Can I "subtract" the sub dataset from a certain dataset?

I made a dataset to use for statistical analysis, and divided it into 2 sub dataset.
One is for calibrating(training) the model and another is for the validation(testing).
Using sampling function, I made a sample that will be used as calibration set. But the problem is, I have no idea how to divide the original dataset with two set, just by subtracting the calibration set from it.
Currently I am ambiguously thinking is just using certain primary key value, doing like this :
original dataset - calibration dataset = > validation dataset,
But I couldn't figure out how to do like above yet.
I think I can make the dataset if I take some steps, but what I want is not to make many lines to code but making it with just intuitive and simple code(or if there is a function which is exactly fit for this purpose it'll be better). So I would be very appreciate if you help me.

I would suggest using dplyr and filter with an anti_join

Start with the total data. Filter by whatever determines the training or test.
then anti-join the training data from the total data leaving the

To do it randomly, you can use something like

library(dplyr) <- data.frame(x=c(1:5), y=c(6:10)) <- mutate(, Pick.Me = sample(c(0,1), size=dim([[1]], replace=TRUE,  prob=c(0.7,0.3))) <- filter(, Pick.Me==0) <- anti_join(,

Well...thank you very much. But I cannot use this exact code because I already extracted a dataset for training by using "strata" function, because I wanted to eliminate bias of sampling from the data( which can be categorized.
In this situation, how can I apply or combine your code with mine...?

This is my code for random stratafied sampling : <- strata(data =, stratanames = "exposure", size = c(100, 100), method = "srswr") <- getdata(,

"setdiff" seems to be a similar function with my objective, considering it performs complement calculation but this is for vector, not a dataframe I guess...

If is what you say it is, then the last line that I wrote above using anti_join is what is needed.

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I'll try with it.
Thank you again.

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