similar to: (sin asunto)

Hola chic en s, querría construir mi primera función, y tengo una duda respecto al comando par( mfrow =c(3,3)). Primero de todo, tengo una tabla con 10 variables, para cada variable, unos 145 datos. Quiero representar para cada variable su gráfica de dispersión respecto a las demás. Es decir, coger la primera variable y la segunda, y hacer gráfica, coger la primera variable y la tercera, y hacer

Muchas gracias a tod en s por las respuetas! tengo en una variable wb una tabla con 145 filas y 17 columnas. dim(wb) 145 17 Creo una variable Serie16, la cual resulta de dividir la columna 3 y la 12. Serie16 <- c(wb[3]/wb[12]) Serie16 Aquí viene el problema. La dimensión de la nueva columna, pensaba que seria 145 filas y 1 columna. Le doy a dim, y me devuelve el valor NULL. Por qué? he

I'm trying to get the Heaven & Earth program (available at http://www.downport.com/wbd/Traveller_Heaven_&_Earth.ZIP ) to run and am having problems. There's a read-me at http://www.downport.com/wbd/Read%20Me.htm which mentions a couple of possible problems - which I've checked and sorted out, but when I try to run the program I get a dialog with "System Error

Dear R users, Please can you help me with a relatively straightforward problem that I am struggling with? I am simply trying to plot a baseline survivor and hazard function for a simple data set of lung cancer survival where `futime' is follow up time in months and status is 1=dead and 0=alive. Using the survival package: lung.wbs <- survreg( Surv(futime, status)~ 1, data=lung,

Hola chic en s, tengo una pregunta teórica. Tengo la evolución de una variable en función del tiempo. Hay 145 valores. Los primeros 1 son 0, y los demás son crecientes. Calculo la desviacion estandard con R, contemplando las 145 muestras (incluyendo los 0), y las 132 muestras (sin incluir los ceros). Me da que la desviación estandard sin contemplar los 0 es mayor. Como puede ser? no le veo el

Hola chicos, muchas gracias, funcionan vuestros códigos, pero no entiendo porqué el mío no funciona, y me gustaría aprender R bien. Quiero sacar números de 5 en 5, desplazándome una posición cada vez para la derecha: a <- c(8,10,4,1,7,2,4,6,3,8) b <- rep(0,5) i=1 while (i<=6) { b <- a[i:i+4] print(b) i=i+1 } quiero que en la primera iteración me saque en

Hola chic en s, tengo un archivo .csv con 145 filas y 17 columnas. He importado en R el archivo con la funcion de leer .csv. //especificar ruta del directori. setwd("/home/albert/Documentos/R/PAC1") wb <- read.csv("Dades_PAC1Des96_Des08.csv",header=TRUE) wbs1<-(wb[,1:3]) quiero meter en una variable una sub-tabla de la tabla 145x17. Se hacer una matriz 145*3, es

Hola, ¿qué tal? Lo que te pasa no es tan raro: set.seed(1234) muestra <- abs(rnorm(100)) sd(muestra) #[1] 0.5811866 muestra.ceros <- c(muestra, rep(0, 100000)) sd(muestra.ceros) #[1] 0.03196273 En una muestra de números positivos, añadir un cero (sobre todo si está lejos de la media) sube la varianza. Si añado otro, posiblemente también. Pero cuando añado muchísimos ceros, la varianza

> > >> >> Hola a todos, >> >> es la primera pregunta que hago a esta lista, no se si estoy en el sitio correcto. >> >> Tengo el siguiente vector: >> >> a <- c(8,10,4,1,7,2,4,6,3,8) >> >> y quiero conseguir en pantalla 5 valores, empezando por el final, y recorriendo el vector de derecha i izquierda cada vez, es decir,

Hello, I have a problem with the memory of one xen server (debian sarge). After booting I can see something like this (with xm list): Domain-0 0 195 0 r---- 21.4 server1 70 79 0 -b--- 62.8 9670 server2 11 79 0 -b--- 83.5 9611 server3 13 79 0 -b--- 33.7 9613 server4 14 79 0 -b---

scenario: server_old - Redhat 8 with samba 2.2.8a server_new - Debian 3.1 Sarge with samba 3.0.14a domain to be moved/migrated : DOMINIO smbpasswd -X server & domain in server_old SID for domain DOMINIO is: S-1-5-21-2465226537-159659893-3376541719 SID for domain server_old is: S-1-5-21-2465226537-159659893-3376541719 samba in server_old stopped scp server_old@/etc/samba to

Hello, I have a dataframe w/ 3 variables of interest: transaction,date(tdate) & time(event_tim). How could I create a 4th variable (last_trans) that would flag the last transaction of the day for each day? In SAS I use: proc sort data=all6; by tdate event_tim; run; /*Create last transaction flag per day*/ data all6; set all6; by tdate event_tim; last_trans=last.tdate; Thanks

Hello, I am trying to re-code all my programs from SAS into R. In SAS I use the following code: proc sort data=upper; by tdate stock_symbol expire strike; run; data upper1; set upper; by tdate stock_symbol expire strike; if first.expire then output; rename strike=astrike; run; on the following data set: tdate stock_symbol expiration strike 9/11/2012 C 9/16/2012

I'm attempting to create an array of treatment comparisons for modelling data generation. This involves comparison of one treatment (c2) with another (c3), relative to a common comparator (c1). Attached code gives me the correct array but need to remove duplicates. Duplicates relate only to c2 and c3 such that I need to remove r3 because c2 and c3 are same as r1 with c2 and c3 swapped r5

1)I am wandering how the following SQL statement can be written in R language w/o using sqldf: create table detail2 as select a.* from detail a, pdetail b where a.TDATE=b.TDATE and (a.STIM >= b.STIM and a.STIM <=b.MAXTIM) 2) when try if then in R it only applies to the 1st row & not to whole dataset like in SAS. How do you get round that? in SAS: data summary; set all1;

Hi all .., I have a list of 7000 dataframes with similar column headers and I wanted to add a new column to each dataframe based on a certain condition which is the same for all dataframes. When I extract one dataframe and apply my code it works very well as follows :- First suppose this is my first dataframe in the list > OneDF <- Mylist[[1]] > OneDF ID Pdate

Hello all: I wrote a function: my.bastimeToSynoptic <- function(x) { f<-unlist(strsplit(as.character(x), " ")) hr<-unlist(strsplit(f[2], ":")) if(as.numeric(hr[1])<6) { synoptic<-"00" } else { synoptic<-as.integer(as.numeric(hr[1])/6)*6 } tdate<-paste(c(f[1],"

I'm trying to get 'Heaven and Earth' running, but it gives me an error Run-time error '481': Invalid picture The program is available for download from http://www.downport.com/wbd/HEAVEN_&_EARTH.htm Installation is not straightforward - you need to unpack, configure Wine to run as Win98 in an emulated desktop, install and finally unpack and manually register mfc42.dll,

I have a dataset w/ 184K obs & 16 variables. In SAS I proc sort nodupkey it in seconds by 11 variables. I tried to do the same thing in R using both the unique & then the !duplicated functions but it just hangs there & I get no output. Does anyone know how to solve this? This is how I tried to do it in R: detail3 <-

> On Apr 15, 2018, at 4:08 AM, Allaisone 1 <Allaisone1 at hotmail.com> wrote: > > > Hi all .., > > > I have a list of 7000 dataframes with similar column headers and I wanted to add a new column to each dataframe based on a certain condition which is the same for all dataframes. > > > When I extract one dataframe and apply my code it works very well as