如何在 R 中为多列创建分层采样

Question

my data set has got 821049 variables and 18 columns. I would like to take 9 columns for the stratified sampling. These are "BASKETS_NZ", "PIS", "PIS_AP" "PIS_DV", "PIS_PL", "PIS_SDV", "PIS_SHOPS" "PIS_SR", "QUANTITY". My stratification variable is ID = 1:821049. How do I choose the intervals for my variables? How do I set the size of the sampling?

dpt(rbind(head(WKA_ohneJB, 10), tail(WKA_ohneJB, 10)))

structure(list(X = c(1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 

821039L, 821040L, 821041L, 821042L, 821043L, 821044L, 821045L, 

821046L, 821047L, 821048L), BASKETS_NZ = c(1L, 1L, 1L, 1L, 1L, 

1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), 

LOGONS = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 0L, 1L, 

1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), PIS = c(71L, 39L, 50L, 4L, 

13L, 4L, 30L, 65L, 13L, 31L, 111L, 33L, 3L, 46L, 11L, 8L, 

17L, 68L, 65L, 15L), PIS_AP = c(14L, 2L, 4L, 0L, 0L, 0L, 

1L, 0L, 2L, 1L, 13L, 0L, 0L, 2L, 1L, 0L, 3L, 8L, 0L, 1L), 

PIS_DV = c(3L, 19L, 4L, 1L, 0L, 0L, 6L, 2L, 2L, 3L, 38L, 

8L, 0L, 5L, 2L, 0L, 1L, 0L, 3L, 2L), PIS_PL = c(0L, 5L, 8L, 

2L, 0L, 0L, 0L, 24L, 0L, 6L, 32L, 8L, 0L, 0L, 4L, 0L, 0L, 

0L, 0L, 0L), PIS_SDV = c(18L, 0L, 11L, 0L, 0L, 0L, 0L, 0L, 

0L, 1L, 6L, 0L, 0L, 13L, 0L, 0L, 1L, 15L, 1L, 0L), PIS_SHOPS = c(3L, 

24L, 13L, 3L, 0L, 0L, 6L, 28L, 2L, 11L, 71L, 16L, 2L, 5L, 

6L, 0L, 1L, 0L, 3L, 2L), PIS_SR = c(19L, 0L, 14L, 0L, 0L, 

0L, 2L, 23L, 0L, 3L, 6L, 0L, 0L, 20L, 0L, 0L, 3L, 32L, 1L, 

0L), QUANTITY = c(13L, 2L, 18L, 1L, 14L, 1L, 4L, 2L, 5L, 

1L, 5L, 2L, 2L, 4L, 1L, 3L, 2L, 8L, 17L, 8L), WKA = c(1L, 

1L, 1L, 1L, 1L, 1L, 0L, 0L, 1L, 0L, 1L, 1L, 1L, 1L, 1L, 1L, 

0L, 0L, 1L, 1L), NEW_CUST = c(0L, 0L, 0L, 0L, 0L, 0L, 0L, 

0L, 0L, 0L, 1L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L), EXIST_CUST = c(1L, 

1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 0L, 1L, 1L, 1L, 1L, 1L, 

1L, 1L, 1L, 1L), WEB_CUST = c(1L, 0L, 0L, 0L, 1L, 1L, 0L, 

1L, 1L, 1L, 1L, 1L, 1L, 0L, 0L, 0L, 0L, 0L, 0L, 1L), MOBILE_CUST = c(0L, 

1L, 1L, 1L, 0L, 0L, 1L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 

1L, 0L, 1L, 0L), TABLET_CUST = c(0L, 0L, 0L, 0L, 0L, 0L, 

0L, 0L, 0L, 0L, 0L, 0L, 0L, 1L, 1L, 1L, 0L, 1L, 0L, 0L), 

LOGON_CUST_STEP2 = c(0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 

0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L)), row.names = c(1L, 

2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 821039L, 821040L, 821041L, 

821042L, 821043L, 821044L, 821045L, 821046L, 821047L, 821048L

), class = "data.frame") 

Answer 1

Here is a solution to perform a stratified sampling based on multiple columns. Before implementing this, consider that your data is continuous and a sufficiently large that just a random sampling is adequate.

To solve this problem is to take a stratified sample from each group. The potential approaches to group the data together is by either pasting the 9 columns together or using dplyr's groupby function.

Using the solution is this question How to get around error "factor has new levels" in cross-validation glm? and updating with dplyr style.

This dplyr_stratified function will take the desired sampling ration and an arbitrary number of column and will return a data frame with the sampled rows. See the example below for taking 2 columns.

set.seed(1)
x <- rnorm(n = 100)
y <- rep(x = c("A","B"), times = c(50,50))
z <- rep(x = c("D","E","F"), times = c(33,33,34))
data <- data.frame(x, y=sample(y, replace = TRUE), z=sample(z, replace=TRUE))

library(dplyr)
#optional tag row for later identification: 
data$rowid<-1:nrow(data)
dplyr_stratified <- function(df, percent, ...){
  columns<-enquos(...)
   #group then sample each group
  out<-df %>% group_by(!!!columns)  %>% slice( sample(1:n(), percent*n())) 
}

testgroup<-dplyr_stratified(data, 0.8, z, y)
testgroup

Note: this is assuming each grouping will have a sufficient number of sample in order to select a representative sample. (If the groups are too small then this approach may not meet expectations)