![]() ![]() if each vector element of the first vector is equal to the corresponding vector element in the second vector. In Example 1, I’ll illustrate how to test whether our two vectors are exactly the same, i.e. Display the second element in the vector. Example 1: Check If Two Vectors are Exactly the Same Using identical () Function. create a vector vec <- c(10, 20, 30, 40) remove the first element vec <- vec-1 display the vector print(vec) Output: 1 20 30 40. In general, avoid switching from logical to integer subsetting unless you want, for example, the first or last TRUE value. It allows you to mix R code, notes and outputs with nice formatting and is a nice way to record and present R. As you can see, the previous R syntax returned every 3rd letter of the alphabet. For this task, we need to use a combination of the seq and length functions. You’ll get no values, instead of all values. In this example, we’ll extract every 3rd element of the LETTERS vector. Which(y) will be integer(0) and -integer(0) is still integer(0), so By combining vector comparisons, logical operators, and vector indexing, you can perform powerful selections with very little R code. There was no match for the first element of our. The match function returns NA when no match is found. As you can see, the match function returned a vector with two NA’s and three values. ![]() X is not equivalent to x: if y is all FALSE, vec <- c (4, 5, 1, 3, 7) Now, we can identify matches for this whole vector as follows: match ( vec, tab) Apply match with input vector NA 2 5 NA 3. To use which() for this side-effect, but I don’t recommend it: nothingĪbout the name “which” implies the removal of missing values. Values with NA while which() simply drops these values. When the logical vector contains NA, logical subsetting replaces these In more general cases, there are two important differences. For example, let’s remove elements 3 and 5 from a vector of integers 1 to 5. Here the which() achieves nothing: it switches from logical to integer subsetting but the result is exactly the same. When first learning subsetting, a common mistake is to use x instead of x. ( x1 FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE ( x2 2 4 6 8 10 ( y1 FALSE FALSE FALSE FALSE TRUE FALSE FALSE FALSE FALSE TRUE ( y2 5 10 # X & Y intersect(x, y) x1 & y1 #> FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE TRUE intersect ( x2, y2 ) #> 10 # X | Y union(x, y) x1 | y1 #> FALSE TRUE FALSE TRUE TRUE TRUE FALSE TRUE FALSE TRUE union ( x2, y2 ) #> 2 4 6 8 10 5 # X & !Y setdiff(x, y) x1 & ! y1 #> FALSE TRUE FALSE TRUE FALSE TRUE FALSE TRUE FALSE FALSE setdiff ( x2, y2 ) #> 2 4 6 8 # xor(X, Y) setdiff(union(x, y), intersect(x, y)) xor ( x1, y1 ) #> FALSE TRUE FALSE TRUE TRUE TRUE FALSE TRUE FALSE FALSE setdiff ( union ( x2, y2 ), intersect ( x2, y2 ) ) #> 2 4 6 8 5 ![]()
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