R: How to calculate mean for each row with missing values using dplyr

Question

I want to calculate means over several columns for each row in my dataframe containing missing values, and place results in a new column called \'means.\' Here\'s my datafra

Answer 1

df %>% 
  mutate(means=rowMeans(., na.rm=TRUE))

The . is a "pronoun" that references the data frame df that was piped into mutate.

  A B  C    means
1 3 0  9 4.000000
2 4 6 NA 5.000000
3 5 8  1 4.666667

You can also select only specific columns to include, using all the usual methods (column names, indices, grep, etc.).

df %>% 
  mutate(means=rowMeans(.[ , c("A","C")], na.rm=TRUE))

  A B  C means
1 3 0  9     6
2 4 6 NA     4
3 5 8  1     3

Answer 2

It is simple to accomplish in base R as well:

cbind(df, "means"=rowMeans(df, na.rm=TRUE))
  A B  C    means
1 3 0  9 4.000000
2 4 6 NA 5.000000
3 5 8  1 4.666667

The rowMeans performs the calculation.and allows for the na.rm argument to skip missing values, while cbind allows you to bind the mean and whatever name you want to the the data.frame, df.

Answer 3

Regarding the error in OP's code, we can use the concatenate function c to get those elements as a single vector and then do the mean as mean can take only a single argument.

df %>%
    rowwise() %>% 
    mutate(means = mean(c(A, B, C), na.rm = TRUE))
#     A     B     C    means 
#  <dbl> <dbl> <dbl>    <dbl>
#1     3     0     9 4.000000
#2     4     6    NA 5.000000
#3     5     8     1 4.666667

---

Also, we can use rowMeans with transform

transform(df, means = rowMeans(df, na.rm = TRUE))
#  A B  C    means
#1 3 0  9 4.000000
#2 4 6 NA 5.000000
#3 5 8  1 4.666667

Or using data.table

library(data.table)
setDT(df)[, means := rowMeans(.SD, na.rm = TRUE)]