R-Studio Filtering Data - r

Question

I have this data table as model:
ID PRODUCT_TYPE OFFER INENTORY
1 BED Y Y
2 TABLE N Y
3 MOUSE Y N
4 CELLPHONE Y Y
5 CAR Y Y
6 BED N N
7 TABLE N Y
8 MOUSE Y N
9 CELLPHONE Y Y
10 CAR Y Y
.....
I have to extract a sample of 50% of the total population and the sample must consist on appearance of the values ​​of the variables at least once (product_type == bed, cellphone, car, table, mouse, offer = Y, N, etc).
I used this to extract the sample:
subset1<- data2 %>% sample_frac(.5)
but I don't know how to integrate these conditions, can anyone help me with an advice?

Answer 1

It's unclear from the content of the original post whether the question it asks is How does one generate a stratified random sample based on combinations of a set of grouping variables? A stratified random sample is an appropriate approach in this situation because it ensures that each combination of grouping variables is proportionally represented in the sampled data frame.
A tidyverse solution
Since the question does not include a minimal reproducible example, we'll generate some data and illustrate how to split or group it and then randomly sample each of the subgroups.
To begin, we reset the seed for the random number generator and build a data frame containing 10,000 rows of products, where 50% of the products are on offer, and 70% are in inventory.
set.seed(1053807)
df <- data.frame(
productType = rep(c("Bed","Mouse","Table","Cellphone","Laptop","Car","Chair","Blanket",
"Sofa","Bicycle"),1000),
offer = ifelse(runif(10000) > .5,"Y","N"),
inventory = ifelse(runif(10000) > .3,"Y","N"),
price = rnorm(10000,200,10)
)
Given the three grouping variables in the original post, the df object contains 40 unique combinations of productType, offer, and inventory.
The original code attempts to use the dplyr package to sample the data. It was very close to a workable solution. To stratify the sample we use group_by() to group the data by split variables, and then use the sample_frac() function on the grouped data to generate the stratified sample.
library(dplyr)
df %>%
group_by(productType,offer,inventory) %>%
sample_frac(0.5) -> sampledData
Verifying results
A 50% sample from a 10,000 row data frame should have about 5,000 observations.
> nrow(sampledData)
[1] 5001
So far, so good.
We can then verify the results by counting numbers of rows in each stratum of the sample, and comparing them to the original counts for each subgroup in the input data frame.
# check results
originalCounts <- df %>%
group_by(productType,offer,inventory) %>%
summarise(OriginalCount = n())
sampledData %>%
group_by(productType,offer,inventory) %>%
summarise(SampledCount = n()) %>%
full_join(originalCounts,.) %>%
mutate(SampledPct = round(SampledCount / OriginalCount * 100,2))
...and the output:
# A tibble: 40 x 6
# Groups: productType, offer [20]
productType offer inventory OriginalCount SampledCount SampledPct
<chr> <chr> <chr> <int> <int> <dbl>
1 Bed N N 161 80 49.7
2 Bed N Y 371 186 50.1
3 Bed Y N 132 66 50
4 Bed Y Y 336 168 50
5 Bicycle N N 154 77 50
6 Bicycle N Y 349 174 49.9
7 Bicycle Y N 147 74 50.3
8 Bicycle Y Y 350 175 50
9 Blanket N N 134 67 50
10 Blanket N Y 349 174 49.9
# … with 30 more rows
By inspecting the data, we see that data frames with even numbers of observations result in an exact 50% sample, whereas data frames with odd numbers of observations are slightly above or below 50%.
A Base R solution
We can also solve the problem with Base R. This approach uses the three variables in the original post, product type, offer, and inventory to split the data into subgroups based on the combinations of values for these variables, take a random sample from each subset, and combine the result into a single data frame.
First, we set the seed for the random number generator and build a data frame containing 10,000 rows of products, where 50% of the products are on offer, and 70% are in inventory.
set.seed(1053807)
df <- data.frame(
productType = rep(c("Bed","Mouse","Table","Cellphone","Laptop","Car","Chair","Blanket",
"Sofa","Bicycle"),1000),
offer = ifelse(runif(10000) > .5,"Y","N"),
inventory = ifelse(runif(10000) > .3,"Y","N"),
price = rnorm(10000,200,10)
)
Since we want to separately sample each combination of product, offer, and inventory, we create a combined split variable, and then use it to split the data.
splitvar <- paste(df$productType,df$offer,df$inventory,sep="-")
dfList <- split(df,splitvar)
Given the input data frame parameters of 10 products, 2 levels of offer (Y / N), and 2 levels of inventory (Y / N), this creates a dfList object that is a list of 40 data frames, each with varying numbers of observations.
We then use lapply() to randomly select about 50% of each data frame, using the number of rows for each data frame to drive the sample() function.
sampledDataList <- lapply(dfList,function(x){
x[sample(nrow(x),size = round(.5 * nrow(x))),]
})
At this point the sampledDataList object is a list of 40 data frames, each of which has approximately 50% of the rows as the original list.
To create the final data frame, we use do.call() as follows.
sampledData <- do.call(rbind,sampledDataList)
When we check the number of observations in the resulting data frame, we see that it is approximately 50% of the original data size (10,000).
> # this should be approximately 5,000 rows
> nrow(sampledData)
[1] 5001
We can further verify that each data frame is approximately a 50% sample with the following code.
# verify sample percentage by stratum
stratum <- names(sampledDataList)
OriginalCount <- sapply(dfList,nrow)
SampledCount <- sapply(sampledDataList,nrow)
SamplePct <- round(SampledCount / OriginalCount * 100,2)
head(data.frame(stratum,OriginalCount,SampledCount,SamplePct,row.names = NULL),10)
...and the output:
> head(data.frame(stratum,OriginalCount,SampledCount,SamplePct,row.names = NULL),10)
stratum OriginalCount SampledCount SamplePct
1 Bed-N-N 161 80 49.69
2 Bed-N-Y 371 186 50.13
3 Bed-Y-N 132 66 50.00
4 Bed-Y-Y 336 168 50.00
5 Bicycle-N-N 154 77 50.00
6 Bicycle-N-Y 349 174 49.86
7 Bicycle-Y-N 147 74 50.34
8 Bicycle-Y-Y 350 175 50.00
9 Blanket-N-N 134 67 50.00
10 Blanket-N-Y 349 174 49.86
As was the case with the dplyr solution, we see that strata with odd numbers of rows either sample one more or one less than an exact 50% of the original data.