acos（1）返回NaN的一些值，而不是其他值 [英] acos(1) returns NaN for some values, not others

问题描述

我有一个经度和纬度值的列表，我试图找到他们之间的距离。使用一个标准的大圆方法，我需要找到：

$ $ $ $ $ $ $ $ c $（ac（sin（lat1）* sin（lat2）+ cos lat1）* cos（lat2）* cos（long2-long1））

地球，在我使用的单位。只要我们采取的值在[-1,1]的范围内，这是有效的。如果他们甚至略微超出这个范围，即使差异是由于四舍五入，也会返回 NaN 。

<我有的问题是，有时，当两个纬度/长度值是相同的，这给了我一个 NaN 错误。并不总是，即使是同一对数字，但总是在列表中相同的数字。例如，我有一个人停在沙漠的路上：

 时间| lat | long 
 1： 00 PM|35.08646|-117.5023 
 1:01 PM|35.08646|-117.5023 
 1:02 PM|35.08646|-117.5023 
 1:03 PM|35.08646|-117.5023 
 1:04 PM| 35.08646 | -117.5023 
  

当我计算连续点之间的距离时，第三个值，将始终是 NaN ，即使其他人不是。这似乎是R舍入一个奇怪的错误。

解决方案

如果没有看到您的数据， $ c> dput ），但这大多是 FAQ 7.31 。

>（x1 <-1）
## [1] 1
（x2 <-1 + 1e-16）
## [1] 1
（x3 < ; - 1 + 1e-8）
## [1] 1
acos（x1）
## [1] 0
acos（x2）
## [1] 0
acos（x3）
## [1] NaN

就是说，即使你的值是如此相似以至于它们的打印表示是相同的，它们可能仍然不同：一些将在 .Machine $ double.eps 中， '...

确保输入值受[-1,1]限制的一种方法是使用 pmax 和 pmin ： acos（pmin（pmax（x，-1.0），1.0）） p>

I have a list of latitude and longitude values, and I'm trying to find the distance between them. Using a standard great circle method, I need to find:

acos(sin(lat1)*sin(lat2) + cos(lat1)*cos(lat2) * cos(long2-long1))

And multiply this by the radius of earth, in the units I am using. This is valid as long as the values we take the acos of are in the range [-1,1]. If they are even slightly outside of this range, it will return NaN, even if the difference is due to rounding.

The issue I have is that sometimes, when two lat/long values are identical, this gives me an NaN error. Not always, even for the same pair of numbers, but always the same ones in a list. For instance, I have a person stopped on a road in the desert:

Time  |lat     |long
1:00PM|35.08646|-117.5023
1:01PM|35.08646|-117.5023
1:02PM|35.08646|-117.5023
1:03PM|35.08646|-117.5023
1:04PM|35.08646|-117.5023

When I calculate the distance between the consecutive points, the third value, for instance, will always be NaN, even though the others are not. This seems to be a weird bug with R rounding.

解决方案

Can't tell exactly without seeing your data (try dput), but this is mostly likely a consequence of FAQ 7.31.

(x1 <- 1)
## [1] 1
(x2 <- 1+1e-16)
## [1] 1
(x3 <- 1+1e-8)
## [1] 1
acos(x1)
## [1] 0
acos(x2)
## [1] 0
acos(x3)
## [1] NaN

That is, even if your values are so similar that their printed representations are the same, they may still differ: some will be within .Machine$double.eps and others won't ...

One way to make sure the input values are bounded by [-1,1] is to use pmax and pmin: acos(pmin(pmax(x,-1.0),1.0))

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