similar to: Data manipulation problem

Dear R?ers I?m trying to get a summary table using latex and summary in the rms package to no avail. I?m running R 2.10.1, Mac OS X snow leopard and I have the mactex 2009 distribution installed. Any obvious things I?m missing? //M options(digits=3) set.seed(173) sex <- factor(sample(c("m","f"), 500, rep=TRUE)) age <- rnorm(500, 50, 5) treatment <-

Dear all, I was intrigued by the ctree command and wanted to check it out. I first ran the demo with example(ctree) and did get the survival graphs in the end. Upon doing this with my own data and yielding a "Invalid operation on a survival time" I tried to rerun example(ctree) and now I also get "Invalid operation on a survival time" after the example runs plot(GBSG2ct)...

Dear Rui and UseRs,[a text file has also been attached, in case the format of my email is difficult to get]I am extremely sorry that I am bothering you once again, but I?ll have to get to the bottom of it. The following command sp <- lapply(split(agg, agg$st), function(x) x[order(x$year, x$month), ]) gave me an output with monthly mean of population(as under). i am not able to apply

Hi all: I'm struggling with getting my data re-formatted using functions in reshape/reshape2 to get from: 1957 0.862500000 1958 0.750000000 1959 0.300000000 1960 0.287500000 1963 0.675000000 1964 0.937500000 1965 0.025000000 1966 0.387500000 1969 0.087500000 1970 0.275000000 1973 0.500000000 1974 0.362500000 1976 0.925000000 1978 0.712500000 1979 0.337500000 1980 0.700000000 1981 0.425000000

Dear R´ers.. In this mock dataset how can I generate a logical variable based on whether just tes or tes3 are NA in each row?? test<-sample(c("A",NA,"B"),100,replace=T) test2<-sample(c("A",NA,"B"),100,replace=T) test3<-sample(c("A",NA,"B"),100,replace=T) tes<-cbind(test,test2,test3)

This does not use reshape/reshape2, but it is pretty straightforward. Assuming X is your example data: > Y <- split(X[, 2], X[, 1]) > vals <- sapply(Y, length) > pad <- max(vals) - vals > Y2 <- lapply(seq_along(Y), function(x) c(Y[[x]], rep(NA, pad[x]))) > names(Y2) <- names(Y) > X2 <- do.call(cbind, Y2) > X2[, 1:6] 1957 1958 1959

The reason it doesn't work easily with reshape/reshape2 is that the order of the rows is not determined. Your answer could be 1957 1958 ... 1985 1986 0.8625000 0.7500000 ... 0.7307692 0.23750000 0.0733945 0.6435644 ... NA 0.05769231 0.5096154 NA ... NA 0.65137615 or 1957 1958 ... 1985 1986 0.0733945 0.6435644 ... NA

Hi Tom, Or perhaps: #assume the data frame is named "tadf" library(prettyR) stretch_df(tadf,1,2) Jim On Thu, Jul 6, 2017 at 6:50 AM, Ista Zahn <istazahn at gmail.com> wrote: > The reason it doesn't work easily with reshape/reshape2 is that the > order of the rows is not determined. Your answer could be > > 1957 1958 ... 1985 1986 >

Hi everybody, If I've a data frame like this: dataframe a X0 X2 X4 X6 X8 X10 X12 X14 X16 1957 0 0 0 0 0 0 0 0 0 1958 0 0 0 0 0 0 0 0 0 1959 831 0 0 0 0 0 0 0 0 1960 544 282 0 0 0 0 0 0 0 1961 446 365 0 0 0 0 0 0 0 1962 442 473 0 0 0 0 0 0 0 1963 595 468 0 0 0 0 0 0 0 1964

Dear R-users, I have a complex line by xy-values (ordered by z). And I would like to get interpolated y-values on the positions of x = 0:600. How do I get the correct points? x=c(790,790,790,790,790,786,783,778,778,766,763,761,761,761,715,628,521,350,160,134,134,129,108,101,93,111,161,249,288,243,139,45,7)

