similar to: building a data.frame from a matrix: preventing conversion to factors

Hi all, I was trying to convert a data frame to a zoo object so I can use some time series functions like lag(). But it seems then everything became a factor, so I have to convert it back to numeric to run the correct regressions. Is there a way to avoid it? Here is an example: ############################# a <- data.frame(nn =as.character(c("a", "b", "c",

I have a two question regarding the "aggregate.data.frame" method of the "aggregate" function. My situation: a. My "x" variable is a data.frame ("mydf") with two columns, both columns of type/format "numeric". b. My "by" variable is a data.frame("mybys") with two columns, both columns of type/format "character". c.

Dear All, I'm not sure if I understand the parameter stringsAsFactors correctly. I'm trying to convert the string columns in aframe1 to factors. But it seems stringsAsFactors=T in as.data.frame() doesn't do anything. Could anybody let know what is the correct way to converting strings to factors? > aframe1=data.frame(x=LETTERS[1:10], y=LETTERS[1:10], stringsAsFactors=F) >

Hi, I have two lists (c and h - see below) containing matrices with similar cases but different values. I want to split these matrices into multiple matrices based on the values in h. So, I did the following: years<-c(1997:1999) for (t in 1:length(years)) { year=as.character(years[t]) h[[year]]<-sapply(colnames(h[[year]]), function(var)

I have a data frame with columns which draw on the same underlying universe, so I want them to be factors with the same level set: --8<---------------cut here---------------start------------->8--- > z <- data.frame(a=c("a","b","c"),b=c("b","c","d"),stringsAsFactors=FALSE) > str(z) 'data.frame': 3 obs. of 2

Hi list, Using the script below, I have generated two lists (c and h) containing yearly matrices. Now I would like to divide the matrices in c into multiple matrices based on h. The number of matrices should be equal to: length(unique(DF1$B))*length(h). So each unique value in DF1$B get's a yearly matrix. Each matrix should contain all values from c where element cij is 1. An example for

After a lot of processing I get a matrix into M. I expected each row and column to be a vector. But it is a list. R-Inferno says... "Arrays (including matrices) can be subscripted with a matrix of positive numbers. The subscripting matrix has as many columns as there are dimensions in the array—so two columns for a matrix. The result is a vector (not an array) containing the selected

Hi, I have a question concerning the selection of data. Let's say that given list h created below, I would like to select a section of the 1999 matrix. For a case (rownames and colnames) I would like to select the cells that have a value > 0. So for case 8025 8025 8026 8027 8025 1 1 1 8026 1 1 1 8027 1 1 1 And for case 8028 8028 8029 8028 1

HI Jim and all, I want to put one more condition. Include col2 and col3 if they are not in col1. Here is the data mydat <- read.table(textConnection("Col1 Col2 col3 K2 X1 NA Z1 K1 K2 Z2 NA NA Z3 X1 NA Z4 Y1 W1"),header = TRUE,stringsAsFactors=FALSE) The desired out put would be Col1 Col2 col3 1 X1 0 0 2 K1 0 0 3 Y1 0 0 4 W1 0 0 6 K2 X1

> On Jul 7, 2017, at 7:03 PM, John Kane <jrkrideau at yahoo.ca> wrote: > > Thanks Marc. > It never occurred to me that I would need a ""stringsAsFactors" expression in a data.frame. I could have sworn I never did before when mocking up some data but clearly I was wrong or there has been a change in R v. 3.4.1 which seems unlikely. Welcome John. Going back to

I would like to generate a vector having the same length as the number of rows in a matrix. The vector should contain an integer indicating the "group" of the row, where identical matrix rows are in a group, and a unique row has a unique integer. Thus, for a <- c(1,2) b <- c(1,3) c <- c(1,2) d <- c(1,2) e <- c(1,3) f <- c(2,1) mat <- rbind(a,b,c,d,e,f) I would

Jim has been exceedingly patient (and may well continue to be so), but this smells like "failure to launch". At what point will you start showing your (failed) attempts at solving your own problems so we can help you work on your specific weaknesses and become self-sufficient? -- Sent from my phone. Please excuse my brevity. On February 25, 2018 7:55:55 AM PST, Val <valkremk at

Thanks Marc. It never occurred to me that I would need a ""stringsAsFactors" expression in a data.frame.? I could have sworn I never did before when mocking up some data but clearly I was wrong or there has been a change in R v. 3.4.1 which seems unlikely. On Friday, July 7, 2017, 10:37:29 AM EDT, Marc Schwartz <marc_schwartz at me.com> wrote: > On Jul 7, 2017, at 6:03

Hi, I am trying to use a mantel test on two distance matrices. The code I have entered for each is: Gen_dists <- read.csv(file.choose(), stringsAsFactors = FALSE, na.strings = c(" "), fill = T,

Hi Val, My fault - I assumed that the NA would be first in the result produced by "unique": mydat <- read.table(textConnection("Col1 Col2 col3 Z1 K1 K2 Z2 NA NA Z3 X1 NA Z4 Y1 W1"),header = TRUE,stringsAsFactors=FALSE) val23<-unique(unlist(mydat[,c("Col2","col3")])) napos<-which(is.na(val23)) preval<-data.frame(Col1=val23[-napos],

hi all another simple question. i've written a dummy program so that you get the concept. (the code could be simplfied such that there are no loops. but lets leave the loops in for now.) z1<-function(w) { for (i in 1:w) { set.seed(i+6) ss<-0 for (j in 1:5) { set.seed(j+1+(i-1)*6) r<-rnorm(1) ss<-ss+r } list(ss=ss) } } check.1<-z1(3) check.1 the results is: $ss [1]

Hi, I formed a 49 by 3 data frame by reading in a text file using read.table(), and combining it with a matrix that I made by using unlist() on a list of character strings. I would like to do some simple arithmetic operations on the elements in the data frame columns (e.g. column 3/column2) but the values in the data frame are stored as factors and using stringsAsFactors=FALSE did not work. I

> On Jul 7, 2017, at 6:03 AM, John Kane via R-help <r-help at r-project.org> wrote: > > This is not serious problem but I just wonder if someone can explain what is happening. > The same command within a dataframe is giving me a factor and as a plain vector is giving me a character. It's probably something simple that I have read and forgotten but I thought I'd ask.

Thank you Jim, I read the data as you suggested but I could not find K1 in col1. rbind(preval,mydat) Col1 Col2 col3 1 <NA> <NA> <NA> 2 X1 <NA> <NA> 3 Y1 <NA> <NA> 4 K2 <NA> <NA> 5 W1 <NA> <NA> 6 Z1 K1 K2 7 Z2 <NA> <NA> 8 Z3 X1 <NA> 9 Z4 Y1 W1 On Sat, Feb 24, 2018 at 6:18 PM, Jim

Hi list, I would like to use the following data.frame to generate matrices over a 3 year moving window: DF = data.frame(read.table(textConnection(" A B C 80 8025 1995 80 8026 1995 80 8029 1995 81 8026 1996 82 8025 1997 82 8026 1997 83 8025 1997 83 8027 1997 84 8026 1999 84 8027 1999 85 8028 1995 85 8029 1998"),head=TRUE,stringsAsFactors=FALSE)) Function to be