Take randomly sample based on groups

Question

I have a df made by almost 50,000 rows spread in 15 different IDs (every ID has thousands of observations). df looks like:

        ID  Year    Temp    ph
1           


        

Answer 1

Although this is not very elegant solution, but it may work.

library(data.table)
df <- data.table(df)
f <- list()
for(i in unique(df1$ID)){
 f[[i]] <- df1[id == i][sample(.N,(500))]
  }
 dfnew <- rbindlist(f)

Answer 2

library(data.table) #1
df <- data.table(df) #2
df[,group_num := sample(2,.N,replace = TRUE,prob = c(500,.N-500)/.N),by = "ID"] #3
df_sample = df[group_num == 1,] #4

or you can change line #3 and #4 to:

df[,random_num := sample(.N,.N),by="ID"]
df_sample  = df[random_num <=500,]

Answer 3

In case you have big datasets, a data.table solution could go like this:

library(data.table)

# Generate 26 mil rows random data
set.seed(2019)
dt <- data.table(c1 = sample(length(LETTERS)*10^6), 
                 c2 = sample(LETTERS, replace = TRUE))

# For each letter, sample 500 rows
dt_sample <- dt[, .SD[sample(x = .N, size = 500)], by = c2]

# We indeed sampled 500 rows for each letter
dt_sample[, .N, by = c2][order(c2)]
#>     c2   N
#>  1:  A 500
#>  2:  D 500
#>  3:  G 500
#>  4:  I 500
#>  5:  M 500
#>  6:  N 500
#>  7:  O 500
#>  8:  P 500
#>  9:  Q 500
#> 10:  R 500
#> 11:  S 500
#> 12:  T 500
#> 13:  U 500
#> 14:  V 500
#> 15:  W 500
#> 16:  Y 500
#> 17:  Z 500

^{Created on 2019-04-23 by the reprex package (v0.2.1)}

In case your data is unbalanced in the sense that some groups happen to be smaller (as number of rows) than your desired sample size, then you need to set a defensive trick like sample size should be min(500, .N) - see sample random rows within each group in a data.table. So like:

dt[, .SD[sample(x = .N, size = min(500, .N))], by = c2]

Answer 4

Try this:

library(plyr)
ddply(df,.(ID),function(x) x[sample(nrow(x),500),])

Answer 5

An approach if on of the IDs is < 500. Here I used the mtcars set:

n <- 8
df <- mtcars
df$ID <- df$cyl

FUN <- function(x, n) {
    if (length(x) <= n) return(x)
    x[x %in% sample(x, n)]
}

df[unlist(lapply(split(1:nrow(df), df$ID), FUN, n = 8)), ]

Answer 6

This is available as the sample_n function in dplyr:

library(dplyr)
new_df <- df %>% group_by(ID) %>% sample_n(500)