3,2] and A[3,3] would all be set to one, and

I realize this is probably pretty basic but I can't figure it out. I'm looping through an array, doing various calculations and producing a resulting data frame in each loop iteration. I need to give each data frame a different name. Although I can easily create a new character string for writing each frame to an output file, I cannot figure out how to convert such strings to

I continue to have great frustrations with NA values--in particular making summary calculations on rows or cols of a matrix containing them. For example, why does: > a = matrix(1:30,nrow=5) > is.na(a[c(1:2),c(3:4)]);a [,1] [,2] [,3] [,4] [,5] [,6] [1,] 1 6 NA NA 21 26 [2,] 2 7 NA NA 22 27 [3,] 3 8 13 18 23 28 [4,] 4 9 14 19 24 29

Hola everyone, Solving this issue would be a huge time saver for analyzing my thesis data, so I'll try to be as concise as possible and hope that someone can give me a hand. I am trying to run a function that I created, which works fine until I try to use it with the by function. That function is listed at the bottom (f3) just in case that might be the issue. Essentially I have compiled a

Hello, Try this. ss1 <- "z:f(5, a=3, b=4, c='1:4', d=2)" ss2 <- "f(5, a=3, b=4, c=\"1:4\", d=2)*z" fun <- function(s) sub("(\\().*(\\))", "\\1\\2", s) fun(ss1) #[1] "z:f()" fun(ss2) #[1] "f()*z" Hope this helps, Rui Barradas On 5/21/2018 2:33 PM, Vito M. R. Muggeo wrote: > dear all, > I am stuck

I want to perform linear regression on groups of consecutive rows--say 5 to 10 such--of two matrices. There are many such potential groups because the matrices have thousands of rows. The matrices are both of the form: > shp[1:5,16:20] SL495B SL004C SL005C SL005A SL017A -2649 1.06 0.56 NA NA NA -2648 0.97 0.57 NA NA NA -2647 0.46 0.30 NA NA

I admit that I've not done a thorough search on this topic, but from the several instructional manuals and/or tutorials I've looked at, I don't see any mention of relational database capabilities in R? Have I missed something, and if so, can someone point me in the right direction to get started? Thanks! Jim Bouldin, PhD Research Ecologist Department of Plant Sciences, UC Davis

I've got a couple of SPARCstation 2s (about as fast as a fast 486, for most thing) that I'm going to be using for some testing. I realize that these machines are a bit slow, but when connecting via OpenSSH, it's MUCH slower than connecting to my 486-DX 50. The point where it waits is just after "debug: Sent encrypted session key.". The pause is for about 10 seconds, while