search for: 0.89

Dear R users,   I have a dataframe with lot of duplicate cases and I want to delete duplicate ones which have low rank and keep that case which has highest rank. e.g   > df1   cno      rank 1  1342    0.23 2  1342    0.14 3  1342    0.56 4  2568    0.15 5  2568    0.89   so I want to keep 3rd and 5th  cases with highest rank (0.56 & 0.89) and delete rest of the duplicate cases. Could

Dear Sir or Madam, I have a question like this. I want to extract the following data from the square bracket: 1 band band [0.86] 2 band band [0.93] 3 noband noband [0.95] 4 noband noband [0.91] 5 noband noband [0.89] 6 noband noband [0.84] 7 noband

OK, so maybe last night was a little too much at one throw, so I have reduced the data to two stations- one that has precipitation and one that does not. This is going to be in the context of a larger data set. I would like to be able to issue a ggplot command and have cum sum just act on the facets (factors) to apply this. library(chron) library(ggplot2) DF <- structure(list(date_time =

Hi, I have the following table of odds ratios (or), lower limits(ll) and upper limits(ul), which I would like to plot as horizontal lines beginning at the lower limit, ending at the upper limits and with a dot at the odds ratio on an x-axis on a log10 scale. The y axis would be the study sites. >From what I can figure out, it looks like the plotCI function will do everything except give me an

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

Hello, since today I could not update my CentOS 5.3 system with yum cause of the following error message. # yum update [...] Resolving Dependencies --> Running transaction check ---> Package exiv2.i386 0:0.18.2-1.el5.rf set to be updated ---> Package perl-Params-Util.i386 0:1.00-1.el5.rf set to be updated ---> Package perl-Moose.noarch 0:0.86-1.el5.rf set to be updated -->

Hello, I?m wondering if anyone can offer advice on the out-of-memory error I?m getting. I?m using R2.12.2 on Windows XP, Platform: i386-pc-mingw32/i386 (32-bit). I am using the quantreg package, trying to perform a quantile regression on a dataframe that has 11,254 rows and 5 columns. > object.size(subsetAudit.dat) 450832 bytes > str(subsetAudit.dat) 'data.frame': 11253 obs.

Full_Name: Willa Chen Version: 2.7.2 OS: Window XP Submission from: (NULL) (128.122.182.70) The match function does not return value properly. See an example below. > a<-seq(0.6,1,by=0.01) > match(0.88,a) [1] 29 > match(0.89,a) [1] NA > match(0.90,a) [1] NA > match(0.91,a) [1] NA > match(0.92,a) [1] NA > match(0.93,a) [1] NA > match(0.94,a) [1] 35

hello, I wanna use some mathematics formula and to do this I tried several way in paricular using strsplit textconnection scan setdiff but I think that it's a lil hard the data frame that I'm working on is as follow > donCalcara2 Id_Cara Form_C 1 743

Dear R Users,i am very new to R. I want your help on an issue regarding distance matrix and cluster analysis i had discharge data of 4 rivers(a,b,c,d) in 4 vectors each having 364 values > dput(qmu)structure(list(a = c(0.26, 0.25, 0.25, 0.25, 0.24, 0.23, 0.22, 0.21, 0.21, 0.21, 0.2, 0.19, 0.19, 0.19, 0.19, 0.18, 0.18, 0.18, 0.17, 0.17, 0.17, 0.17, 0.17, 0.17, 0.17, 0.17, 0.17, 0.17, 0.17, 0.17,

Dear R people, I notice that the confidence intervals of a very small sample (e.g. n=6) derived from the one-sample wilcox.test are just the maximum and minimum values of the sample. This only occurs when the required confidence level is higher than 0.93. Example: > sample <- c(1.22, 0.89, 1.14, 0.98, 1.37, 1.06) > summary(sample) Min. 1st Qu. Median Mean 3rd Qu. Max.

#this is a daily series of precipitation data. I would like to condense it into weekly means. How can I do this #as a side note I would like to do this same thing to two years worth of fifteen minute interval data and make it into #a series of daily averages (there are 96 readings per day) #is aggregate the right function? or... y <- c(1.23, 0, 0, 0, 0, 0, 0, 0, 0, 0.27, 0, 0.29, 0, 0, 0,

The loop is correct, you just need to make sure that your result is computed and stored as the n-th element that is returned by the loop. Pick up any manual of R, and looping will be explained there. Also, I would recommend that you draw a random number for every iteration of the loop. Defining the random vectors outside the loop make sense to me only if they are the same length as n.

Hi there, I'm trying to make a scatterplot of removed versus duration for each type of bee. No matter what I try, I can't seem to get my code to work. Any help would be appreciated. Thanks! My r-code: dat$BEE <- with(dat, factor(BEE, c(1,2))) plot(REMOVED~DURATION,pch=BEE, col=BEE) REMOVED DURATION BEE 1 0.07 2 QUEEN 2 0.10 5 QUEEN 3 0.11 7 QUEEN 4 0.12 11 QUEEN 5 0.15 12 QUEEN 6 0.19

hi, I don't know what I am doing wrong, but with that code; x1 <- c(1.60, 0.27, 0.17, 1.63, 1.37, 2.00, 0.90, 1.07, 0.89, 0.43, 0.37, 0.59, 0.47, 1.83, 1.79, 0.90, 0.72, 1.83, 0.23, 1.97, 2.03, 2.19, 2.03, 0.86) x2 <- c(1.30, 0.24, 0.20, 0.50, 1.33, 1.87, 1.30, 0.75, 1.07, 0.43, 0.37, 0.87, 1.40, 1.37, 1.63, 0.80, 0.57, 1.60, 0.39, 2.03, 1.90, 2.07, 1.93, 0.93) model <-

aa <- (structure(list(X.0.85 = c(-1.02, -1.17, -1.29, -1.39, -1.46, -1.5, -1.52, -1.5, -1.46, -1.39, -1.3, -1.19, -1.07, -0.93, -0.79, -0.65, -0.5, -0.36, -0.22, -0.08, 0.05, 0.18, 0.3, 0.41, 0.52, 0.62, 0.72, 0.81, 0.89, 0.98, 1.05, 1.13, 1.19, 1.25, 1.29, 1.31, 1.31, 1.29, 1.24, 1.16, 1.06, 0.93, 0.77, 0.58, 0.38, 0.16, -0.07, -0.31, -0.89, -1.05, -1.19, -1.31, -1.41, -1.47, -1.51, -1.51,

I have the following data which I tried to draw the probability density plot. Here is the code I have: x <- read.table("mydat.txt"); d <- rep(x$V2,times=x$V3); hist(d,probability=T, xlab="FlowSignal"); But why the y-axis range from 0 to 6, instead of 0 to 1? What's the correct way to plot it? #id flowsignal frequency 1 0.67 1 1 0.70 1 1

Hi, I'm proposing to make vectorizer-maximize-bandwidth on by default for loop vectorizer because it should generally help performance. I've tested the performance impact on Intel sandybridge machine with speccpu benchmarks: Benchmark Base:Reference (1) ------------------------------------------------------- spec/2006/fp/C++/444.namd 26.84

Dear R users: In the example of meta-analysis (cochrane, package rmeta), I can not found the p-value of Test for overall effect, and some other indices (Z, I, weight and et al). How can I get the these indices listed? > library(rmeta) > data(cochrane) > cochrane name ev.trt n.trt ev.ctrl n.ctrl 1 Auckland 36 532 60 538 2 Block 1 69 5

Hi All , Could you please tell using which library ,Dickey-Fuller Test can be run? Thanks a lot __________________________________________________ [[alternative HTML version deleted]]