search for: 0.2500000

Hello, I am trying to get a linear model of y ~ log(x). *> lm (y~log(x))* However, I always get an error report: /Error in model.frame.default(formula = y ~ log(x), drop.unused.levels = TRUE) : variable lengths differ (found for 'log(x)')/ *Here was my y:* > y [1] 0.4500000 0.0500000 0.5000000 0.4000000 0.0000000 0.5000000 0.4000000 [8] 0.0500000

Hello R-Users, its again me with a question. I´m using R 2.5.0 on Mac Power Book running Mac OS X 10.4.10 I try to calculate distances betweeen two data tables looking like this V1 V2 V3 V4 V5 V6 V7 V8 V9 V10 V11 V12 V13 V14 V15 V16 V17 V18 V19 V20 V21 V22 V23 V24 V25 V26 V27 V28 V29 V30 V31 V32 V33 V34 V35 V36 1 1 0 0 0 1 1 0 1 0 0 0 1 0 0 0 0 0 1 0 0

Hi, Assuming that you provided the sample data from the file. temp <- readLines(textConnection("#Hogd/met, Temp, 005[M], Value #Hogd/met, Difftemp, 051[M], Value BA0+ 1 MTEMP005 1 [deg.C] 2 MDTMP051 1 [deg.C] EOH 891231, 2400, -1.5, -0.21, 900101, 0100, -1.4, -0.25, 900101, 0200, -1.6, -0.28, 900101, 0300, -1.7, -0.25, 900101, 0400, -2.1, -0.0999999, 900101, 0500, -2.3, -0.0899999,

below is my data frame. I would like to compute summary statistics for mgl for each river mile (mean, median, mode). My apologies in advance- I would like to get something like the SAS print out of PROC Univariate. I have performed an ANOVA and a tukey LSD and I would just like the summary statistics. thanks stephen RM mgl 1 215 0.9285714 2 215 0.7352941 3 215 1.6455696 4 215

Hi Patrick Hausmann <c18g at uni-bremen.de> napsal dne 15.11.2007 18:59:06: > Hello Petr, > > one question solved, the next is standing in front of me... If you > have a minute to look .... great! > > > x > V1 V2 F1 > 1 A 2 0.1552277 > 2 A 3 0.1552277 > 3 A 4 0.1552277 > 4 B 3 0.8447723 > 5 B 2 0.8447723 > 6 C 6 0.2500000

Thanks for the proposition. As you see bellow, par("usr") is the same before and after the points() (the full code is bellow): .... > par("usr") [1] -0.2500000 1.2500000 -0.1666667 1.1666667 > # if you remove this points() function, axis will show nothing. > > points(1.5, 1.5, type="p") > p2 <- par(no.readonly=TRUE) > par("usr")

>>>>> Marc Girondot via R-help <r-help at r-project.org> >>>>> on Mon, 24 Jul 2017 09:35:06 +0200 writes: > Thanks for the proposition. As you see bellow, par("usr") is the same > before and after the points() (the full code is bellow): > .... >> par("usr") > [1] -0.2500000 1.2500000 -0.1666667

Hello! In the example below the Reduce() function by default assigns "a" to be the last accumulated value and "b" to be the current value in "x". I could not find this documented anywhere as the default settings for the Reduce() function. Does any sort of documentation for this behavior exist? x <- c(0,0,0,0,0,1,0,0,0,5,0,0,0,7,0,0,0,8,5,10)

Hello, might be rather easy for R pros, but I've been searching to the dead end to ... twsource.area <- table(twsource, area, useNA="ifany") gives me a nice cross tabulation of frequencies of two factors, but now I want to convert to pecentages of those absolute values. In addition I'd like an extra column and an extra row with absolute sums. I know, Excel or the

Thanks... I agree that the problem was explained in the documentation but I can't find a way to have axis() working even manipulating par("plt") or with graphics.reset = TRUE: - adding graphics.reset=TRUE does not allow axis() to be shown; - I see that par()$plt is involved but it is the not sufficient to explain why axis() works because if it is changed by hand, axes are not

