search for: 44.5

It looks like the 'seq' variable to 'for' can be altered from within the loop, leading to incorrect answers. E.g., in the following I'd expect 'sum' to be 1+2=3, but R 2.10.0 (svn 48686) gives 44.5. > x = c(1,2); sum = 0; for (i in x) { x[i+1] = i + 42.5; sum = sum + i }; sum [1] 44.5 or, with a debugging cat()s, > x = c(1,2); sum = 0; for (i in x) {

Hi Jeff, Thank you for the suggestions -- I appreciate your help. Unfortunately, the result2 has two problems... (1) there are now 3 date columns (it looks like 2 cols are merged into 1 col) (2) the output rows should not have any of the basistime dates repeated (maybe I misstated the problem); I need the max fcst value by basistime, but also list the date value for that row; for example:

Thanks for the dput... #### reproducible example of split-apply-combine ### dta <- structure(list(date = structure(c(1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 11L, 12L, 13L, 14L, 15L, 16L, 17L, 18L, 19L, 20L, 5L, 6L, 7L, 8L, 9L, 10L, 11L, 12L, 13L, 14L, 15L, 16L, 17L, 18L, 19L, 20L, 21L, 22L, 23L, 24L, 7L, 8L, 9L, 10L, 11L, 12L, 13L, 14L, 15L, 16L), .Label = c("2012-01-25 18:00:00",

On Thu, 26 Oct 2017, Thomas Adams wrote: > Hi Jeff, > > Thank you for the suggestions -- I appreciate your help. Unfortunately, the > result2 has two problems... > > (1) there are now 3 date columns (it looks like 2 cols are merged into 1 > col) No, there are two date columns. Result2 includes the grouping value as a row name (pulled from the names of the dta2list items

Hello all! I've been struggling with is for many hours today; I'm close to getting what I want, but not close enough... I have a dataframe consisting of two date-time columns followed by two numeric columns. what I need is the max value (in the first numeric column) based on the 2nd date-time column, which is essentially a factor. But, I want the result to provide both date-time values

I need to create 100 variable ,whose name is id.1,id.2~~~~id.100 then I need to let a vector say id<-c(id.1,id.2....id.100) any easy way to do this? thanks a lot~ [[alternative HTML version deleted]]

Hello, It's an alternative to use SAS by function in R? I want to plot d histograms by plot.from example bellow: Thank you! plot d 1 1 16.3 2 1 25.0 3 1 57.8 4 1 17.0 5 2 10.8 13 2 96.4 17 3 76.0 18 3 32.0 19 3 11.0 20 3 11.0 24 3 106.0 25 3 12.5 21 4 19.3 22 4 12.0 26 4 15.0 27 5 99.3 32 7 11.0 36

Hi all, I am trying to merge several data sets and end up with a long data format by date & time so I can run correlations and plots. I am using Deducer as an R GUI but can just use the R console if easier. The data sets are weather with wind speed, relative humidity and temperatures by date and minute and bat activity with date, time, label, and an activity index number. The bat

Dear Community, I have been using two types of survival programs to analyse a data set. The first one is an R function called aftreg. The second one an STATA function called streg. Both of them include the same analyisis with a weibull distribution. Yet, results are very different. Shouldn't the results be the same? Kind regards, J -- View this message in context:

I am developing an R package which includes datasets. The build and install works correctly. However, when I access the dataset ("BowRiver"), I get: > data(BowRiver) > BowRiver Error: object "BowRiver" not found. However, I can access the dataset from > data Example R datasets (such as USArrests) are loaded and can be accessed by the dataset name: >

  Hi r-users,   I would like to use gev and my data (annual rainfall ) is as follows:   > head(dat,20) A B C D E F G H I J 1 45.1 41.5 58.5 50.1 46.0 49.1 37.7 49.1 59.8 54.0 2 50.3 39.8 49.4 56.4 49.4 48.8 42.1 49.8 49.4 58.3 3 41.7 39.3 44.6 39.1 35.7 41.5 40.8 40.8 38.5 45.6 4 50.7 33.9 48.4 28.2 35.5 39.1 61.4 17.0 30.7 38.3 5 39.3 30.6 46.9 23.8 25.8

Should I be able to use axis() on a barplot? i have a data.frame, the first 3 values of which are: > c[1:3,] median mean A1 56.5 58.50000 A61 73.0 73.00000 A62 63.0 63.00000 > str(c) `data.frame': 19 obs. of 2 variables: $ median: num 56.5 73 63 161 51 55 44.5 22 54 49 ... $ mean : num 58.5 73.0 63.0 161.0 47.5 ... if I do barplot(median) and then try

Hi, everyone I got some problems when trying to subscript the value of vector and list, by using calculated indices. Here is the vector I am generated lon<-rep(0,886);lat<-rep(0,691) for (i in 1:886){ lon[i]<-112+0.05*(i-1) } for (i in 691:1){ lat[i]<--44.5+0.05*(691-i) } For a given location of lon(xp) and lat(yp), I would like to calculate the position of them,

Hi, I'm estimating a loglogistic aft (accelerated failure time) model, just a simple plain vanilla one (without time dependent covariates), I'm comparing the results that I obtain between aftreg (eha package) and survreg(surv package). If I don't use any covariate the results are identical , if I add covariates all the coefficients are the same until a precision of 10^4 or 10^-5 except

Hi - I am not R expert and I would appreciate your time if you can help me about my xyplot question. I would like to add text (p-value) in a 4 panels xyplot. I thought panel = function{} should work but I am not sure where I did it wrong. The error message from the following code is "Argument subscripts is missing with no default values" xyplot(GLG ~ PD | factor(TRT) , groups =

Hello. I get a bit confused by the output from the predict function when used on an object from coxph in combination with p-spline, e.g. fit <- coxph(Surv(time1, time2, status)~pspline(x), Data) predict(fit, newdata=data.frame(x=1:2)) It seems like the output is somewhat independent of the x-values to predict at. For example x=1:2 gives the same result as x=21:22. Does the result span the

Dear all, I cannot seem to get the R functions step or stepAIC to perform forward or stepwise regression as I expect. I have enclosed the example data in a dataframe at the end of this mail. Note rubbish is and rnorm(17) variable which I have deliberately added to the data to test the stepwise procedure. I have used wateruse.lm<-lm(waterusage~.,data=wateruse) # Fit full model

Dear all, For estimating Cobb-Douglad production Function [ Y = ALPHA * (L^(BETA1)) * (K^(BETA2)) ], i want to use nls function (without linearizing it). But how can i get initial values? ------------------------------------ > options(prompt=" R> " ) R> Y <- c(59.6, 63.9, 73.5, 75.6, 77.3, 82.8, 83.6, 84.9, 90.3, 80.5, 73.5, 60.3, 58.2, 64.4, 75.4, 85, 92.7, 85.4,

Hi Team Can someone advice me on how i can lower the load average on my asterisk server? dahdi-linux-2.1.0.4 dahdi-tools-2.1.0.2 libpri-1.4.10.1 asterisk-1.4.25.1 2 X TE412P Digium cards on ISDN PRI Im using the system as an IVR without any transcoding or bridging ************************************** top - 10:27:57 up 199 days, 5:18, 2 users, load average: 67.75, 62.55, 55.75 Tasks: 149

Hi, set.seed(45) test1<-data.frame(columnA=rnorm(7,45),columnB=rnorm(7,10)) #used an example probably similar to your actual data apply(test1,2,function(x) sprintf("%.1f",median(x))) #columnA columnB # "44.5"? "10.2" par(mfrow=c(1,2)) lapply(test1,function(x) {b<-