I have data set with two factors: Environments (4 levels), Individuals (500 in each environment) and response variable YD. As part of my analysis I have to randomly sample 100 individuals from each Environment in the following way:
I already solved this problem using several lines of code, however I hope someone will help me generate an R function to do that, which will be very useful in other situations.
Here is example data set with similar structure:
library(BLR)
data (wheat)
Data <- melt(Y)
colnames(Data) <- c('Individuals','Environments','YD')
updated answer:
I just wrapped the answer in a function. Note, that is valid only for exactly 4 levels.
colnames(Data)<-c("Individuals","Environments","YD") #removed spaces from names
myfun <- function(DF, samplefrom, samplelevels, sampletype, samplesize)
{
if(sampletype == "per1")
{
Env1 = sample(unique(DF[[samplefrom]]), samplesize)
Env2 <- Env3 <- Env4 <- Env1
}
if(sampletype == "per4")
{
Env1 = sample(unique(DF[[samplefrom]]), samplesize)
Env2 = sample(unique(DF[[samplefrom]])[!unique(DF[[samplefrom]]) %in% Env1], samplesize)
Env3 = sample(unique(DF[[samplefrom]])[!unique(DF[[samplefrom]]) %in% c(Env1, Env2)], samplesize)
Env4 = sample(unique(DF[[samplefrom]])[!unique(DF[[samplefrom]]) %in% c(Env1, Env2, Env3)], samplesize)
}
if(sampletype == "per2")
{
Env1 = sample(unique(DF[[samplefrom]]), samplesize)
Env2 <- Env1
Env3 = sample(unique(DF[[samplefrom]])[!unique(DF[[samplefrom]]) %in% Env1], samplesize)
Env4 <- Env3
}
ret = do.call(rbind, mapply(function(ind, env) {df <- Data[DF[[samplelevels]] == env,];
df[df[[samplefrom]] %in% ind,]},
env = as.list(sample(unique(DF[[samplelevels]]))), ind = list(Env1, Env2, Env3, Env4),
SIMPLIFY = F)) #in `env = ` added `sample` to select the environments
#in random order and assign them the individuals
return(ret)
}
myfun(Data, "Individuals", "Environments", "per1", 2)
# Individuals Environments YD
#21 13954 1 0.6658681
#345 457982 1 -1.1022770
#620 13954 2 -0.4888968
#944 457982 2 0.6026167
#1219 13954 4 -0.7183965
#1543 457982 4 0.4881141
#1818 13954 5 0.2660623
#2142 457982 5 -2.0626073
myfun(Data, "Individuals", "Environments", "per2", 2)
# Individuals Environments YD
#25 15292 1 -1.1272386
#248 373045 1 -0.6659416
#624 15292 2 -0.2362053
#847 373045 2 0.5778210
#1260 62150 4 1.2077921
#1654 1541043 4 1.1406084
#1859 62150 5 -0.3358584
#2253 1541043 5 0.3897426
myfun(Data, "Individuals", "Environments", "per4", 2)
# Individuals Environments YD
#106 85786 1 1.4480500
#567 3830162 1 -1.8052577
#1029 1301802 2 0.2737786
#1043 1410845 2 1.0617118
#1630 1302304 4 0.6673241
#1678 1766332 4 -0.0451913
#1871 65315 5 -0.0597450
#2336 2621166 5 2.5590801
update 2 some comments
mapply applies a function sequentially to multiple arguments. Here, the function takes two arguments: ind and env. The function 1) subsets the dataframe by env and 2) subsets the subsetted dataframe by ind. env is an environment and ind is the sample of individuals (Env1, ...) previously calculated in myfun. The multiple arguments of the function to be mapplied are env: [1, 2, 3, 4] and ind: [Env1, Env2, Env3, Env4]. mapply takes sequentially env = 1 and ind = Env1, env = 2 and ind = Env2 etc, and gives the result (the necessary subsets) in a list. do.call(rbind,) joins the list in a dataframe output.
P.S. Note that because sample is used env can be [1, 2, 3, 4] or [2, 4, 3, 1] or whatever and so the sequential combination of the function's (to be mapplied) arguments is not only env = 1 and ind = Env1 but env = 1 or 2 or 3 or 4 and ind = Env1, and so on.
update 3 and 4 function with different No levels
No_different_samples is the number of different samples you wish to take; I made it to default to the number of samplelevels (i.e. a different sample for every level). I made the function to give an error if the No_different_samples can't fit inthe No levels (i.e. if you want 3 different samples from a population with 4 levels (as your Data), it throws an error; you have to select either 1 or 2 or 4.
myfun2 <- function(DF, samplefrom, samplelevels,
No_different_samples = NULL, grouping = NULL, samplesize)
{
samp <- sample(unique(DF[[samplefrom]]))
levs <- unique(DF[[samplelevels]])
if(is.null(No_different_samples)) No_different_samples <- length(levs)
if(is.null(grouping)) grouping <- c(1, 1, 1, 1)
if(length(levs) %% No_different_samples) stop("an error message here")
if(length(samp) < No_different_samples * samplesize)
stop("can't take a sample this large from the population")
ls_diffr_samps <- vector("list", length = No_different_samples)
for(i in 1:No_different_samples)
{
ls_diffr_samps[[i]] <- samp[(i * samplesize - (samplesize - 1)) : (i * samplesize)]
}
list_samples <- rep(ls_diffr_samps, times = grouping)
ret = do.call(rbind, mapply(function(ind, env) {df <- DF[DF[[samplelevels]] == env,];
df[df[[samplefrom]] %in% ind,]},
env = as.list(sample(levs)), ind = list_samples,
SIMPLIFY = F))
return(ret)
}
myfun2(Data, "Individuals", "Environments", 1, 4, 2) #same sample for all
myfun2(Data, "Individuals", "Environments", 2, c(2, 2), 2) #same sample per 2
myfun2(Data, "Individuals", "Environments", 2, c(3, 1), 2) #same sample for 3
myfun2(Data, "Individuals", "Environments", 4, c(1, 1, 1, 1), 2) #different sample for all