r get mean of n columns by row

问题:

I have a simple data.frame

> df <- data.frame(a=c(3,5,7), b=c(5,3,7), c=c(5,6,4))
> df
  a b c
1 3 5 5
2 5 3 6
3 7 7 4

Is there a simple and efficient way to get a new data.frame with the same number of rows but with the mean of, for example, column a and b by row? something like this:

  mean.of.a.and.b c
1               4 5
2               4 6
3               7 4

回答1:

Use rowMeans() on the first two columns only. Then cbind() to the third column.

cbind(mean.of.a.and.b = rowMeans(df[-3]), df[3])
#   mean.of.a.and.b c
# 1               4 5
# 2               4 6
# 3               7 4

Note: If you have any NA values in your original data, you may want to use na.rm = TRUE in rowMeans(). See ?rowMeans for more.

回答2:

Another option using the dplyr package:

library("dplyr")

df %>%
  rowwise()%>%
  mutate(mean.of.a.and.b = mean(c(a, b))) %>%
  ## Then if you want to remove a and b:
  select(-a, -b)

回答3:

I think the best option is using rowMeans() posted by Richard Scriven. rowMeans and rowSums are equivalent to use of apply with FUN = mean or FUN = sum but a lot faster. I post the version with apply just for reference, in case we would like to pass another function.

data.frame(mean.of.a.and.b = apply(df[-3], 1, mean), c = df[3])

Output:

  mean.of.a.and.b c
1               4 5
2               4 6
3               7 4

Very verbose using SQL with sqldf

library(sqldf
sqldf("SELECT (sum(a)+sum(b))/(count(a)+count(b)) as mean, c 
      FROM df group by c")

Output:

  mean c
1    7 4
2    4 5
3    4 6