search for: 0.06

Both of the approx functions work correctly individually, but they are not being distinguished in the for loop by the if statments. Any help would be appreciated. for loop of interest is below x <- (structure(list(Site = structure(c(2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,

I have a question concerning the Wilcoxon signed-rank test, and specifically, which R subroutine I should use for my particular dataset. There are three different commands in R (that I'm aware of) that calculate the Wilcoxon signed-rank test; wilcox.test, wilcox.exact, and wilcoxsign_test. When I run the three commands on the same dataset, I get different p-values. I'm hoping that

R Users, I am new to R and have tried to figure out how to automate this process instead of using excel. I have read in this dataframe into r with read.table. I need to reshape the data from the first table into the format of the second table. TractID StandID Species CruiseDate DBHClass TreesPerAcre Carbon Stand 1 Loblolly Pine 5/20/2010 10 1.2 Carbon Stand 1 Loblolly Pine

Dear Group, I have the following data matrix which is a timeseries. > dput(tData) structure(list(A = c(0.2, 0.13, 0.05, 0.1, 0.02, 0.18, 0.09, 0.06, 0.13), B = c(0.15, 0.06, 0.09, 0.02, 0.03, 0.12, 0.01, 0.15, 0.06), C = c(-0.1, 0, -0.07, -0.06, -0.05, -0.05, -0.06, -0.08, -0.07), D = c(-0.15, -0.05, -0.1, -0.03, -0.13, -0.04, -0.1, -0.04, -0.15), E = c(-0.17, -0.16, -0.08, -0.07, -0.09,

Hi R-experts I have many text files to read and combined them into one into R that are output from other programs. My textfile have unbalanced number of rows for example: ;this is example ; r help Var1 Var2 Var3 Var4 Var5 0 0.05 0.01 12 1 0.04 0.06 18 A 2 0.05 0.08 14 3 0.01 0.06

Hello everyone,           I have a basic question regarding logical operators. > x<-seq(-1,1,by=0.02) > x   [1] -1.00 -0.98 -0.96 -0.94 -0.92 -0.90 -0.88 -0.86 -0.84 -0.82 -0.80 -0.78  [13] -0.76 -0.74 -0.72 -0.70 -0.68 -0.66 -0.64 -0.62 -0.60 -0.58 -0.56 -0.54  [25] -0.52 -0.50 -0.48 -0.46 -0.44 -0.42 -0.40 -0.38 -0.36 -0.34 -0.32 -0.30  [37] -0.28 -0.26 -0.24 -0.22 -0.20 -0.18 -0.16

hello, i'm pete ,how can i order rows of matrix by max to min value? I have a matrix of membership degrees, with 82 (i) rows and K coloumns, K are clusters. I need first and second largest elements of the i-th row. for example 1 0.66 0.04 0.01 0.30 2 0.02 0.89 0.09 0.00 3 0.06 0.92 0.01 0.01 4 0.07 0.71 0.21 0.01 5 0.10 0.85 0.04 0.01 6 0.91 0.04 0.02 0.02 7 0.00 0.01 0.98 0.00 8 0.02

Dear All, I need to do some very basic R profiling, something along the lines of: run this whole script five times and return the average completion time. I do not want (at this stage) delve into the details of the percentage of the time spent in which function and doing what. Which tools should I use? Any recommendation is welcome. Best Regards Lorenzo

Hi, I have this symmetric matrix, at least I think so. col1 col2 col3 [1,] 0.20 0.05 0.06 [2,] 0.05 0.10 0.03 [3,] 0.06 0.03 0.08 or structure(c(0.2, 0.05, 0.06, 0.05, 0.1, 0.03, 0.06, 0.03, 0.08 ), .Dim = c(3L, 3L), .Dimnames = list(NULL, c("var1", "var2", "var3"))) But isSymmetric() doesn't agree. Any comment? I am on R 2.10.1 Thanks. Jun

