## How to identify what are the exact elements of a vector?

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I have a vector x of the form:

x=c(601, 602, 603, 604, 605, 606, 607, 608, 609, 610, 611, 612, 613, 614,
615, 616, 617, 618, 619, 620, 621, 622, 623, 624, 625, 626, 627, 628, 629, 630,
631, 632, 633, 634, 635, 636, 637, 638, 639, 640, 641, 642, 643, 644, 645, 646,
647, 648, 649, 650)


If length (x) == 50: I would like to test (TRUE or FALSE) if x is exactly equal to one of the intervals 1:50 or 51:100 or 101:150 or 151:200 .... or 951:1000.

Or if length (x)> 50, I would like to test (TRUE or FALSE) if x is exactly equal to the union of intervals composed by ((1:50 U 51: 100) or (1:50 U 101: 150) or (51 : 100 U 151: 200) ....).

My attempt:

all(
x == c(1:50) |
x == c(51:100) |
x == c(101:150) |
x == c(151:200) |
x == c(201:250) |
x == c(251:300) |
x == c(301:350) |
x == c(351:400) |
x == c(401:450) |
x == c(451:500) |
x == c(501:550) |
x == c(551:600) |
x == c(601:650) |
x == c(651:700) |
x == c(701:750) |
x == c(751:800) |
x == c(801:850) |
x == c(851:900) |
x == c(901:950) |
x == c(951:1000)
)


I would like to optimize this code.

PS: I'm not trying to have a frequency table of the x elements and intervals like this question. I want to know if x corresponds exactly to one or the union of those intervals.

a data.table solution:

test <- function(u){
ifelse(all(as.data.table(u)[,
.N,
by = cut(u,
breaks = seq(0, 1000, 50))][, unique(N)] == 50),
TRUE,
FALSE)
}

##### Tests:
x <- 1:50 # TRUE
y <- 2:51 # FALSE
z <- 1:100 # TRUE
w <- 2:101 # FALSE

test(x)
> TRUE

test(y)
> FALSE

test(z)
> TRUE

test(w)
> FALSE


Quantitative Methods for Finance and Investments, In a financial sense, where elements within a vector represent payoffs of a given For example, how does one determine the exact element to place in the third  If we want to identify all elements of our vectors, which are existent in each of the vectors, we can use a combination of the Reduce (), intersect (), and list () functions. Have a look at the following R code: Reduce ( intersect, list ( x1, x2, x3)) # Identify common elements # "A" "D".

You can use cut, i.e.

unique(cut(x, breaks = seq(0, 1000, by = 50)))
# (600,650]


If you want a boolean If x is included in one those intervals, then you can do,

unique(cut(x, breaks = seq(0, 1000, by = 50))) != ''
# TRUE

#or If you only want to be in 1 group, then as suggested by Ronak,
length(unique(cut(x, breaks = seq(0, 1000, by = 50)))) == 1
# TRUE


Structural Analysis: A Unified Classical and Matrix Approach, Exact element matrices can be generated only for bars. Assemblage of stiffness matrices and load vectors of individual elements, Problems 17.1 Determine any element S* for the bar of Example 17.5, using the shape functions in Fig. 17.4. The typical element of $${\bf A}$$ is $$a_{ij}$$, denoting the element of row $$i$$ and column $$j$$. Matrix addition and subtraction Matrices are added and subtracted on an element-by-element basis.

Another option:

test <- function(x) length(x)%%50==0 & x[length(x)]%%50==0 & all(diff(x)==1)


length(x)%%50==0 ensures that the vector is of length that are multiples of 50.

x[length(x)]%%50==0 checks that the last element is divisible by 50.

all(diff(x)==1) checks that vector is a sequence in steps of 1. (e.g. x <- rep(50,50) will fail this part)

check:

> test(1:50)
# TRUE
> test(2:51)
# FALSE
> test(1:100)
# TRUE
> test(2:101)
# FALSE


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Find Array Elements That Meet a Condition, This example shows how to filter the elements of an array by applying conditions to of the array elements that meet a condition rather than their actual values. MATLAB ® treats the array as a single column vector with each column appended to the bottom of the previous column. Thus, linear indexing numbers the elements in the columns from top to bottom, left to right. For example, consider a 3-by-3 matrix. You can reference the A(2,2) element with A(5), and the A(2,3) element with A(8).

Advances in Information Retrieval: 29th European Conference on IR , The two less similar vectors of A define the area(a1,a2), which is the So we can decide whether there is an exact name in two Finding the best lower name means finding a query q that ranks first the biggest number of elements of A, i.e.  A colleague came to my office the other day with an interesting question: Is there a way in R to find the closest number to X in a list? I knowing full well the power the power of R, I naturally said that surely there is such a function, but I have never used it.

• you need to library(data.table) before defining the function