Question
Using R
and rmongodb
, how do I create a mongodb document from two data frames, the second of which will be an array element of the first?
Data
My first data.frame is always one row. e.g.
df_1 <- data.frame(myVar1 = 1,
myVar2 = 2,
myVar3 = 3)
My second data.frame is always one or more rows e.g.
df_2 <- data.frame(arrVar1 = c(1,2),
arrVar2 = c(1,2))
Required Solution
my goal is to have a document in a collection that structured like:
# {
# "_id" : ObjectId("565a939aa30fff2d67bfd492"),
# "vars" : {
# "myVar1" : 1.0000000000000000,
# "myVar2" : 2.0000000000000000,
# "myVar3" : 3.0000000000000000,
# "myArr" : [
# {
# "arrVar1" : 1,
# "arrVar2" : 1
# },
# {
# "arrVar1" : 2,
# "arrVar2" : 2
# }
# ]
# }
# }
How can I achieve this?
Edit
(removed all my attempts)
Thanks to Dmitriy for the answer and showing me what structure I needed to achieve.
As such, I've benchmarked a few different ways of getting the solution.
library(microbenchmark)
fun_1 <- function(df){
list(myArr = unname(split(df, seq(nrow(df)))))
}
fun_2 <- function(df){
list('myArr' = Map(function(i, d) d[i, ],
seq_len(nrow(df)),
MoreArgs = list('d' = df)
))
}
fun_3 <- function(df){
list(myArr = (lapply(as.list(1:dim(df)[1]), function(x) df[x[1],])))
}
microbenchmark(fun_1(df_2), fun_2(df_2), fun_3(df_2), times = 1000)
Unit: microseconds
expr min lq mean median uq max neval
fun_1(df_2) 162.135 176.7315 197.8129 187.7065 201.0385 1555.802 1000
fun_2(df_2) 84.770 92.2840 102.3595 96.3135 108.8165 1441.410 1000
fun_3(df_2) 85.052 93.8675 103.7496 97.9310 109.4090 1422.860 1000
There is nothing rmongodb special here. As I wrote everywhere: rmongodb will convert unnamed lists into arrays and named lists into objects. So you just should to convert your second data.frame into correct list:
df2_transformed <- list('myArr' = Map(function(i, df) df[i, ],
seq_len(nrow(df_2)),
MoreArgs = list('df' = df_2)
))
df1_df2_comb <- c(df_1, df2_transformed)
str(df1_df2_comb)
mongo.insert(mongo, paste0(db,".",coll), df1_df2_comb)
You can use Map
, lapply
, mapply
- depends on your preference.