similar to: Use a variable of a list in a loop

Displaying 20 results from an estimated 50000 matches similar to: "Use a variable of a list in a loop"

2012 May 05
1
Correct use of ddply with own function
Hi, I am really confused how ddply work, so maybe you can help me. I created a function that sorts a vector etc. fn <- function(x){ x1 <- sort(x) x2 <- seq(length(x)) x3 <- x2/max(x2) df <- data.frame(x1,x2,x3) df } Probably this is not the best form of the function, but at least it produces what I want (data to plot a cumulative count curve). This function works on a
2003 Aug 26
2
Simple simulation in R
Hello all I have a feeling this is very simple......but I am not sure how to do it My boss has two variables, one is an average of 4 numbers, the other is an average of 3 of those numbers i.e var1 = (X1 + X2 + X3 + X4)/4 var2 = (X1 + X2 + X3)/3 all of the X variables are supposed to be measuring similar constructs not surprisingly, these are highly correlated (r = .98), the question is how
2010 Dec 02
2
Hmisc label function applied to data frame
Hello, I'm attempting to create a data frame with correlations between every pair of variables in a data frame, so that I can then sort by the value of the correlation coefficient and see which pairs of variables are most strongly correlated. The sm2vec function in the corpcor library works very nicely as shown here: library(Hmisc) library(corpcor) # Create example data x1 = runif(50) x2 =
2008 Dec 07
2
subsetting large data frames.
Hi all, I have a question regarding subsetting of large data frames. I have two data frames ?catches? and ?tows? and they both have the same 30 variables (columns). I would like to select rows in the data frame ?tows? where all 5 specific variables are NOT matched in ?catches. That is to say, the combination of these 5 variables is unique. One or more of the variables could be the same but the
2010 Jan 27
1
selecting significant predictors from ANOVA result
Dear all,   I did ANOVA for many response variables (Var1, Var2, ....Var75000), and i got the result of p-value like below. Now, I want to select those predictors, which have pvalue less than or equal to 0.05 for each response variable. For example, X1, X2, X3, X4, X5 and X6 in case of Var1, and similarly, X1, X2.......X5 in case of Var2, only X1 in case of Var3 and none of the predictors in case
2010 Jan 18
1
problem of data manipulation
Hello, See my problem below. a<-data.frame(c("s","c","c","n","n","n"),c(rep(1,3),rep(2,3)),c(rep(2,3),rep(1,3)),c("01/01/1999","10/02/2000","13/02/2000","11/02/2000","15/02/2000","23/02/2000"))
2009 Nov 11
1
fitting a glm with matrix of responses
hi all - quick question: i have a matrix 'y' of response values, with two explanatory variables 'x1' and 'x2'. tested values of 'x1' and 'x2' are sitting in two vectors 'x1' and 'x2'. i want to learn model parameters without "unrolling" the matrix of response values. example below: # some fake data for the example x1 <- 1:5 x2
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
2008 Oct 10
1
Correlation among correlation matrices cor() - Interpretation
Hello, If I have two correlation matrices (e.g. one for each of two treatments) and then perform cor() on those two correlation matrices is this third correlation matrix interpreted as the correlation between the two treatments? In my sample below I would interpret that the treatments are 0.28 correlated. Is this correct? > var1<- c(.000000000008, .09, .1234, .5670008, .00110011002200,
2012 Jun 28
4
Printing a variable in a loop
Dear R Users: I'm a STATA user converting to R, and I'd like to be to do the following. #Assign var_1 and var_2 a value 10->var1 20->var2 #Now I'd like to print the values of var_1 and var_2 by looping through var_1 and var_2 in such a manner: while(y<3){ print(var_y) y+1->y } In STATA, the "y" appended to " var_" is called the local variable and
2011 Oct 19
1
using a dictionary in R
Hi all, Is there a way to create a "dictionary" in R, such that it has <key,value> pairs? Something to the effect of: x=dictionary(c("Hi","Why","water") , c(1,5,4)) x["Why"]=5 In truth I'm looking two categorial variables function. So that if x=dictionary(c("a","b"),c(5,2)) x val 1 a 5 2 b 2 I want to compute
2012 Sep 11
1
plotting smoother function on raw data
Hi, I have used the mgcv library to generate a simple additive model. I want to know how to plot the function on the raw data with confidence intervals whan I have TWO variables in the model. I get it to work with one variable but not with two. I am on the limit for what I understand in R, so be gentle. I have read the help file on predict.gam, but did not get any help out of it. #My model:
2010 Dec 20
2
Turning a Variable into String
I would like to know how to turn a variable into a string. I have tried as.symbol and as.name but it doesnt work for what I'd like to do Essentially, I'd like to feed the function below with two variables. This works fine in the bit working out number of elements in each variable. In the print(sprintf("OK with %s and %s\n", var1, var2)) line I would like var1 and var2 to be
2013 Jul 18
1
Bland Altman summary stats for all column combinations
Hello, I have the following data.frame structure(list(Study = structure(c(1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 11L, 12L, 13L, 14L, 15L, 16L, 17L, 18L, 19L, 1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 11L, 12L, 13L, 14L, 15L, 16L, 17L, 18L, 19L, 1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 11L, 12L, 13L, 14L, 15L, 16L, 17L, 18L, 19L, 1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 11L, 12L, 13L, 14L,
2002 Jul 09
3
Assignment converts variable to factor
Hello, I would like to know if this behaviour in R is as expected. I have a data frame 'dat' with column var1 being character (and not factor). Then I create a new column 'var2' by: > dat[,"var2"] <- dat$var1; Column var2 is now a factor. But if I do: > dat$var2 <- dat$var1; Then column var2 is character (and not factor). I don't want to have
2002 Nov 29
2
Obtaining the variable names of a glm object
Is names(model1$coef) what you're looking for? -----Original Message----- From: Kenneth Cabrera [mailto:krcabrer at epm.net.co] Sent: 29 November 2002 10:36 Cc: R-help at stat.math.ethz.ch Subject: [R] Obtaining the variable names of a glm object Hi, R users! Suppose I make a model like this:
2014 Apr 11
2
crear variable en base a nombre de columnas que tienen un 1
Carlos, en principio si sería algo así, sólo que en vez de quedarme con todas las columnas var1 a var4 tuviera sólo 3, ya que en mis datos no hay ningún caso que tenga el valor 1 en más de 3 variables.. Había llegado a una solución (mucho menos elegante que usando reshape), que implicaba un for sobre las filas. Jorge, creo que tu solución me vale. Muchas gracias a los dos.. Saludos El
2008 Dec 10
2
how to merge panel data stored by variable?
Hi, I have two datasets stored in tab-separated format in the following way file1: country year1 year2 Germany var1 var1 Hungary var1 var1 file2: country year1 year2 Germany var2 var2 Hungary var2 var2 I can easily read in these files, but how can I merge them as a panel dataset? Thanks, Viktor
2008 Jul 16
2
Group level frequencies
Dear List, I have Multi-level Data i= Indivitual Level g= Group Level var1= First Variable of interest var2= Second Variable of interest and I want to count the frequency of "var1" and "var2" on the group level. I found a way, but there must be a much simpler way. data.ml <- data.frame(i=c(1:8),g=as.factor(c(1,1,1,2,2,3,3,3)),var1=c(3,3,3,4,4,4,4 ,4),
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