similar to: expand.grid() function

Displaying 20 results from an estimated 20000 matches similar to: "expand.grid() function"

2013 Mar 25
2
Faster way of summing values up based on expand.grid
Hello! # I have 3 vectors of values: values1<-rnorm(10) values2<-rnorm(10) values3<-rnorm(10) # In real life, all 3 vectors have a length of 25 # I create all possible combinations of 4 based on 10 elements: mycombos<-expand.grid(1:10,1:10,1:10,1:10) dim(mycombos) # Removing rows that contain pairs of identical values in any 2 of these columns: mycombos<-mycombos[!(mycombos$Var1
2013 Feb 01
2
expand.grid on contents of a list
Hello! I have a list of variable length. One example is: X=vector("list",3) X[[1]]=1:2 X[[2]]=1:2 X[[3]]=1:2 How could I run expand.grid on the elements of X so that the results would be the same as expand.grid(1:2,1:2,1:2)? Thank you! Dimitri -- Dimitri Liakhovitski gfk.com <http://marketfusionanalytics.com/> [[alternative HTML version deleted]]
2003 May 08
2
Expanding upon expand.grid()
Hello All: The function expand.grid() does nearly exactly what I want for permutation tests I wish to carry out, and it does so quickly when the number is kept small as in the example below: expand.grid(rep(list(c(-1, 1)), 3)) Var1 Var2 Var3 1 -1 -1 -1 2 1 -1 -1 3 -1 1 -1 4 1 1 -1 5 -1 -1 1 6 1 -1 1 7 -1 1 1 8 1 1 1 Understandably,
2009 Sep 21
5
More elegant way of excluding rows with equal values in any 2 columns?
Hello, dear R-ers! I built a data frame "grid" (below) with 4 columns. I want to exclude all rows that have equal values in ANY 2 columns. Here is how I am doing it: index<-expand.grid(1:4,1:4,1:4,1:4) dim(index) # Deleting rows that have identical values in any two columns (1 line of code):
2007 Sep 16
1
Putting column names in some automated way
Dear all, I have following codes: colnames(data) = c("var", "var", "var") i = c(1,2,3) Now I want construct a "for" loop starting from 1 to 3 to give the new names of columns for dataframe "data" like below colnames(data) > c("var1", "var2", "var3") Definitely I could do this manually, however I want to put
2012 May 01
2
Question about expand.grid function in R
Hi, I am extremely new to R, and was wondering if someone would be able to help me with a question regarding the expand.grid function. When I input expand.grid.rep <- function(x, n=1) do.call(expand.grid, rep(list(x),n)) expand.grid.rep(c("a", "b", "c"), 3) my output is as follows, Var1 Var2 Var3 1     a    a    a 2     b    a    a 3     c    a    a 4     a  
2009 Sep 22
3
converting a character vector to a function's input
Hi all, I have been trying to solve this problem and have had no luck so far. I have numeric vectors VAR1, VAR2, and VAR3 which I am trying to cbind. I also have a character vector "VAR1,VAR2,VAR3". How do I manipulate this character vector such that I can input a transformed version of the character vector into cbind and have it recognize that I'm trying to refer to my numeric
2004 Aug 17
5
Bug in colnames of data.frames?
