This probably didn't exist when the answers were written, but since I came here with the same question and found a better solution, here it is for future googlers:
What you want is the coalesce() function from dplyr:
y <- c(1, 2, NA, NA, 5)
z <- c(NA, NA, 3, 4, 5)
coalesce(y, z)
[1] 1 2 3 4 5
You can also do: with(d,ifelse(is.na(A),B,A))
where d is your data frame.
Another very simple solution in this case is to use the rowSums function.
df$New<-rowSums(df[, c("A", "B")], na.rm=T)
df<-df[, c("ID", "New")]
Update: Thanks @Artem Klevtsov for mentioning that this method only works with numeric data.
You can use unite from tidyr:
library(tidyr)
df[is.na(df)] = ''
unite(df, new, A:B, sep='')
# ID new
#1 1 3
#2 2 2
#3 3 4
#4 4 1
You could try
New <- do.call(pmax, c(df1[-1], na.rm=TRUE))
Or
New <- df1[-1][cbind(1:nrow(df1),max.col(!is.na(df1[-1])))]
d1 <- data.frame(ID=df1$ID, New)
d1
# ID New
#1 1 3
#2 2 2
#3 3 4
#4 4 1
Assuming either A or B have a NA, that would work just fine:
# creating initial data frame (actually data.table in this case)
library(data.table)
x<- as.data.table(list(ID = c(1,2,3,4), A = c(3, NA, NA, 1), B = c(NA, 2, 4, NA)))
x
# ID A B
#1: 1 3 NA
#2: 2 NA 2
#3: 3 NA 4
#4: 4 1 NA
#solution
y[,New := na.omit(c(A,B)), by = ID][,c("A","B"):=NULL]
y
# ID New
#1: 1 3
#2: 2 2
#3: 3 4
#4: 4 1
This question's been around for a while, but just to add another possible approach that does not depend on any libraries:
df$new = t(df[-1])[!is.na(t(df[-1]))]
# ID A B new
# 1 1 3 NA 3
# 2 2 NA 2 2
# 3 3 NA 4 4
# 4 4 1 NA 1