Hi, again i am stuck in my presentation, and i have never learn R before in my life but need this to be done, so please help me out for a favour: http://www.nabble.com/file/p20333155/kew.dat kew.dat run this in R and these comes up: Month Year Rain 1 Jan 1900 74.400000 2 Feb 1900 80.500000 3 Mar 1900 23.600000 4 Apr 1900 23.600000 5 May 1900 25.100000 6

Dear R Helpers: I have a data set where the unit of observation is country-year. I would like to generate a new data set based on some inclusionary (exclusionary) criteria. Here is an example of the type of data that I have. df<-data.frame(cbind("country"=c(rep("Angola", 9), rep("Burundi", 7), rep("Chad", 13)), "year"=c(1975:1983, 1989:1995,

Hi, I have a panel data set that I am trying to subset. I am trying to keep values for years >=1960. The full set is from 1940 to 2000. I tried a few things, but none worked. Here are a couple that I am trying to use. TableAPS1 <- subset(TableAP, year => 1959) TableAPS1 <- TableAP[ which(year > 1959),] It would be really nice if someone can point out what I am doing wrong

Hello all! I have a problem to extract values greater that for example 1820. I try this code: x[x[,1]>1820,]->x1 Please help me! Thank you! The data structure is: structure(c(2.576, 1.728, 3.434, 2.187, 1.928, 1.886, 1.2425, 1.23, 1.075, 1.1785, 1.186, 1.165, 1.732, 1.517, 1.4095, 1.074, 1.618, 1.677, 1.845, 1.594, 1.6655, 1.1605, 1.425, 1.099, 1.007, 1.1795, 1.3855, 1.4065, 1.138, 1.514,

Dear R´ers, How can I create one single factor variable from two variables incorporating all possible combinations of the values?? test<-sample(c("A",NA,"B"),100,replace=T) test2<-sample(c("E",F,"A"),100,replace=T) tes<-cbind(test,test2) pseduocode: r<-function(test,test2) r AE AF AA NAE NAF NAA BE BF BA //M [[alternative HTML

Hello all! I have a problem with R. I try to merge data like this: structure(c(2.1785, 1.868, 2.1855, 2.5175, 2.025, 2.435, 1.809, 1.628, 1.327, 1.3485, 1.4335, 2.052, 2.2465, 2.151, 1.7945, 1.79, 1.6055, 1.616, 1.633, 1.665, 2.002, 2.152, 1.736, 1.7985, 1.9155, 1.7135, 1.548, 1.568, 1.713, 2.079, 1.875, 2.12, 2.072, 1.906, 1.4645, 1.3025, 1.407, 1.5445, 1.437, 1.463, 1.5235, 1.609, 1.738, 1.478,

Hello all! I have a problem to group my data (years) in 10 years classes. For example for year year decade 1598 1590-1600 1599 1590-1600 1600 1590-1600 1601 1600-1610 --- my is like this> [1] 1598 1599 1600 1601 1602 1603 1604 1605 1606 1607 1608 1609 1610 1611 1612 [16] 1613 1614 1615 1616 1617 1618 1619 1620 1621 1622 1623 1624 1625 1626 1627 [31] 1628 1629 1630 1631 1632 1633

Dear all, I just received a file from a colleague in spss. The read.spss could not finish the file due to an error (Unrecognized record type 7, subtype 18 encountered in system file) so instead I converted the file using stat-transfer. Looking at my data I see that most labels are in the attributes and I?d love to access them and assign the pertinent variables to factors without doing the whole

On 3/12/16 12:33 AM, peter dalgaard wrote: >> On 12 Mar 2016, at 00:05 , Mick Jordan <mick.jordan at oracle.com> wrote: >> >> This is definitely obscure but we had a unit test that called .Internal(strptime, "1942/01/01", %Y/%m/%d") with timezone (TZ) set to CET. > Umm, that doesn't even parse. And fixing the typo, it doesn't run: > >>

Dear all, Given rr<-data.frame(r1<-rnorm(1000,10,5),r2<-rnorm(1000,220,5)) How can I add a column (rr$p) for the joint probability of each r1 & r2 pair? I know how to add the column.. I just dont know how to compute the p value for joint probabilities given the two samples. //M