具有不同条件的示例 data.table 行

Question

I have a data.table with multiple columns. One of these columns currently works as a 'key' ( keyb for the example). Another column (let's say A ), may or may not have data in it. I would like to supply a vector that randomly sample two rows per key, -if this key appears in the vector, where 1 row contains data in A , while the other does not.

MRE:

#data.table
trys <- structure(list(keyb = c("x", "x", "x", "x", "x", "y", "y", "y", 
"y", "y"), A = c("1", "", "1", "", "", "1", "", "", "1", "")), .Names = c("keyb", 
"A"), row.names = c(NA, -10L), class = c("data.table", "data.frame"
))
setkey(trys,keyb)

#list with keys
list_try <- structure(list(a = "x", b = c("r", "y","x")), .Names = c("a", "b"))

I could, for instance subset the data.table based on the elements that appear in list_try :

trys[keyb %in% list_try[[2]]]

My original (and probably inefficient idea), was to try to chain a sample of two rows per key, where the A column has data or no data, and then merge. But it does not work:

#here I was trying to sample rows based on whether A has data or not
#here for rows where A has no data
trys[keyb %in% list_try[[2]]][nchar(A)==0][sample(.N, 2), ,by = keyb]
#here for rows where A has data
trys[keyb %in% list_try[[2]]][nchar(A)==1][sample(.N, 2), ,by = keyb]

In this case, my expected output would be two data.tables (one for a and one for b in list_try ), of two rows per appearing element: So the data.table from a would have two rows (one with and without data in A), and the one from b , four rows (two with and two without data in A).

Please let me know if I can make this post any clearer

Answer 1

You could add A to the by statement too, while converting it to a binary vector by modifying to A != "" , combine with a binary join (while adding nomatch = 0L in order to remove non-matches) you could then sample from the row index .I by those two aggregators and then subset from the original data set

For a single subset case

trys[trys[list_try[[2]], nomatch = 0L, sample(.I, 1L), by = .(keyb, A != "")]$V1]
#    keyb A
# 1:    y 1
# 2:    y  
# 3:    x 1
# 4:    x  

---

For a more general case, when you want to create separate data sets according to a list of keys, you could easily embed this into lapply

lapply(list_try, 
       function(x) trys[trys[x, nomatch = 0L, sample(.I, 1L), by = .(keyb, A != "")]$V1]) 
# $a
# keyb A
# 1:    x 1
# 2:    x  
# 
# $b
# keyb A
# 1:    y 1
# 2:    y  
# 3:    x 1
# 4:    x