问题描述
我试图在下面的向量中找到两个值，它们接近10.预期值为10.12099196和10.63054170.您的输入将不胜感激.
I tried to find two values in the following vector, which are close to 10. The expected value is 10.12099196 and 10.63054170. Your inputs would be appreciated.
[1] 0.98799517 1.09055728 1.20383713 1.32927166 1.46857509 1.62380423 1.79743107 1.99241551 2.21226576 2.46106916 2.74346924 3.06455219 3.42958354 3.84350238 4.31005838
[16] 4.83051356 5.40199462 6.01590035 6.65715769 7.30532785 7.93823621 8.53773241 9.09570538 9.61755743 10.12099196 10.63018180 11.16783243 11.74870531 12.37719092 13.04922392
[31] 13.75661322 14.49087793 15.24414627 16.00601247 16.75709565 17.46236358 18.06882072 18.51050094 18.71908344 18.63563523 18.22123225 17.46709279 16.40246292 15.09417699 13.63404124
[46] 12.11854915 10.63054170 9.22947285 7.95056000 6.80923943 5.80717982 4.93764782 4.18947450 3.54966795 3.00499094 2.54283599 2.15165780 1.82114213 1.54222565 1.30703661
[61] 1.10879707 0.94170986 0.80084308 0.68201911 0.58171175 0.49695298 0.42525021 0.36451350 0.31299262 0.26922281 0.23197860 0.20023468 0.17313291 0.14995459 0.13009730
[76] 0.11305559 0.09840485 0.08578789 0.07490387 0.06549894 0.05735864
推荐答案
另一种选择是允许用户控制公差"以设置紧密度"，这可以通过使用一个简单的函数来完成:
Another alternative could be allowing the user to control for the "tolerance" in order to set what "closeness" is, this can be done by using a simple function:
close <- function(x, value, tol=NULL){
if(!is.null(tol)){
x[abs(x-10) <= tol]
} else {
x[order(abs(x-10))]
}
}
其中x是值的向量，value是紧密度的比较值，而tol是逻辑的，如果它是NULL，则它将所有按"closeness"排序的"close"值返回给value，否则它仅返回满足tol中给出的条件的值.
Where x is a vector of values, value is the value of comparison for closeness, and tol is logical, if it's NULL it returns all the "close" values ordered by "closeness" to value, otherwise it returns just the values meeting the condition given in tol.
> close(x, value=10, tol=.7)
[1] 9.617557 10.120992 10.630182 10.630542
> close(x, value=10)
[1] 10.12099196 9.61755743 10.63018180 10.63054170 9.22947285 9.09570538 11.16783243
[8] 8.53773241 11.74870531 7.95056000 7.93823621 12.11854915 12.37719092 7.30532785
[15] 13.04922392 6.80923943 6.65715769 13.63404124 13.75661322 6.01590035 5.80717982
[22] 14.49087793 5.40199462 4.93764782 15.09417699 4.83051356 15.24414627 4.31005838
[29] 4.18947450 16.00601247 3.84350238 16.40246292 3.54966795 3.42958354 16.75709565
[36] 3.06455219 3.00499094 2.74346924 2.54283599 17.46236358 17.46709279 2.46106916
[43] 2.21226576 2.15165780 1.99241551 18.06882072 1.82114213 1.79743107 18.22123225
[50] 1.62380423 1.54222565 18.51050094 1.46857509 18.63563523 1.32927166 1.30703661
[57] 18.71908344 1.20383713 1.10879707 1.09055728 0.98799517 0.94170986 0.80084308
[64] 0.68201911 0.58171175 0.49695298 0.42525021 0.36451350 0.31299262 0.26922281
[71] 0.23197860 0.20023468 0.17313291 0.14995459 0.13009730 0.11305559 0.09840485
[78] 0.08578789 0.07490387 0.06549894 0.05735864
在第一个示例中，我将紧密度"定义为在value与x中的每个元素之间最多相差0.7.在第二个示例中，函数close返回一个值向量，其中第一个最接近value中给出的值，而最后一个是距value最远的值.
In the first example I defined "closeness" to be at most a difference of 0.7 between value and each elements in x. In the second example the function close returns a vector of values where the firsts are the closest to the value given in value and the lasts are the farest values from value.
由于我的解决方案没有像@Arun所指出的那样提供一种简单的(实用的)方法来找到tol，因此，找到最接近的值的一种方法是设置tol=NULL并要求精确的闭合值数量，如:
Since my solution does not provide an easy (practical) way to find tol as @Arun pointed out, one way to find the closest values would be seting tol=NULL and asking for the exact number of close values as in:
> close(x, value=10)[1:3]
[1] 10.120992 9.617557 10.630182
这显示了x中最接近10的三个值.
This shows the three values in x closest to 10.
这篇关于r查找向量中的两个最接近的值的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！