You may try pmax df$c <- pmax(df$a, df$b) df # a b c # 1 dog <NA> dog # 2 mouse <NA> mouse # 3 <NA> cat cat # 4 bird <NA> bird …or ifelse: df$c <- ifelse(is.na(df$a), df$b, df$a) For more general solutions in cases with more than two columns, you find several ways … Read more
Use rowSums. To remove rows from a data frame (df) that contain precisely n NA values: df <- df[rowSums(is.na(df)) != n, ] or to remove rows that contain n or more NA values: df <- df[rowSums(is.na(df)) < n, ] in both cases of course replacing n with the number that’s required
It didn’t work because status was a factor. When you mix factor with numeric then numeric is the least restrictive. By forcing status to be character you get the results you’re after and the column is now a character vector: TEST$UNIT[is.na(TEST$UNIT)] <- as.character(TEST$STATUS[is.na(TEST$UNIT)]) ## UNIT STATUS TERMINATED START STOP ## 1 ACTIVE ACTIVE 1999-07-06 2007-04-23 … Read more
You can do: x[, 1:2][is.na(x[, 1:2])] <- 0 or better (IMHO), use the variable names: x[c(“a”, “b”)][is.na(x[c(“a”, “b”)])] <- 0 In both cases, 1:2 or c(“a”, “b”) can be replaced by a pre-defined vector.
Here’s a data table approach that uses na.omit() across the columns, grouped by ID. library(data.table) setDT(df)[, lapply(.SD, na.omit), by = ID] # ID Col1 Col2 Col3 Col4 # 1: 1 5 10 15 20 # 2: 2 25 30 35 40
For the purpose of a “true-NA”: Seems the most direct route is just to modify the value returned by paste2 to be NA when the value is “” paste3 <- function(…,sep=”, “) { L <- list(…) L <- lapply(L,function(x) {x[is.na(x)] <- “”; x}) ret <-gsub(paste0(“(^”,sep,”|”,sep,”$)”),””, gsub(paste0(sep,sep),sep, do.call(paste,c(L,list(sep=sep))))) is.na(ret) <- ret==”” ret } val<- paste3(c(“a”,”b”, “c”, … Read more
Not my own technique I saw it on the boards a while back: dat <- read.table(text = “id taxa length width 101 collembola 2.1 0.9 102 mite 0.9 0.7 103 mite 1.1 0.8 104 collembola NA NA 105 collembola 1.5 0.5 106 mite NA NA”, header=TRUE) library(plyr) impute.mean <- function(x) replace(x, is.na(x), mean(x, na.rm = … Read more
This: df[ df == “foo” ] <- NA
See my comment in @gsk3 answer. A simple example: > m <- matrix(sample(c(NA, 1:10), 100, replace = TRUE), 10) > d <- as.data.frame(m) V1 V2 V3 V4 V5 V6 V7 V8 V9 V10 1 4 3 NA 3 7 6 6 10 6 5 2 9 8 9 5 10 NA 2 1 7 2 … Read more