CODEWITHSUNDEEP
rrandomdataframe
Math
Math

Reputation: 1294

Sample random rows in dataframe with probability

How can I take a random sample (with or without replacement) but with given probabilities?

I am trying to extract a random sample of rows in iris data frame but with this condition of species: 80% versicolor and 20% virginica

> head(iris)
  Sepal.Length Sepal.Width Petal.Length Petal.Width Species
1          5.1         3.5          1.4         0.2  setosa
2          4.9         3.0          1.4         0.2  setosa
3          4.7         3.2          1.3         0.2  setosa
4          4.6         3.1          1.5         0.2  setosa
5          5.0         3.6          1.4         0.2  setosa
6          5.4         3.9          1.7         0.4  setosa

Upvotes: 3

Views: 2765

Answers (3)

akrun
akrun

Reputation: 887961

We can make use of the quosures from the devel version of dplyr (soon to be released 0.6.0) for creating the function

library(tidyverse)
f.sample <- function(dat, colN, value, perc){
       colN <- enquo(colN)
       value <- quo_name(enquo(value))
       dat %>%
            filter(UQ(colN) == UQ(value)) %>%
            sample_frac(perc) %>%
            droplevels
}

f.sample(iris, Species, versicolor, 0.8)
f.sample(iris, Species, virginica, 0.2)
#Sepal.Length Sepal.Width Petal.Length Petal.Width   Species
#20          6.0         2.2          5.0         1.5 virginica
#9           6.7         2.5          5.8         1.8 virginica
#15          5.8         2.8          5.1         2.4 virginica
#10          7.2         3.6          6.1         2.5 virginica
#12          6.4         2.7          5.3         1.9 virginica
#49          6.2         3.4          5.4         2.3 virginica
#22          5.6         2.8          4.9         2.0 virginica
#34          6.3         2.8          5.1         1.5 virginica
#2           5.8         2.7          5.1         1.9 virginica
#44          6.8         3.2          5.9         2.3 virginica

The enquo does similar functionality as substitute by taking the input arguments and convert it to quosure, while quo_name convert to string, and within the filter/group_by/summarise/mutate the quosures are evaluated by unquoting (!! or UQ)

Based on the comments below, we modified the function so that it would work for other cases

f.sample2 <- function(dat, colN, values, perc){
          colN <- enquo(colN)
          dat %>%
              filter(UQ(colN) %in% values) %>%
              droplevels %>%
              nest(-UQ(colN)) %>%
              .$data %>%
              setNames(values) %>%
              Map(sample_frac, ., perc) %>%
              bind_rows(.id = quo_name(colN))               

        } 


res <- f.sample2(iris, Species, c("versicolor", "virginica"), c(0.8, 0.2))
prop.table(table(res$Species))
#versicolor  virginica 
#      0.8        0.2

Upvotes: 3

talat
talat

Reputation: 70336

I seem to have a different understanding than the other answerers.

The following function should produce a 80/20 dataset regardless of the group sizes in the original data set.

foo <- function(DF, n = 50, group_var, groups, probs, replace = FALSE) {

  # subset relevant groups & split
  DF <- DF[DF[[group_var]] %in% groups, ]
  DF <- split(DF, as.character(DF[[group_var]]))
  DF <- DF[match(names(DF), groups)]

  # sample number of observations per group (this requires replace= TRUE)
  smpl <- sample(groups, size = n, replace = TRUE, prob = probs)
  # subset random rows per group according to group size
  DF <- Map(function(x,y) x[sample(1:nrow(x), y, replace = replace),], DF, c(table(smpl)))

  # combine and clean up
  DF <- do.call(rbind, DF)
  DF <- DF[sample(nrow(DF)),]  # not really necessary  
  row.names(DF) <- NULL        # not really necessary  
  DF
}


foo(iris, 50, "Species", c("versicolor", "virginica"), c(0.8, 0.2))

Upvotes: 3

989
989

Reputation: 12935

You could try this in base R:

f.sample <- function(a, percent) a[sample(nrow(a), nrow(a)*percent, replace = TRUE),]

f.sample(iris[iris$Species=="versicolor",], 0.8)
f.sample(iris[iris$Species=="virginica",], 0.2)

You can set the replace argument accordingly.

Upvotes: 3

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