We want to two-sample independent test using rsample-bootstrap() of tidymodels.
But the data will be split into train and test (Analysis/Assess), right?
How can you draw a bootstrap distribution?
group_1 <- rnorm(20,250,1)
group_2 <- rnorm(20,260,1)
group_1_2 <- group_1 - group_2
samples <- bootstrap(group_1_2,apparent=FALSE)
mean_list <- map(samples$splits,function(x){mean(x)})
hist(mean_list)
Do you have an idea of how to run bootstrap() or later?
Hi @Rsky,
This is probably how I'd do it:
library(rsample)
library(tibble)
library(purrr)
group_1 <- rnorm(20,250,1)
group_2 <- rnorm(20,260,1)
dd <- tibble(
gp = c(rep(1, 20), rep(2, 20)),
x = c(group_1, group_2)
)
boots <- bootstraps(dd, times = 100)
results <- map(boots$splits, ~t.test(x ~ gp, data = analysis(.x)))
results
#> [[1]]
#>
#> Welch Two Sample t-test
#>
#> data: x by gp
#> t = -39.723, df = 37.979, p-value < 2.2e-16
#> alternative hypothesis: true difference in means is not equal to 0
#> 95 percent confidence interval:
#> -10.387307 -9.379903
#> sample estimates:
#> mean in group 1 mean in group 2
#> 249.9325 259.8161
#>
#>
#> [[2]]
#>
#> Welch Two Sample t-test
#>
#> data: x by gp
#> t = -39.95, df = 32.403, p-value < 2.2e-16
#> alternative hypothesis: true difference in means is not equal to 0
#> 95 percent confidence interval:
#> -10.636565 -9.605016
#> sample estimates:
#> mean in group 1 mean in group 2
#> 249.8622 259.9830
#>
#>
#> [[3]]
#>
#> Welch Two Sample t-test
#>
#> data: x by gp
#> t = -38.852, df = 30.97, p-value < 2.2e-16
#> alternative hypothesis: true difference in means is not equal to 0
#> 95 percent confidence interval:
#> -10.496965 -9.449833
#> sample estimates:
#> mean in group 1 mean in group 2
#> 249.9768 259.9502
#>
#>
#> [[4]]
#>
#> Welch Two Sample t-test
#>
#> data: x by gp
#> t = -36.184, df = 37.634, p-value < 2.2e-16
#> alternative hypothesis: true difference in means is not equal to 0
#> 95 percent confidence interval:
#> -10.140577 -9.065691
#> sample estimates:
#> mean in group 1 mean in group 2
#> 250.1526 259.7558
#>
#>
#> [[5]]
#>
#> Welch Two Sample t-test
#>
#> data: x by gp
#> t = -36.379, df = 37.399, p-value < 2.2e-16
#> alternative hypothesis: true difference in means is not equal to 0
#> 95 percent confidence interval:
#> -9.570911 -8.561376
#> sample estimates:
#> mean in group 1 mean in group 2
#> 250.5136 259.5797
#>
#>
#> [[6]]
#>
#> Welch Two Sample t-test
#>
#> data: x by gp
#> t = -34.786, df = 37.945, p-value < 2.2e-16
#> alternative hypothesis: true difference in means is not equal to 0
#> 95 percent confidence interval:
#> -10.246793 -9.119708
#> sample estimates:
#> mean in group 1 mean in group 2
#> 250.2552 259.9385
#>
#>
#> [[7]]
#>
#> Welch Two Sample t-test
#>
#> data: x by gp
#> t = -45.541, df = 33.333, p-value < 2.2e-16
#> alternative hypothesis: true difference in means is not equal to 0
#> 95 percent confidence interval:
#> -10.154724 -9.286519
#> sample estimates:
#> mean in group 1 mean in group 2
#> 250.0092 259.7298
#>
#>
#> [[8]]
#>
#> Welch Two Sample t-test
#>
#> data: x by gp
#> t = -48.884, df = 33.506, p-value < 2.2e-16
#> alternative hypothesis: true difference in means is not equal to 0
#> 95 percent confidence interval:
#> -10.502344 -9.663542
#> sample estimates:
#> mean in group 1 mean in group 2
#> 249.7921 259.8750
#>
#>
#> [[9]]
#>
#> Welch Two Sample t-test
#>
#> data: x by gp
#> t = -35.212, df = 30.62, p-value < 2.2e-16
#> alternative hypothesis: true difference in means is not equal to 0
#> 95 percent confidence interval:
#> -10.171712 -9.057381
#> sample estimates:
#> mean in group 1 mean in group 2
#> 250.2398 259.8543
#>
#>
#> [[10]]
#>
#> Welch Two Sample t-test
#>
#> data: x by gp
#> t = -36.841, df = 33.874, p-value < 2.2e-16
#> alternative hypothesis: true difference in means is not equal to 0
#> 95 percent confidence interval:
#> -10.197028 -9.130719
#> sample estimates:
#> mean in group 1 mean in group 2
#> 250.0272 259.6911
#>
#>
#> [ reached getOption("max.print") -- omitted 90 entries ]
results is a list of t.test objects that has as many list-entries as number of bootstrap replicates. results can be inspected in whatever way you desire.
hi, @mattwarkentin
Thanks for the reply.
Does bootstrap generate the same number of lines(row) of data as the original data?
boots$splits
<Analysis/Assess/Total>
that mean split analysis & assess data, right?
If bootstrap() are splitting, I want to stop splitting.
rsample performs bootstrapping the way you'd expect (i.e. sampling with replacement). Yes, each bootstrap has as many rows as the number of rows in the original data.
For a given resample, some rows in the original data will be included multiple times while some rows will not be included at all. So each split in boots$splits has two parts, the analysis part, which is the standard bootstrap, and also the assessment part, which is the rows that did not get included.
So pulling out the analysis part of a bootstrap split will be a data frame with as many rows as the original data, created by sampling with replacement.
analysis(boots$splits[[1]])
The analysis parts will always be the same size (nrow(original_data)), while the assessment part will vary in size for the various bootstrap replicates.
Yes, exactly! The assessment split is the OOB observations.
I just want to ask one more question about your code.
Should 'strate' be specified in the case of the above 'difference comparison'?
bootstrap(dd,strata=gp) ?
No, it should not be used for testing purposes.
thank you.
How do you decide if you should use 'strate' or not?
Since the ABtest group is likely to affect the distribution, I was wondering if I should do 'sampling' in equal proportions.
this page told
strate = This can help ensure that the number of data points in the bootstrap sample is equivalent to the proportions in the original data set.
rsample/R/boot.R at main · tidymodels/rsample · GitHub
Is it a case of excessive imbalance?
and Learn to use this
Bootstrap Sampling — bootstraps • rsample
tidymodels - Bootstrap resampling and tidy regression models
strata is often used when you are creating a predictive model and what the training and test sets to have similar outcome distributions. For testing, it would probably bias the results.
Thank you!
I was able to understand.
Thank you for introducing me to a nice web site.
I will study with it.
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