similar to: Excluding all missing values with dcast ("reshape2" package)

>>>>> Evan Cortens <ecortens at mtroyal.ca> >>>>> on Thu, 17 Nov 2016 15:51:03 -0700 writes: > I wonder if this could be related to the issue that I > submitted to bugzilla about two months ago? ( > https://bugs.r-project.org/bugzilla3/show_bug.cgi?id=17159) > That is to say, could it be that it's treating the first >

The packages "readxl" and "haven" (and possibly others) no longer access files on shared network drives. The problem appears to be in the normalizePath() function. The file can be read from a local drive or by functions that don't call normalizePath(). The error thrown is Error: path[1]="\\Hzndhhsvf2/data/OCPH/EPI/BHSDM/Group/17.xls": The system cannot find the

I found this as well. At our institution, our home directories are on network shares that are mapped to local drives. The default, it appears, is to set the location for libraries (etc) to the network share name (//computer//share/director/a/b/user) rather than the local drive mapping (H:/). Given the issue with dir.create(), this means it's impossible to install packages (since it tries to

HI, Dear R community, I am using the xtable to create the table, but how can I see the table? The following is the codes I used: > data(tli) > tli.table <- xtable(tli[1:10, ]) > digits(tli.table)[c(2, 6)] <- 0 > print(tli.table, floating = FALSE) % latex table generated in R 2.11.0 by xtable 1.5-6 package % Thu Jul 1 20:43:43 2010 \begin{tabular}{rrlllr} \hline &

Dear R Users: I have race-ethnicity groups identified in the factor variable Ethnic_G. I need to collapse Ethnic_G into a new variable with only two factors, 1 (White, non-Hispanic) and 2 (Minority). As seen in the code and output below, the recoded race-ethnicity variable is put into logical format, not factor format. I've used library(car) and the package was updated. Any ideas on

Hi all, When I try the following with pkg Survey it returns the error below: ftable(svyby(~INCOME, ~AGECL+RACECL, svymean, design=q50), rownames=list(AGECL=c("<35", "35-44", "45-54", "55-64", "65-74", ">=75"), RACECL=c("white non hispanic", "non white or

Hi, I am new to R. My problem is with the ordered logistic model. Here is my question: Generate an order discrete variable using the variable wrwage1 = wages in first full calendar quarter after benefit application in the following way: * wage*1*Ordered *= 1 *if*0 *· wrwage*1 *< *1000 2 *if*1000 *· wrwage*1 *< *2000 3 *if*2000 *· wrwage*1 *< *3000 4 *if*3000 *· wrwage*1 *<

I have a function (see below) that does some bootstrapping (I am happy to expand offline why I could use existing functions.) I put my results into and empty matrix and add a row of results with each iteration. My problem is i am a new user to R and I don't understand data frames, matrices, elements, and vectors well. What I would like is to have a data frame I can manipulate outside of the

Newbie here. Many apologies in advance for using the incorrect lingo. I'm new to statistics and VERY new to R. I'm attempting to "group" or "bin" data together in order to analyze them as a combined group rather than as discrete set. I'll provide a simple example of the data for illustrative purposes. Patient ID | Charges | Age | Race 1 |

Hi R-ers: I'm drawing a plot and have used different line types (lty) for different race/ethnicity groups. I want a legend that explains what line types correspond to the different race/ethnicity groups. I used the following code: legend( 1992 , 42 , c("Hispanic" , "non-Hispanic white (NHW)" , "non-Hispanic black" , "AI/AN" , "Asian" ) ,

try resetting your factor levels and re-run? q50 <- update( q50 , INCOME = factor( INCOME ) , AGECL = factor( AGECL ) , RACECL = factor( RACECL ) ) On Sun, Jul 9, 2017 at 2:59 PM, Orsola Costantini via R-help < r-help at r-project.org> wrote: > Hi all, > > When I try the following with pkg Survey it returns the error below: > > ftable(svyby(~INCOME, ~AGECL+RACECL,

My data frame consists of character variables, factors, and proportions, something like c1 <- c("A", "B", "C", "C") c2 <- factor(c(1, 1, 2, 2), labels = c("Y","N")) x <- c(0.5234, 0.6919, 0.2307, 0.1160) y <- c(0.9251, 0.7616, 0.3624, 0.4462) df <- data.frame(c1, c2, x, y) pct <- function(x) round(100*x, 1) I want to

I am trying to use tree to partition a data set. The data set has 3924 observations. Partitioning seems to work for small subsets of the data, but when I use the entire data set, no partitioning occurs. The variables are: RESP respondent to a survey (0 = not a respondent, 1 = respondent) AGE_P Age (continuous) ORIGIN_I Hispanic Ethnicity (1 = Hispanic, 2 = non-Hispanic) RACRECI2 Race

Hello all, I have a .csv file like below. Tool,Step_Number,Data1,Data2... etc up to 100 columns. A,1,0,1 A,2,3,1 A,3,2,1 . . B,1,3,2 B,2,1,2 B,3,3,2 . . ...... so on upto 50 rows where the column "*Tool*" has distinct steps in second column "*Step_Number*",but both have same entries in Step_Number column. I want the output like below.

Hello, I'm trying to run a loop that will subset my data into specific sets by regions and by race/ethnicity. I'm trying to do this fairly compactly, and I cannot get this to work. A "simple" version of the code that I am trying to run is: names <- c("white", "black", "asian", "hispanic") for(j in names){ for(i in 1:9){

I'm having difficulty overlaying maps when writing to a file graphics device. My command sequence has the structure plot(map1) par(new = T) plot(map2) On the screen device, it works fine. When I attempt something like png(file = "map.png") plot(map1) par(new = T) plot(map2) dev.off() only the last map appears, the previous ones having been cleared. Can someone clarify? Thanks,

HI there, i know this is a basic question, though i need some help because this is somewhat away from my current issue, but nevertheless interesting to me... Lets assume i have some estimated probabilities, say estimated by a logit model. i know i can also state them as an odds ratio. Now i?d like to state these odds ratios as a reference to a specific outcome of my investigated

I'm trying to use sqldf to query for the earliest date of a blood test when patients have had multiple tests in a given year. My query looks like this: test11 <- sqldf("select CHILD_ID, min(SAMP_DATE) from lab group by CHILD_ID having extract (year from SAMP_DATE) = 2011") SAMP_DATE has class "date." I get the error

I'm trying to use the survey package to calculate a risk difference with confidence interval for binge drinking between sexes. Variables are X_RFBING2 (Yes, No) and SEX. Both are factors. I can get the group prevalences easily enough with result <- svyby(~X_RFBING2, ~SEX, la04.svy, svymean, na.rm = TRUE) and then extract components from the svyby object with SE() and coef() to do the

Buenas tardes, Quiero aplicar la función rbind y necesito tener los mismos nombres de columnas. Como tengo unas 195 variables en cada dataframe, necesito hacerlo de una forma rápida. Tengo 9 bases de datos y tengo que fusionar todas. ¿Como puedo comprobar que los nombres de las variables son los mismos? Y de lo contrario, ¿como detecto las diferencias? He probado con