To apply two functions to multiply variables with aggregate() in r

Question

There is a data frame x with 5753 observations of 4 variables.

The column names are: date, Depth, var1, and var2. I converted date and Depth to factor before performing aggregate().

I wanted to calculate average and standard deviation to 2 variables with grouping by date and Depth.

When applying aggregate(x[,3:4], by = list(x$date, x$Depth), FUN = function(x) c(avg = mean(x, na.rm = TRUE), SD= sd)), I got average of var1 and average of var 2 grouping by date and Depth, but I did not get SD.

When applying aggregate(. ~ date+Depth, data = x, FUN = function(x) c(avg = mean(x, na.rm = TRUE), SD= sd)), I got an error message: "Error in aggregate.data.frame(lhs, mf[-1L], FUN = FUN, ...) : no rows to aggregate".

After counting NA in two column, I found out that there are 5622 NA in var1, 5049 NA in var2. I donot want to remove NA before applying aggregate() yet.

My questions are:

1. why I did not get sd by applying the first syntax?

2. why is the second syntax not workable? I learned this syntax from stackoverflow, and it worked with the following data frame,

   x3 <- read.table(text = "  id1 id2 val1 val2
   1   a   x    1    9
   2   a   x    2    4
   3   a   y    3    NA
   4   a   y    4    NA
   5   b   x    1    NA
   6   b   y    4    NA
   7   b   x    3    9
   8   b   y    2    8", header = TRUE)
   

Answer 1

We can use dplyr, where we pass the grouping columns in group_by and the columns to summarise in summarise with across

library(dplyr) #1.0.0
x3 %>%
   group_by(id1, id2) %>% 
   summarise(across(starts_with('val'),  
        list(mean = ~ mean(., na.rm = TRUE) , sd = ~sd(., na.rm = TRUE))))
# A tibble: 4 x 6
# Groups:   id1 [2]
#  id1   id2   val1_mean val1_sd val2_mean val2_sd
#  <chr> <chr>     <dbl>   <dbl>     <dbl>   <dbl>
#1 a     x           1.5   0.707       6.5    3.54
#2 a     y           3.5   0.707     NaN     NA   
#3 b     x           2     1.41        9     NA   
#4 b     y           3     1.41        8     NA   

If the version of dplyr is < 1.0.0, we can use summarise_at

x3 %>%
   group_by(id1, id2) %>%  
   summarise_at(vars(-group_cols()), list(mean = ~ mean(., na.rm = TRUE), 
              sd = ~ sd(., na.rm = TRUE)))

---

With aggregate, the error we get because of the NA elements and it uses by default na.action = na.drop removing the row if there is any NA in that row. Either specify na.action = na.pass or NULL and this would resolve that issue. But, having multiple functions to be applied with c, it will result in a matrix column. Inorder to have normal data.frame, columns, we can wrap with data.frame in do.call

do.call(data.frame, aggregate(. ~ id1 + id2, data = x3, FUN = function(x) 
   c(avg = mean(x, na.rm = TRUE), SD= sd(x, na.rm = TRUE)), na.action = NULL))