r dplyr sample_frac using seed in data - r

Question

I have a grouped data frame, in which the grouping variable is SEED. I want to take the groups defined by the values of SEED, set the seed to the value of SEED for each group, and then shuffle the rows of each group using dplyr::sample_frac. However, I cannot replicate my results, which indicates that the seed isn't being set correctly.
To do this in a dplyr-ish way, I wrote the following function:
> library(dplyr)
> ss_sampleseed <- function(df, seed.){
> set.seed(df$seed.)
> sample_frac(df, 1)
> }
I then use this function on my data:
> dg <- structure(list(Gene = c("CAMK1", "ARPC4", "CIDEC", "CAMK1", "ARPC4",
> "CIDEC"), GENESEED = c(1, 1, 1, 2, 2, 2)), class = c("tbl_df",
> "tbl", "data.frame"), row.names = c(NA, -6L), .Names = c("Gene",
> "GENESEED"))
> dg2 <- dg %>%
> group_by(GENESEED) %>%
> ss_sampleseed(GENESEED)
> dg2
Source: local data frame [6 x 2]
Groups: GENESEED
Gene GENESEED
1 ARPC4 1
2 CIDEC 1
3 CAMK1 1
4 CIDEC 2
5 ARPC4 2
6 CAMK1 2
However, when I repeat the above code, I cannot replicate my results.
> dg2
Source: local data frame [6 x 2]
Groups: GENESEED
Gene GENESEED
1 ARPC4 1
2 CAMK1 1
3 CIDEC 1
4 CAMK1 2
5 ARPC4 2
6 CIDEC 2

Answer 1

The problem here is that dollar sign will not substitute for the parameter you are passing. See this minimal example:
df <- data.frame(x = "x", GENESEED = "GENESEED")
h <- function(df,x){
df$x
}
h(df, GENESEED)
[1] x
Levels: x
See that h returns x even though you asked for GENESEED. So your function is actually trying to get df$seed which does not exist so it returns NULL.
But there is another problem. Even correcting this and passing directly the seed, it seems that it would not work as you want, because, if you look at the code of sample_frac, dplyr will eventually run the following line:
sampled <- lapply(index, sample_group, frac = TRUE, tbl = tbl,
size = size, replace = replace, weight = weight, .env = .env)
Notice that it runs a lapply after you set the seed, so you will not have defined a different seed for each group according to GENESEED as you wanted.
Taking this into consideration, I came up with this solution, using sample.int and do:
ss_sampleseed <- function(x){
set.seed(unique(x$GENESEED))
x[sample.int(nrow(x)), ]
}
dg %>% group_by(GENESEED) %>% do(ss_sampleseed(.))
This seems to be working as you want.

Answer 2

I think the main thing going here is the use of $ coding like you are inside your function. I certainly had to learn this the hard way. See also:
library(fortunes)
fortune(312)
fortune(343)
Take the simple function from #Carlos Cinelli and try to use it outside of any dplyr functions.
h = function(df, seed.){
df$seed.
}
h(dg, GENESEED)
NULL
It's those darn dollar signs. Now change the function to use [[ instead.
h2 = function(df, seed.){
df[[seed.]]
}
h2(dg, "GENESEED")
[1] 1 1 1 2 2 2
That's more like it, although you did have to put quotes around the variable name in the function.
So where does that leave your original function? You can go two ways. First, you could just change to [[ and use quotes around the variable name in your function.
ss_sampleseed = function(df, seed.){
set.seed(df[[seed.]])
sample_frac(df, 1)
}
dg %>%
group_by(GENESEED) %>%
ss_sampleseed("GENESEED")
Source: local data frame [6 x 2]
Groups: GENESEED
Gene GENESEED
1 CAMK1 1
2 CIDEC 1
3 ARPC4 1
4 CIDEC 2
5 CAMK1 2
6 ARPC4 2
The other option is to use deparse(substitute(seed.)) inside your function to allow for non-standard evaluation. You'll still need [[, though.
ss_sampleseed2 = function(df, seed.){
set.seed(df[[deparse(substitute(seed.))]])
sample_frac(df, 1)
}
dg %>%
group_by(GENESEED) %>%
ss_sampleseed2(GENESEED)
Source: local data frame [6 x 2]
Groups: GENESEED
Gene GENESEED
1 CAMK1 1
2 CIDEC 1
3 ARPC4 1
4 CIDEC 2
5 CAMK1 2
6 ARPC4 2
I get replicated results with either of these, although I didn't check if the seed is specifically set to what you want it to be.