similar to: A installation problem that needs your supports.

Dear friends, are there enough R users in Cyprus to form a club? jason Dr. Iasonas Lamprianou Department of Education The University of Manchester Oxford Road, Manchester M13 9PL, UK Tel. 0044 161 275 3485 iasonas.lamprianou at manchester.ac.uk ----- Original Message ---- From: "r-help-request at r-project.org" <r-help-request at r-project.org> To: r-help at r-project.org

I can't make R on a Solaris 9 box. Does anyone have any suggestions? Thanks in advance. <3>-> make `Makedeps' is up to date. `libbz2.a' is up to date. `Makedeps' is up to date. `libpcre.a' is up to date. `Makedeps' is up to date. `libz.a' is up to date. ../../../src/include/libintl.h is unchanged ../../../include/libintl.h is unchanged `localecharset.h'

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

R version 2.5.0, under gentoo linux. This may be just my ignorance about naming conventions inside loops and subsets, but the following appears like a bug to me. y = c( 1963, 1963, 1964, 1964, 1965, 1965 ); r1= rnorm(6); d= data.frame ( y=y, r1=r1 ); ## note: I am not attach()ing anything anywhere ## this should give me two results, which it does ahw.y= subset(d, d$y==1963);

Hello all! I have a problem with ggplot2 library. I want to do an heat map and the y variables are the year months. If I use the following code, he y values are in alphabetical order, but I want it in month order. The code is: library(reshape) library(ggplot2) library(scales) p <- ggplot(data.m, aes(variable, Month)) + geom_tile(aes(fill = value),

https://bugzilla.mindrot.org/show_bug.cgi?id=1963 Bug #: 1963 Summary: IPQoS not honoured Classification: Unclassified Product: Portable OpenSSH Version: 5.8p1 Platform: amd64 OS/Version: Linux Status: NEW Severity: normal Priority: P2 Component: ssh AssignedTo: unassigned-bugs at

Estimados miembros de la comunidad de R Tengo el siguiente formato en un fichero csv que corresponde a datos de la población para un conjunto de países y para un rango amplio de años. Pais 1960 1961 1962 1963 Albania vvvvv vvvv Algeria vvvvv vvvv Me gustaría pasarlo a la siguiente forma Pais Año Poblacion Albania 1960 vvv Albania 1961 vvvv Albania

https://bugzilla.mindrot.org/show_bug.cgi?id=1963 --- Comment #5 from martin f. krafft <bugzilla.mindrot.org at pobox.madduck.net> --- With reference to http://bugs.debian.org/650512, which I just reopened, I am sorry to say that the bug persists in OpenSSH 6.0. -- You are receiving this mail because: You are watching the assignee of the bug. You are watching someone on the CC list of the

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

Hi, I want to remove the entire row from the dataset if any of the row contains blank or NA my dataset look like State Year Y X2 X3 X4 X5 X6 State1 1960 27.8 397.5 42.2 50.7 78.3 65.8 State2 1960 29.9 413.3 38.1 52 79.2 66.9 State1 1961 29.8 439.2 40.3 54 79.2 67.8 State2 1961 30.8 459.7 39.5 79.2 69.6 State1 1962 31.2 492.9 37.3 54.7 77.4 68.7 State2 1962 528.6 38.1 80.2 73.6 State1 1963

Hello, I would like to switch this dataframe: > k country 1960 1961 1962 1963 99 ARG 7493 7733 7581 7108 246 AUS 10484 10342 10809 11357 295 AUT 7438 7808 7938 8212 393 BDI 587 502 555 608 442 BEL 8223 8638 9021 9311 in this structure: year ARG AUS AUT BDI BEL 1960 7493 10484 7438 587 8223 1961 7733 10342 7808 502 8638 1962 7581 10809 7938 555

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

> gather(pobla, key = year, value = totpop, year60:year63) Country year totpop 1 Afghanistan year60 8996351 2 Albania year60 1608800 3 Algeria year60 11124888 4 Andorra year60 13411 Gracias Carlos Antonio On Tue, 19 Feb 2019 at 12:54, Carlos Ortega <cof en qualityexcellence.es> wrote: > Sí, tienes varias formas. > > Mira la función

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

This one showed up while looking at one of Ripley's other reports: > data(freeny) > model.frame(y~1,data=freeny,subset=1:10) y 1962.25 8.79236 1962.5 8.79137 1962.75 8.81486 1963 8.81301 1963.25 8.90751 1963.5 8.93673 1963.75 8.96161 1964 8.96044 1964.25 9.00868 1964.5 9.03049 > model.frame(y~1,data=freeny,subset=1:10)$y Warning: Replacement length not a

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 >

Después del "gather()" puedes hacer un "arrange()" que es una ordenación. Y dentro de "arrange()" le indicas la variable por la que ordenas (no hacen falta comillas)... Lo ordenará alfabéticamente. Saludos, Carlos Ortega www.qualityexcellence.es El mar., 19 feb. 2019 a las 13:47, Antonio Rodriguez Andres (< antoniorodriguezandres70 en gmail.com>) escribió:

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

Hello, I am sure this is a trivial question, but I don't get it... Having a dataframe (b.70) like that: > b.70[1:4,2:3] ARG AUS 1960 19041 25949 1961 19675 25451 1962 19302 26463 1963 18121 27644 Now I want to calculate the log for each year and save the result in a new dataframe called 'x'. x<- data.frame(w=NULL) for(i in 1:3)