How to test whether a vector contains repetitive elements?

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how do you test whether a vector contains repetitive elements in R?

I think I found the answer. Use duplicated() function:

a=c(3,5,7,2,7,9)
b=1:10
any(duplicated(a)) #True
any(duplicated(b)) #False

Check if there are repeated elements in a vector, Check if there are repeated elements in a vector. Learn more about check, same, repeted, vector, elements. The java.util.vector.contains() method is used to check whether a specific element is present in the Vector or not. So basically it is used to check if a vector contains any particular element or not. Syntax: Vector.contains(Object element) Parameters: This method takes a mandatory parameter element which is of the type of vector. This is the element that needs to be tested if it is present in the vector or not.

Also try rle(x) to find the lengths of runs of identical values in x.

Check if a given array contains duplicate elements within k distance , Recommended Posts: Check if it is possible to sort an array with conditional swapping of elements at distance K � Remove duplicate elements in� 1. std::count. The simplest solution is to count number of elements in the vector having specified value. If count is non-zero, we have found our element. This can be easily done using std::count function. #include <iostream> #include <vector> #include <algorithm> int main () { std::vector<int> v = { 4, 7, 5, 2, 6, 9 }; int key = 6; if (std::count (v.begin (), v.end (), key)) std::cout << "Element found"; else std::cout << "Element not found"; return 0; }

If you're looking for consecutive repeats you can use diff.

a <- 1:10
b <- c(1:5, 5, 7, 8, 9, 10)
diff(a)
diff(b)

Or anywhere in the vector:

length(a) == length(unique(a))
length(b) == length(unique(b))

Determine Duplicate Elements, Description. duplicated() determines which elements of a vector or data frame are duplicates of elements with smaller subscripts, and returns a logical vector indicating which elements (rows) are duplicates. Usage. duplicated(x, incomparables = FALSE, ) Arguments. x. Details. Value. Warning. References. See Also. R Programming: Vector Exercise-10 with Solution. Write a R program to test whether a given vector contains a specified element. Sample Solution:. R Programming Code :

check this:

> all(diff(c(1,2,3)))
[1] TRUE
Warning message:
In all(diff(c(1, 2, 3))) : coercing argument of type 'double' to logical
> all(diff(c(1,2,2,3)))
[1] FALSE
Warning message:
In all(diff(sort(c(1, 2, 4, 2, 3)))) : coercing argument of type 'double' to logical

You can add some casting to get rid of warnings.

vsetequal: Check whether two vectors contain exactly the same , Unlike the base::setequal function, if the vectors have repeated elements in common, the count of these elements is checked. As a result, vectors of different� test if vector contains elements of another vector (disregarding the position) Hi, I have the following problem: I have two vectors: i <- c('a','c','g','h','b','d','f','k','l','e','i') j <- c('a', 'b', 'c') now I would like to generate a vector with the length of i that has zeros where i[x] != any element of j and 1 where i[x] == any element of j.

As mentioned in the comment section by Hadley:

anyDuplicated will be a bit faster for very long vectors - it can terminate when it finds the first duplicate.

Example

a=c(3,5,7,2,7,9)
b=1:10
anyDuplicated(b) != 0L # TRUE
anyDuplicated(b) != 0L # FALSE

Benchmark with 1 million observations:

set.seed(2011)
x <- sample(1e7, size = 1e6, replace = TRUE)
bench::mark(
  ZNN = any(duplicated(x)),
  RL = length(x) != length(unique(x)),
  BUA = !all(diff(sort(x))),
  AD = anyDuplicated(x) != 0L
)

# A tibble: 4 x 13
  expression      min   median `itr/sec` mem_alloc `gc/sec` n_itr  n_gc total_time result    memory            time     gc                
  <bch:expr> <bch:tm> <bch:tm>     <dbl> <bch:byt>    <dbl> <int> <dbl>   <bch:tm> <list>    <list>            <list>   <list>            
1 ZNN         64.62ms  70.04ms      11.5    11.8MB     0        8     0      693ms <lgl [1]> <df[,3] [2 x 3]>  <bch:tm> <tibble [8 x 3]>  
2 RL          66.95ms  70.67ms      12.5    15.4MB     0        7     0      561ms <lgl [1]> <df[,3] [3 x 3]>  <bch:tm> <tibble [7 x 3]>  
3 BUA         84.66ms  87.79ms      10.6      42MB     3.54     3     1      283ms <lgl [1]> <df[,3] [11 x 3]> <bch:tm> <tibble [4 x 3]>  
4 AD           2.45ms   2.87ms     314.        8MB     5.98   105     2      335ms <lgl [1]> <df[,3] [1 x 3]>  <bch:tm> <tibble [107 x 3]>

Benchmark with 100 observations

set.seed(2011)
x <- sample(1e7, size = 100, replace = TRUE)

bench::mark(
  ZNN = any(duplicated(x)),
  RL = length(x) != length(unique(x)),
  BUA = !all(diff(sort(x))),
  AD = anyDuplicated(x) != 0L
)

# A tibble: 4 x 13
  expression      min   median `itr/sec` mem_alloc `gc/sec` n_itr  n_gc total_time result    memory            time     gc                   
  <bch:expr> <bch:tm> <bch:tm>     <dbl> <bch:byt>    <dbl> <int> <dbl>   <bch:tm> <list>    <list>            <list>   <list>               
1 ZNN          7.14us   8.93us    60429.    1.48KB     6.04  9999     1    165.5ms <lgl [1]> <df[,3] [2 x 3]>  <bch:tm> <tibble [10,000 x 3]>
2 RL           8.03us   9.37us    83754.    1.92KB     0    10000     0    119.4ms <lgl [1]> <df[,3] [3 x 3]>  <bch:tm> <tibble [10,000 x 3]>
3 BUA         54.89us  61.58us     8317.    4.83KB     6.74  3701     3      445ms <lgl [1]> <df[,3] [11 x 3]> <bch:tm> <tibble [3,704 x 3]> 
4 AD            5.8us   6.69us   123838.    1.05KB     0    10000     0     80.8ms <lgl [1]> <df[,3] [1 x 3]>  <bch:tm> <tibble [10,000 x 3]>

Test if Vector Contains Certain Element in R (Example), How to check whether an element exists within a vector - Step by step example code in R - Programming syntax explained - Test for value in R language. The contains () method of Java Vector class is used to check the vector which is in use contains the specified element or not. It returns true if this vector contains the specified element, otherwise returns false.

Check for duplicates in a list using Set & by comparing sizes. To check if a list contains any duplicate element follow the following steps, Add the contents of list in a set. As set contains only unique elements, so no duplicates will be added to the set. Compare the size of set and list.

a = randi ( [1 5],200,1); Find the unique elements in the vector. Return the index vectors ia and ic. [C,ia,ic] = unique (a); Count the number of times each element in C appears in a. Specify ic as the first input to accumarray and 1 as the second input so that the function counts repeated subscripts in ic.

Create vector V that contains the symbolic equation and inequalities as its elements: syms x V = [x ~= x + 1, abs(x) >= 0, x == x]; Use all to test whether all of them are valid for all values of x :

Comments
  • anyDuplicated will be a bit faster for very long vectors - it can terminate when it finds the first duplicate.
  • I think the use of anyDuplicated() is a bit strange. It returns an integer of length one with value the 1-based index of the first duplicate, or 0. So, the equivalent of any(duplicated(a)) should be anyDuplicated(a)==0. It is more intuitive if anyDuplicated() returns TRUE if there is any duplicate.