测试单个数值向量的所有元素之间的相等性 [英] Test for equality among all elements of a single numeric vector

问题描述

我正在尝试测试向量的所有元素是否彼此相等.我想出的解决方案似乎有些round回，都涉及检查 length().

I'm trying to test whether all elements of a vector are equal to one another. The solutions I have come up with seem somewhat roundabout, both involving checking length().

x <- c(1, 2, 3, 4, 5, 6, 1)  # FALSE
y <- rep(2, times = 7)       # TRUE

使用 unique():

length(unique(x)) == 1
length(unique(y)) == 1

使用 rle():

length(rle(x)$values) == 1
length(rle(y)$values) == 1

一个让我包括一个用于评估要素之间平等"的容差值的解决方案将是避免

A solution that would let me include a tolerance value for assessing 'equality' among elements would be ideal to avoid FAQ 7.31 issues.

我是否完全忽略了用于测试类型的内置函数? identical()和 all.equal()比较两个R对象，因此它们在这里不起作用.

Is there a built-in function for type of test that I have completely overlooked? identical() and all.equal() compare two R objects, so they won't work here.

编辑1

以下是一些基准测试结果.使用代码:

Here are some benchmarking results. Using the code:

library(rbenchmark)

John <- function() all( abs(x - mean(x)) < .Machine$double.eps ^ 0.5 )
DWin <- function() {diff(range(x)) < .Machine$double.eps ^ 0.5}
zero_range <- function() {
  if (length(x) == 1) return(TRUE)
  x <- range(x) / mean(x)
  isTRUE(all.equal(x[1], x[2], tolerance = .Machine$double.eps ^ 0.5))
}

x <- runif(500000);

benchmark(John(), DWin(), zero_range(),
  columns=c("test", "replications", "elapsed", "relative"),
  order="relative", replications = 10000)

结果:

          test replications elapsed relative
2       DWin()        10000 109.415 1.000000
3 zero_range()        10000 126.912 1.159914
1       John()        10000 208.463 1.905251

所以看起来像 diff(range(x))<.Machine $ double.eps ^ 0.5 是最快的.

So it looks like diff(range(x)) < .Machine$double.eps ^ 0.5 is fastest.

推荐答案

我使用此方法，该方法将均值除以最小值和最大值:

I use this method, which compares the min and the max, after dividing by the mean:

# Determine if range of vector is FP 0.
zero_range <- function(x, tol = .Machine$double.eps ^ 0.5) {
  if (length(x) == 1) return(TRUE)
  x <- range(x) / mean(x)
  isTRUE(all.equal(x[1], x[2], tolerance = tol))
}

如果您更认真地使用此功能，则可能需要在计算范围和均值之前删除缺失的值.

If you were using this more seriously, you'd probably want to remove missing values before computing the range and mean.

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