na

i hope this one isn´t stupid. I have two dataframes with Variables ID and gender/sex. In df1, there are NAs. In df2, the variable is complete. I want to complete the column in df1 with the values from df2. (In df1 the variable is called "gender". In df2 it is called "sex".) Here is what i tried so far: #example-data ID<-seq(1,30,by=1) df1<-as.data.frame(ID) df2<-df1 df1$gender<-c(NA,"2","1",NA,"2","2","2","2","2","2",NA,"2","1","1",NA,"2","2","2","2","2","1","2","2",NA,"2","2","2","2","2",NA) df2$sex<-c("2","2","1","2","2","2","2","2","2","2","2","2","1","1","2","2","2","2","2","2","1","2","2"

This question already has an answer here: Replace missing values with column mean 11 answers if I have a data frame df df=data.frame(x=1:20,y=c(1:10,rep(NA,10)),z=c(rep(NA,5),1:15)) I know to replace NAs with mean value for a given column is, we can use df[is.na(df$x)]=mean(df$x,na.rm=T) What I am trying to find is a way to use a single command so that it does this for the columns at once instead of repeating it for every column. Suspecting, I need to use sapply and function, I tried something like this but clearly this does not work sapply(df,function(x) df[is.na(df$x)]=mean(df$x,na.rm=T))

I have a large matrix, z, that I removed all values >3 and replaced with NA using: z[z>3]<-NA I have another matrix, y , of identical dimensions that I need to replace values with NA in positions corresponding to the locations where the elements were replaced in element z. That is, if z[3,12] was >3 and replaced with NA, I need y[3,12] to be replaced with NA too. They have the same row names if that helps. Just use is.na on the first matrix to select the values to replace in the second matrix. Example: set.seed(1) m1 <- matrix(sample(5, 25, TRUE), 5) m2 <- matrix(sample(5, 25, TRUE), 5) m1[m1

Brief Dataset description: I have survey data generated from Qualtrics, which I've imported into R as a tibble. Each column corresponds to a survey question, and I've preserved the original column order (to correspond with the order of the questions in the survey). Problem in plain language: Due to normal participant attrition, not all participants completed all of the questions in the survey. I want to know how far each participant got in the survey, and the last question they each answered before stopping. Problem statement in R: I want to generate (using tidyverse): 1) A new column ( lastq

Here is an example: set.seed(123) data<-data.frame(X=rep(letters[1:3], each=4),Y=sample(1:12,12),Z=sample(1:100, 12)) data[data==3]<-NA What I am to realize is to select the unique row of X with minimum Y by ignoring NA s: a 4 68 b 1 4 c 2 64 What's the best way to do that? Using the data.table package, this is trivial: library(data.table) d <- data.table(data) d[, min(Y, na.rm=TRUE), by=X] You can also use plyr and its ddply function: library(plyr) ddply(data, .(X), summarise, min(Y, na.rm=TRUE)) Or using base R: aggregate(X ~ ., data=data, FUN=min) Based on the edits, I would use data.table

From a large data frame, I have extracted a row of numeric data and saved as a vector. Some of the values are missing and marked as NA. I want to impute the missing values with row mean. Thanks Let x be your vector: x <- c(NA,0,2,0,2,NA,NA,NA,0,2) ifelse(is.na(x), mean(x, na.rm = TRUE), x) # [1] 1 0 2 0 2 1 1 1 0 2 Or if you don't care for the original vector, you can modify it directly: x[is.na(x)] <- mean(x, na.rm = TRUE) Ferdinand.kraft Use this: filter <- is.na(myVec) myVec[filter] <- colMeans(myDF[,filter], na.rm=TRUE) Where myVec is your vector and myDF is your data.frame. 来源： https:/