Olly Betts <olly at survex.com> wrote: > On Fri, Aug 21, 2020 at 09:06:59AM +0000, Eric Wong wrote: > > Going back to the "prioritizing aggregated DBs" thread from > > February 2020, I've got 390 Xapian shards for 130 public inboxes > > I want to search against(*). There's more on the horizon (we're > > expecting tens of thousands of public

Going back to the "prioritizing aggregated DBs" thread from February 2020, I've got 390 Xapian shards for 130 public inboxes I want to search against(*). There's more on the horizon (we're expecting tens of thousands of public inboxes). After bumping RLIMIT_NOFILE and running ->add_database a bunch, the actual queries seem to be taking ~30s (not good :x). Now I'm

** Disclaimer: I'm looking for general suggestions ** I'm sorry, but can't send out the file I'm using, so there is no reproducible example. I'm using read.table and it's taking over 30 seconds to read a tiny file. The strange thing is that it takes roughly the same amount of time if the file is 100 times larger. After re-reviewing the data Import / Export manual I think

Hello there, I am not quite sure how to interpret the output of Rprof (in the following the output I was staring at). I was poking around the web a little bit for documentation but without much success. I guess if I want to figure out what takes so long in my code the 2nd table $by.total and the total.pct column (pct = percent) is the most helpful. What does it mean that [ or [.data.frame is

Hello, Please keep this on the list, always cc r-help. One of the files in your attachment is empty: y <- read.csv(file.choose("GT.csv")) Error in read.table(file = file, header = header, sep = sep, quote = quote,? : ? no lines available in input Rui Barradas ? Citando Chaitanya Ganne <Chaitanya.Ganne at jefferson.edu>: > Thank you so much for your input. > > I am

Just out of curiosity, I took the default "iris" example in the RF helpfile... but seeing the admonition against using the formula interface for large data sets, I wanted to play around a bit to see how the various options affected the output. Found something interesting I couldn't find documentation for... Just like the example... > set.seed(12) # to be sure I have

Short: get rid of the loops I use and optimize runtime Dear all, I want to calculate for each row the amount of the month ago. I use a matrix with 2100 rows and 22 colums (which is still a very small matrix. nrows of other matrixes can easily be more then 100000) Table before Year month quarter yearmonth Service ... Amount 2009 9 Q3 092009 A ...

Should I submit this as a bug report? --- reg-tests-3.Rout.save Thu Jul 3 09:55:40 2003 +++ reg-tests-3.Rout Sun Nov 23 13:10:57 2003 @@ -1,17 +1,18 @@ -R : Copyright 2003, The R Development Core Team -Version 1.8.0 Under development (unstable) (2003-07-03) +R : Copyright 2003, The R Foundation for Statistical Computing +Version 1.8.1 (2003-11-21), ISBN 3-900051-00-3 R is free software and

Greeting. Dear Mr/Mrs/Miss, OTU ID Health Disease Bacterial 1 0.29 0.34 Bacterial 2 0.25 0.07 Bacterial 3 0.06 0.06 Bacterial 4 0.07 0.09 Bacterial 5 0.02 0.05 Above show the first 6 data sets, may I ask that the reason of R show the error like "Error in cor(data) : 'x' must be numeric" ? And how to solve it? Besides, isn't this data can conduct correlation matrix?

Dear Guys, I do converting codes from Fortran into R and got stuck in solving LOOPING procedure with R. In FORTRAN, it is (DO and END DO) for looping in the net. In R, it is (FOR with { }). I believe there is something wrong with my coding in R, do hope that you can help me solving following problems. It seems easy, but results are not the same.

Hello, Also, the other file, NPA.csv, is not in tabular form. Can you please reformat it? Rui Barradas Citando ruipbarradas at sapo.pt: > Hello, > > Please keep this on the list, always cc r-help. > One of the files in your attachment is empty: > > y <- read.csv(file.choose("GT.csv")) > Error in read.table(file = file, header = header, sep = sep, quote =