Sum observations from two columns, looping over many columns in R - r

Question

I have searched high and low, but am stuck on how to approach this. I have two sets of columns that I want to sum, row by row, but which I want to loop over many columns. If I were to do this manually, I would want:
df1[1,1]+df2[1,1]
df1[2,1]+df2[2,1]
etc... I've found many helpful examples on how to do something like:
apply(df[,c("a","d")], 1, sum)
though I want to do this over lots of columns. Also, while it's not entirely relevant, I want to phrase my question as close to my reality as possible, so my example below includes NA's, since my actual data contains many missing values.
# make a data frame, df1, with three columns
a <- sample(1:100, 50, replace = T)
b <- sample(100:300, 50, replace = T)
c <- sample(2:50, 500, replace = T)
df1 <- cbind(a,b,c)
# make another data frame, df2, with three columns
x <- sample(1:100, 50, replace = T)
y <- sample(100:300, 50, replace = T)
z <- sample(2:50, 50, replace = T)
df2 <- cbind(x,y,z)
# make another data frame, df2, with three columns
x <- sample(1:100, 50, replace = T)
y <- sample(100:300, 50, replace = T)
z <- sample(2:50, 50, replace = T)
df2 <- cbind(x,y,z)
Make it possible to randomly throw a few NAs in, function from http://www.r-bloggers.com/function-to-generate-a-random-data-set/
NAins <- NAinsert <- function(df, prop = .1){
n <- nrow(df)
m <- ncol(df)
num.to.na <- ceiling(prop*n*m)
id <- sample(0:(m*n-1), num.to.na, replace = FALSE)
rows <- id %/% m + 1
cols <- id %% m + 1
sapply(seq(num.to.na), function(x){
df[rows[x], cols[x]] <<- NA
}
)
return(df)
}
Add the NAs to the frames
NAins(df1, .2)
NAins(df2, .14)
Then, I tried to seq along the columns in each data frame, and used apply setting the index to 1, meaning to sum each row entry. This doesn't work.
for(i in seq_along(df1)){
for(j in seq_along(df2)){
apply(c(df1[,i], col2[j]), 1, function(x) sum(x, na.rm = T))}}
Thanks for any help!

Answer 1

You should be able to just replace NA with 0, and then add with "+":
replace(df1, is.na(df1), 0) + replace(df2, is.na(df2), 0)
# X Y Z
# 1 7 19 6
# 2 11 12 1
# 3 16 14 11
# 4 13 7 13
# 5 10 2 11
Alternatively, if you have more than just two data.frames, you can collect them in a list and use Reduce:
Reduce("+", lapply(mget(c("df1", "df2", "df3")), function(x) replace(x, is.na(x), 0)))
Here's some sample data (and what I think is an easier way to create it):
set.seed(1) ## Set a seed so others can reproduce your sample data
dfmaker <- function() {
setNames(
data.frame(
replicate(3, sample(c(NA, 1:10), 5, TRUE), FALSE)),
c("X", "Y", "Z"))
}
df1 <- dfmaker()
df1
# X Y Z
# 1 2 9 2
# 2 4 10 1
# 3 6 7 7
# 4 9 6 4
# 5 2 NA 8
df2 <- dfmaker()
df2
# X Y Z
# 1 5 10 4
# 2 7 2 NA
# 3 10 7 4
# 4 4 1 9
# 5 8 2 3
df3 <- dfmaker()

Answer 2

You can transform the data.frame to an array and sum them using apply function.
install.package('abind')
library(abind)
df <- abind(list(df1,df2), along = 3)
results <- apply(df, MARGIN = c(1,2), FUN = function(x) sum(x, na.rm = TRUE))
results