#----------------------------------#
   # ORIGINAL

    # Sample DAta
    require (data.table)
    require (plyr)
    dtb <- data.table (expand.grid (month = rep (month.abb[1:3], each = 3),
                                           fac = letters[1:3]),
                              value = rnorm (27))


    # create two copies, since we will be altering the original
    dtb1 <- copy(dtb)
    dtb2 <- copy(dtb)


    # OP's original method
    (agg1 <- unique(dtb1[, value := mean (value), by = list (month, fac)]))

    # @Aruns suggestion
    (agg2 <- dtb2[, list(value = mean(value), newCol), by = list (month, fac)])

#----------------------------------#


The issue (and reasoning) becomes more apparent if you look at a data.table with rows in excess of the ones being used in the computation. 

    # Let's add a new column
    dtb[, newCol := LETTERS[seq(length(value))]

    
Note that if we just want to output the computed value, then expression on the `RHS` as you have it is just fine. 

    # This gives the expected results
    dtb[, mean (value), by = list (month, fac)]

    # This on the other hand assigns the respective values to *each* row
    dtb[, value := mean (value), by = list (month, fac)]




In other words, the data is being subsetted to only return unique values.  
However, if you want to save this value back into the _SAME_ data table (which is what happens when using `:=`) 
then all rows that are identified in `i` (all rows by defualt) will be assigned a value. (which, when you look at the output with additional columns, makes sense)

Then copying this data.table to agg still sends through all the rows. 


Therefore, if you want to copy to a new table, only those rows from your original table that are unique, you can 

   a.  wrap the table inside `unique()` before assigning it
   b.  assign the table, above, that is returned when you are not assigning the RHS output (which is what @Arun suggested)




#----------------------------------#


The following example might help illustrate.  _(You would need to copy + paste this, as the output is ommitted)_


      # SAMPLE DATA, as above
      library(data.table)
      dtb.bak <- data.table (expand.grid (month = rep (month.abb[1:3], each = 3), fac = letters[1:3]), value = rnorm (27))

      #  METHOD 1  # 
      #------------#
      dtb <- copy(dtb.bak)  # restore, from sample data.


      dtb[, value := mean (value), by = list (month, fac)]
      dtb

      # this is what you would like to assign
      unique(dtb)


      #  METHOD 2  # 
      #------------#
      dtb <- copy(dtb.bak)  # restore, from sample data.

      # this is what you would like to assign
      # next two lines are the same, only differnce is column name
      dtb[, mean (value), by = list (month, fac)]
      dtb[, list("mean" = mean (value)), by = list (month, fac)]  # quote marks added for clarity
      
      # dtb is unchanged. 
      dtb



      # NOW COMPARE THE SAME TWO METHODS, BUT IF THERE IS AN ADDITIOANL COLUMN
      dtb.bak[, newCol := rep(c("A", "B", "A"), length(value)/3)]


      dtb1 <- copy(dtb.bak)  # restore, from sample data.
      dtb2 <- copy(dtb.bak)  # restore, from sample data.


      # Method 1
      dtb1[, value := mean (value), by = list (month, fac)]
      dtb1
      unique(dtb1)

      #  METHOD 2  # 
      dtb2[, list("mean" = mean (value)), by = list (month, fac)]  # quote marks added for clarity
      dtb2

      # METHOD 2, WITH ADDED COLUMNS IN list() in `j`
      dtb2[, list("mean" = mean (value), newCol), by = list (month, fac)]  # quote marks added for clarity
      # notice this has more columns thatn 
      unique(dtb1)