Hello, I have a dataframe (tab separated file) which looks like the example below - two values separated by a comma, and tab separation between each of these. [,1] [,2] [,3] [ ,4] [1,] 0,1 1,3 40,10 0,0 [2,] 20,5 4,2 10,40 10,0 [3,] 0,11 1,2 120,10 0,0 I would like to calculate the percentage of the smallest number separated by the comma by: 1) summing the values e.g. for

Hi marc, Try: par("usr") before and after the call to points and see if it changes. Jim On Sat, Jul 22, 2017 at 12:05 AM, Marc Girondot via R-help <r-help at r-project.org> wrote: > It is known (several discussions on internet) that axis() cannot be used > after fields:::image.plot() (axis() shows nothing). > > However, if points(1.5, 1.5, type="p") is

Fellow R-helpers, Suppose we create a histogram as follows (although it could be any vector with zeroes in it): R> lenh <- hist(iris$Sepal.Length, br=seq(4, 8, 0.05)) R> lenh$counts [1] 0 0 0 0 0 1 0 3 0 1 0 4 0 2 0 5 0 6 0 10 0 9 0 4 0 [26] 1 0 6 0 7 0 6 0 8 0 7 0 3 0 6 0 6 0 4 0 9 0 7 0 5 [51] 0 2 0 8 0 3 0 4 0 1 0 1 0 3

It is known (several discussions on internet) that axis() cannot be used after fields:::image.plot() (axis() shows nothing). However, if points(1.5, 1.5, type="p") is inserted before the axis() finctions, it works. I have investigated what points(1.5, 1.5, type="p") is doing to allow axis to work and I don't find a solution. par() options are identical (p1 and p2 are

Hi, I have run into another problem using offsets, this time with the predict function, where there seems to be a contradiction again between the behavior and the help page. On the man page for predict.lm, it says Offsets specified by offset in the fit by lm will not be included in predictions, whereas those specified by an offset term in the formula will be. While it indicates nothings about

Hi, I sometimes play around with extreme parameters for distributions and found that qbeta is not always monotone as the following example shows. I don't know whether this is serious enough to submit a bug report (as this example is near to the limitations of floating point arithmetic). Josef > x <- qbeta((0:100)/100,0.01,5) > x [1] 0.000000e+00 1.253990e-201 1.589622e-171

Hi I believe this is a clipping bug in the 'graphics' package. A simpler version of the problem is this ... plot(1:10) mtext("margin-label", side=2, at=9, las=1, line=1, adj=0) par(mfg=c(1,1)) ## Only the text within the plot region is drawn mtext("margin-label", side=2, at=9, las=1, line=1, adj=0, col="red") I have committed a fix to the development

Dear Mailing-List, I think this is a newbie question. However, i would like to integrate a loop in the function below. So that the script calculates for each variable within the dataframe df1 the connecting data in df2. Actually it takes only the first row. I hope that's clear. My goal is to apply the function for each data in df1. Many thanks in advance. An example is as follows: df1

Hi, Given some example data: dat <- seq(4, 7, by = 0.05) x <- sample(dat, 30) y <- sample(dat, 30) error <- x - y I have broken the rage of x into 10 groups and I can calculate the bias (mean(error)) for each of these 10 groups: groups <- cut(x, breaks = 10) max.bias <- aggregate(error, list(group = groups), mean) max.bias group x 1 (4,4.3] -0.7750000 2

Hi, I have a matrix, say m=matrix(c( 983,679,134, 383,416,84, 2892,2625,570 ),nrow=3 ) i can find its row/col sum by rowSums(m) colSums(m) How do I divide each row/column by its rowSum/colSums and still return in the matrix form? (i.e. the new rowSums/colSums =1) Thanks. Casper -- View this message in context: