小数点-语言R中的概率值为0 [英] Decimal points - Probability value of 0 in Language R

问题描述

如何处理R中的p值？

我期望p值非常低，例如：

I am expecting very low p values like:

1.00E-80

我需要-log10

-log10(1.00E-80)

-log10（0）也是Inf，但在舍入意义上也是Inf。

-log10(0) is Inf, but Inf at sense of rounding too.

但似乎在1.00E-308之后，R产生0。

But is seems that after 1.00E-308, R yields 0.

1/10^308  
[1] 1e-308

 1/10^309 
[1] 0

p值显示的准确性与 lm 函数具有相同的截止点1e-308，或者它的设计目的是我们需要一个截止点，而我需要考虑一个不同的截止点-如1e-100（例如），将0替换为< 1e-100。

Is the accuracy of p-value display with lm function the same as the cutoff point, 1e-308, or it is just designed such that we need a cutoff point and I need to consider a different cutoff point - such as 1e-100 (for example) to replace 0 with <1e-100.

推荐答案

有多种可能的答案-其中最有用的取决于上下文：

There are a variety of possible answers -- which one is most useful depends on the context:

• 在通常情况下，R确实无法存储比接近零的浮点值。Machine $ double.xmin ，具体取决于平台，但通常（如您所见）在 1e-308 的顺序上。如果您确实需要使用这么小的数字，并且找不到直接在对数刻度上工作的方法，则需要在Stack Overflow或R Wiki中搜索用于处理任意/扩展精度值的方法（但您可能应该尝试在对数刻度上工作-麻烦就少了


• 在许多情况下，R实际上会在内部（自然）对数刻度上计算p值，如果需要的话，可以返回日志值，而不是在给出答案之前对它们取幂。例如， dnorm（-100，log = TRUE）给出-5000.919。您可以通过除以 log（10）直接转换为log10刻度（不进行幂运算，然后使用 log10 ）： dnorm（-100，log = TRUE）/ log（10） =-2171，该值太小而无法以浮点数表示。对于 p *** （累积分布函数）函数，请使用 log.p = TRUE 而不是 log = TRUE 。 （这一点在很大程度上取决于您的特定上下文。即使您没有使用内置的R函数，您也可以找到一种以对数刻度提取结果的方法。）


• 在某些情况下，即使已知更精确的值，R也会显示p值结果为 <2.2e-16 ：（t1 <- t.test（rnorm（10,100），rnorm（10,80）））

• R is indeed incapable under ordinary circumstances of storing floating-point values closer to zero than .Machine$double.xmin, which varies by platform but is typically (as you discovered) on the order of 1e-308. If you really need to work with numbers this small and can't find a way to work on the log scale directly, you need to search Stack Overflow or the R wiki for methods for dealing with arbitrary/extended precision values (but you probably should try to work on the log scale -- it will be much less of a hassle)
• in many circumstances R actually computes p values on the (natural) log scale internally, and can if requested return the log values rather than exponentiating them before giving the answer. For example, dnorm(-100,log=TRUE) gives -5000.919. You can convert directly to the log10 scale (without exponentiating and then using log10) by dividing by log(10): dnorm(-100,log=TRUE)/log(10)=-2171, which would be too small to represent in floating point. For the p*** (cumulative distribution function) functions, use log.p=TRUE rather than log=TRUE. (This particular point depends heavily on your particular context. Even if you are not using built-in R functions you may be able to find a way to extract results on the log scale.)
• in some cases R presents p-value results as being <2.2e-16 even when a more precise value is known: (t1 <- t.test(rnorm(10,100),rnorm(10,80)))

打印

....
t = 56.2902, df = 17.904, p-value < 2.2e-16

但是您仍然可以从结果中提取精确的p值

but you can still extract the precise p-value from the result

> t1$p.value
[1] 1.856174e-18

（在许多情况下，这是行为由 format.pval（）函数控制）

(in many cases this behaviour is controlled by the format.pval() function)

如何使用 lm ：

d <- data.frame(x=rep(1:5,each=10))
set.seed(101)
d$y <- rnorm(50,mean=d$x,sd=0.0001)
lm1 <- lm(y~x,data=d)

摘要（lm1）将斜率的p值打印为< 2.2e-16 ，但是如果我们使用 coef（summary（lm1 ））（不使用p值格式），我们可以看到该值为9.690173e-203。

summary(lm1) prints the p-value of the slope as <2.2e-16, but if we use coef(summary(lm1)) (which does not use the p-value formatting), we can see that the value is 9.690173e-203.

A更多极端情况：

set.seed(101); d$y <- rnorm(50,mean=d$x,sd=1e-7)
lm2 <- lm(y~x,data=d)
coef(summary(lm2))

表示p值实际上已经降为零。但是，我们仍然可以获得对数刻度的答案：

shows that the p-value has actually underflowed to zero. However, we can still get an answer on the log scale:

tval <- coef(summary(lm2))["x","t value"]
2*pt(abs(tval),df=48,lower.tail=FALSE,log.p=TRUE)/log(10)

给出-692.62（您可以在前面的示例中检查此方法，其中p值不会溢出，并且看到相同的值答案显示在摘要中。）

gives -692.62 (you can check this approach with the previous example where the p-value doesn't overflow and see that you get the same answer as printed in the summary).

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