Hi, I am using R 1.9.1 on on i686 PC with SuSE Linux 9.0. I have a data.frame, e.g.: > myData <- data.frame( var1 = c( 1:4 ), var2 = c (5:8 ) ) If I add a new column by > myData$var3 <- myData[ , "var1" ] + myData[ , "var2" ] everything is fine, but if I omit the commas: > myData$var4 <- myData[ "var1" ] + myData[ "var2" ] the name
2014 Apr 11
6
crear variable en base a nombre de columnas que tienen un 1
Buenos días. Hoy ando un poco (o bastante) espeso y no doy con la tecla de una cosa que seguro que es muy simple.. Pongo un ejemplo. var1 <- c(rep(0,3),rep(1,2)) var2 <- c(rep(1,2),0,0,1) var3 <- c(rep(1,2),rep(0,3)) var4 <- c(rep(1,2),rep(0,3)) datos <- data.frame(fila=1:5,var1, var2, var3, var4) datos datos fila var1 var2 var3 var4 1 1 0 1 1 1 2 2 0
2010 Nov 08
2
Several lattice plots on one page
Dear all, I am trying (!!!) to generate pdfs that have 8 plots on one page: df = data.frame( day = c(1,2,3,4), var1 = c(1,2,3,4), var2 = c(100,200,300,4000), var3 = c(10,20,300,40000), var4 = c(100000,20000,30000,4000), var5 = c(10,20,30,40), var6 = c(0.001,0.002,0.003,0.004), var7 = c(123,223,123,412), var8 = c(213,123,234,435), all = as.factor(c(1,1,1,1)))
2014 Aug 21
2
pregunta
Estimados Estoy entrenando hacer funciones que respondan a comandos, en esta caso en la salida gráfica se observa que dice : Exposure=var3 y outcome=var 1 quisiéramos que se reflejan los nombres de la base de datos : var1=estado, var2=cake, var3=chocolate Espero haberme explicado adecuadamente Adjunto tabla con datos #################################### #Comando que llama
2006 Jan 13
2
find mean of a list of timeseries
Can someone please give me a clue how to 're'write this so I dont need to use loops. a<-ts(matrix(c(1,1,1,10,10,10,20,20,20),nrow=3),names=c('var1','var2','var3')) b<-ts(matrix(c(2,2,2,11,11,11,21,21,21),nrow=3),names=c('var1','var2','var3'))
2014 Aug 21
2
pregunta
Buenas noches Javier y José, Estoy en contra de usar attach(), asi que propongo la siguiente alternativa con with(): # paquete require(epicalc) # los argumentos en ... pasan de epicalc:::cc # ver ?cc para mas informacion foo <- function(var1, var2, var3, ...){ or1 <- cc(var1, var2, ...) or2 <- cc(var1, var3, ...) list(or1 = or1, or2 = or2) } # datos x <-
2008 Mar 19
4
plot with diffrent colour and plotting symbols
Dear mailing list members, I am a new R user, I would like to plot the follewing data var1 <- c(1,2,1,1,2,1,2,1,2,2) var2 <- round(rgamma(10,2,1)/0.1)*0.1 var3 <- c(0,1,0,1,0,0,0,0,1,0) var4 <- c(1,2,2,2,1,1,1,1,1,1) Var <- data.frame(var1,var2,var3,var4) Var <- Var[sort(Var$var1),] tt <- Var$var1+((runif(length(Var$var1))/6)-(0.5/6)) labelname <- c("time 1",
2011 May 19
2
trouble with summary tables with several variables using aggregate function
Dear all, I am having trouble creating summary tables using aggregate function. given the following table: Var1 Var2 Var3 dummy S1 T1 I 1 S1 T1 I 1 S1 T1 D 1 S1 T1 D 1 S1 T2 I 1 S1 T2 I 1 S1 T2 D 1 S1 T2 D 1 S2
2013 Dec 06
2
Using assign with mapply
I have a data frame whose first colum contains the names of the variables and whose second colum contains the values to assign to them: : kkk <- data.frame(vars=c("var1", "var2", "var3"), vals=c(10, 20, 30), stringsAsFactors=F) If I do : assign(kkk$vars[1], kkk$vals[1]) it works : var1 [1] 10 However, if I try with mapply
2012 Jul 01
2
list to dataframe conversion-testing for identical
HI R help, I was trying to get identical data frame from a list using two methods. #Suppose my list is: listdat1<-list(rnorm(10,20),rep(LETTERS[1:2],5),rep(1:5,2)) #Creating dataframe using cbind dat1<-data.frame(do.call("cbind",listdat1)) colnames(dat1)<-c("Var1","Var2","Var3") #Second dataframe conversion
2009 Nov 29
3
How to z-standardize for subgroups?
Hi folks, I have a dataframe df.vars with the follwing structure: var1 var2 var3 group Group is a factor. Now I want to standardize the vars 1-3 (actually - there are many more) by class, so I define z.mean.sd <- function(data){ return.values <- (data - mean(data)) / (sd(data)) return(return.values) } now I can call for each var z.var1 <- by(df.vars$var1, group,
2010 May 05
3
concatenate values of two columns
Dear list, I'm trying to concatenate the values of two columns but im not able to do it: i have a dataframe with the following two columns: X VAR1 VAR2 1 2 2 1 3 2 4 3 5 4 6 4 what i would like to
2009 Aug 18
1
adding points to a wireframe
A quick question. I'm trying to plot a surface from a fitted model along with the original points, as in the following example: df<-data.frame(expand.grid(100*runif(1:100), 100*runif(1:100))) df$Var3<-rnorm(length(df$Var1), mean=df$Var1*df$Var2, sd=10) my.lm<-lm(Var3 ~ Var1*Var2, data=df) my.fun<-function(x,y) predict(my.lm, data.frame(Var1=x, Var2=y),