search for: jrbouldin

When I try to select only those rows from the following data frame, called "data", in which X > Y X Y V3 2 2 1 8.062258 3 3 1 2.236068 4 4 1 6.324555 5 5 1 5.000000 6 1 2 8.062258 8 3 2 9.486833 9 4 2 2.236068 10 5 2 5.656854 11 1 3 2.236068 12 2 3 9.486833 14 4 3 8.062258 15 5 3 5.099020 16 1 4 6.324555 17 2 4 2.236068 18 3 4 8.062258 20 5 4 5.385165 21 1 5 5.000000

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

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 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

Is there a command for partial matching of character strings? Specifically, I'd like to be able to calculate the mean of the values in any columns in a data frame or matrix that have identity in part of their column names. For example, columns labeled "mpw06a" and "mpw06b" match on the first five characters; their mean would be taken whereas any columns beginning with

I am trying to perform an nls for a valid negative exponential function: zz=nls(y~constant+a.est*2.7183^(b.est*x),start=list(constant=4.0,a.est=-4,b.est = -.005),trace=T) and am getting a number of different error messages, the most problematic of which is "Error in nls(ring.area ~ constant + a.est * 2.7183^(b.est * ba.beg), start = list(constant = 4, : REAL() can only be applied to a

I can't seem to find info on how to unload packages that have been loaded. My goal in doing so is to gain access to functions that have been masked out by those packages. Or is there another way to do so? Thanks in advance. Jim Bouldin, PhD Research Ecologist Department of Plant Sciences, UC Davis Davis CA, 95616 530-554-1740

I realize this should be simple but I'm having trouble subsetting vectors and matrices, for example extracting all values meeting a certain criterion, from a vector. Cannot seem to figure out the correct syntax and help page not very helpful. Or should I be using some other function than subset. Thanks for any help. Jim Bouldin

When I try to run the following non-linear regression with variables index1 and prl3: > beta = 4 > nls(index1~beta*(1/prl3),start = list(beta = 4)) I get this error message: Error in nls(index1 ~ beta * (1/prl3), start = list(beta = 4)) : REAL() can only be applied to a 'numeric', not a 'logical' I've got no clue as to the REAL() to which this is referring. Any

OK, this has to be simple but I've searched through help files, mailing list archives and well, everything I could think of, and still no luck. I simply want to change the x axis labels in a time series graph, from its default numbering (which starts at 1 and increments by 1), to values I have in another vector, "Year". It has to be a time series graph, I don't want to have to

I just upgraded from 2.8.1 to 2.10 on Windows Vista. BIG MISTAKE apparently because now when I type: > help(functionname) or ?functionname I get only a small text window giving some very basic info on the topic, e.g.: base-package package:base R Documentation The R Base Package Description: Base R functions Details: This package contains the

Thanks much Ted. I actually had just tried what you suggest here before you posted, and resolved the problem. Thanks also for the other tips. I wrote x = as.vector(c(1:12)) because I thought that the mode of x might be the problem, the error message pointing to .Random.seed notwithstanding. On a related note, I did a brief test a couple weeks back where I ran a million random samples of 3 from

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

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

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

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

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

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