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Dear R users I have a big matrix like 6021 1188 790 290 1174 1015 1990 6613 6288 100714 6021 1 0.658 0.688 0.474 0.262 0.163 0.137 0.32 0.252 0.206 1188 0.658 1 0.917 0.245 0.331 0.122 0.148 0.194 0.168 0.171 790 0.688 0.917 1 0.243 0.31 0.122 0.15 0.19 0.171 0.174 290 0.474

Hi, In my analysis of impacts of insecticide-treated bednets on malaria, I look at the relationship between malaria incidence and mosquito behaviors. The condensed data set is copied here. Ordinary regression (lm) shows that Incidence was negatively related to Mortality. This makes sense because the latter reflected the strength of killing mosquitoes by insecticide-treated nets. Since the

The editor in R version 1.2.0 returns an error after editing a data frame that has row names. This is on a Windows 2000 system, but I've encountered the same problem under Windows 95. Here's a simple example: ------------------------------------------------------------------------------------ R : Copyright 2000, The R Development Core Team Version 1.2.0 (2000-12-15) R is free

On Sun, 31 Dec 2000 jfox@mcmaster.ca wrote: > The editor in R version 1.2.0 returns an error after editing a data frame > that has row names. This is on a Windows 2000 system, but I've encountered > the same problem under Windows 95. Here's a simple example: Actually, that's not where the error is. If you use traceback() you will see it is in edit.data.frame. That tries

dear R wizards---more ignorance on my part, exacerbated by too few examples in the function documentations. > d <- data.frame( id=rep(1:3,3), x=rnorm(9), y=rnorm(9)) Question 1: how do I work with the output of "by"? for example, > b <- by( d, d$id, function(x) coef(lm( y ~ x, data=x ) )) > b d$id: 1 (Intercept) x 0.2303 0.3618

At present the example data sets in R libraries are to be given as expressions that can be read directly into R. For example, the acid.R file in the main library looks like acid <- data.frame( carb = c(0.1, 0.3, 0.5, 0.6, 0.7, 0.9), optden = c(0.086, 0.269, 0.446, 0.538, 0.626, 0.782), row.names = paste(1:6)) This is great when you have only a few observations. I have one example data

Dear HelpeRs, I have some data: "ice" <- structure(c(0.386, 0.374, 0.393, 0.425, 0.406, 0.344, 0.327, 0.288, 0.269, 0.256, 0.286, 0.298, 0.329, 0.318, 0.381, 0.381, 0.47, 0.443, 0.386, 0.342, 0.319, 0.307, 0.284, 0.326, 0.309, 0.359, 0.376, 0.416, 0.437, 0.548, 41, 56, 63, 68, 69, 65, 61, 47, 32, 24, 28, 26, 32, 40, 55, 63, 72, 72, 67, 60, 44, 40, 32, 27, 28, 33,

Hi, I hope some one can help. I need to compute Thurston's case 5 on a large set of data. I have gotten as far as computing the proportional preference matrix but the next math is beyond me. Here us my matrix 0.500 0.472 0.486 0.587 0.366 0.483 0.496 0.434 0.528 0.500 0.708 0.578 0.633 0.554 0.395 0.620 0.514 0.292 0.500 0.370 0.557 0.580 0.615 0.329 0.413 0.422 0.630 0.500 0.783 0.641 0.731

Hello all, I am working with a 1000x1000 matrix, and I would like to return a 1000x1000 matrix that tells me which value in the matrix is greater than a theshold value (1 or 0 indicator). i have tried mat2<-as.matrix(as.numeric(mat1>0.25)) but that returns a 1:100000 matrix. I have also tried for loops, but they are grossly inefficient. THanks for all your help in advance. Lanre

Hello, I've been looking through ?phantom and ?expression and this forum for examples of how I might be able to manipulate some of the names that appear on the y-axis of the barplot below. For example, the "gw" in "ECgw" would appear as a subscript...or "qr" would be the theta symbol followed by subscript "r". My attempts haven't even come close

I have a data matrix with genes as columns and samples as rows. I want to create all possible gene ratios.Is there an elegant and fast way to do it in R and write it to a dataframe? Thanks for any help. Som. -- View this message in context: http://r.789695.n4.nabble.com/Calculating-all-possible-ratios-tp4627405.html Sent from the R help mailing list archive at Nabble.com. [[alternative HTML

Hi, I have configured samba to authenticate against a windows 2003 AD. There is no error joining to the domain .. If I try to access any share using smbclient using an existing user account from the AD, it returns an error "NT_STATUS_LOGON_FAILURE". On adding the same user to local machine (/etc/passwd), the smbclient logs in and displays all information properly. Is this the usual

Dear list, I am facing an unusual situation where I need to create two sets of legends based on the color mapping. Can't get exactly what I want and really appreciate any advice from ggplot experts. Let's say I have the first dataset "df1" that draws some points and based on which a "loess" line with confidence interval is added. Then the second dataset

I am encountering difficulty with NextMethod in 0.50-a4. We created a class of groupedData objects which are data.frames with additional attributes. The most important attribute is a formula describing roles of some of the variables in the experimental design. The class of such objects ends in "groupedData", "data.frame". The print method for the groupedData class simply

Hello: I have a dataframe, where the first column in a month: Month CATCHMENT_ med min max 1 Apr Alknda01 71.68080 11.9406600 196.2753 2 Apr Alknda02 69.44580 12.1065600 196.2753 3 Apr ArialKhan 39.89310 5.8598700 175.4433 4 Apr Arjun01 4.79652 0.0093336 55.7136 5 Apr Arjun02 4.79652 0.0093336 55.7136 6 Apr Arun01 52.42320 0.1571928

Hi r-users,   I would like to extract the data that match.  Attached is my data: I'm interested in matchind the value in column 'intg' with value in column 'rand_no' > cbind(z=z,intg=dd,rand_no = rr)             z  intg rand_no    [1,]  0.00 0.000   0.001    [2,]  0.01 0.000   0.002    [3,]  0.02 0.000   0.002    [4,]  0.03 0.000   0.003    [5,]  0.04 0.000   0.003    [6,] 

I have got pairs of time series, where one usually is shorter (n typically about 5400) than the other (n typically about 52000). I would like to calculate the ccf for these series, but I haven't found a smart way to let the shorter "slide" along the longer one in steps. Manually splitting the longer series into shorter ones of the same length as the shorter is possible, but tedious.

Hi- I am having trouble with the syntax of looping through the rows and columns of a data frame. I have a table with 17 observations for 84 lines at n=5-10 per line. So the table is ~700x17. I want to pull out the median and stdev for each line and put it in a dataframe with rowname = linename. So I have tried the following.... #read in the table input.table <- read.table(file =

Dear R gurus, I am interested in permutations-based cpu-intensive methods so I had to pay a little attention to Random Number Generators (RNG). For my needs, RNGs have to: 1) be fast. I profiled my algorithms, and for some the bottleneck was the RNG. 2) be scalable. Meaning that I want the RNG to remain fast as I add threads. 3) offer a long cycle length. Some basic generators have a

... and the mailing list is picky about attachments... whatever you attached did not conform to the stringent requirements mentioned in the Posting Guide. Pasting the code right into the email is usually safest, though you DO have to post using plain text (as the Posting Guide indicates) or your code may get mangled by the automatic html format removal. On May 15, 2018 7:04:31 AM PDT, Bert Gunter