2013-02-11 18 views
3

我不知道在R中是否可以使用大於.Machine$double.xmax~1.79e308)的整數。 Rmpfrgmp庫中的R您可以分配任何大小的值,直到系統上RAM的限制?我認爲這大於.Machine$double.xmax,但顯然不是。在gmp和機器限制下使用大整數

> require(gmp) 
> as.bigz(.Machine$double.xmax) 
Big Integer ('bigz') : 
[1] 179769313486231570814527423731704356798070567525844996598917476803157260780028538760589558632766878171540458953514382464234321326889464182768467546703537516986049910576551282076245490090389328944075868508455133942304583236903222948165808559332123348274797826204144723168738177180919299881250404026184124858368 
> as.bigz(1e309) 
Big Integer ('bigz') : 
[1] NA 
> 

是否有可能有人解釋爲什麼使用電腦的64位內存尋址不能存儲值大於1.79e308更大?對不起 - 我沒有計算機科學背景,但我正在努力學習。

謝謝。

+1

您可能會發現閱讀有用的文檔,特別是'?as.bigz'的Note部分。 – joran 2013-02-11 18:22:43

+0

謝謝喬蘭。我錯過了最後一行。令人煩惱的是,我現在不能使用科學記數法! – 2013-02-11 18:42:31

+2

是的,但你可以做'as.bigz(10)^ 309'。事實上,你可以這樣做:''%e%'< - function(x,y)as.bigz(x)* 10^as.bigz(y); 1%e%309' – 2013-02-11 19:18:15

回答

3

Rmpfr可以做到使用mpfr_set_str字符串轉換...

val <- mpfr("1e309") 

## 1 'mpfr' number of precision 17 bits 
## [1] 9.999997e308 

# set a precision (assume base 10)... 
est_prec <- function(e) floor(e/log10(2)) + 1 

val <- mpfr("1e309", est_prec(309)) 

## 1 'mpfr' number of precision 1027 bits 
## [1]1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 

.mpfr2bigz(val) 

## Big Integer ('bigz') : 
## [1] 1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 

# extract exponent from a scientific notation string 
get_exp <- function(sci) as.numeric(gsub("^.*e",'', sci)) 

# Put it together 
sci2bigz <- function(str) { 
    .mpfr2bigz(mpfr(str, est_prec(get_exp(str)))) 
} 

val <- sci2bigz(paste0(format(Const("pi", 1027)), "e309")) 

identical(val, .mpfr2bigz(Const("pi",1027)*mpfr(10,1027)^309)) 

## [1] TRUE 

## Big Integer ('bigz') : 
## [1] 3141592653589793238462643383279502884197169399375105820974944592307816406286208998628034825342117067982148086513282306647093844609550582231725359408128481117450284102701938521105559644622948954930381964428810975665933446128475648233786783165271201909145648566923460348610454326648213393607260249141273724587004 

至於爲什麼上存儲的數大於.Machine$double.xmax,上浮點編碼在IEEE規範的文檔時,R常見問題和維基百科進入所有的行話,但我發現它有助於只定義條款(使用?'.Machine')...

double.xmax(最大標準化浮點數)=
(1 - double.neg.eps) * double.base^double.max.exp其中

  1. double.neg.eps(一個小的正浮點數x使得1 - X = 1)!= double.base^double.neg.ulp.digits其中
    • double.neg.ulp.digits =最大負整數,使得1 - double.base^i != 1
  2. double.max.exp = double.base溢出的最小正功率和
  3. double.base(浮點數表示的基數)= 2(二進制數)。

考慮有限浮點數可以與另一個區別開來嗎?在IEEE規範告訴我們,對於一個binary64數11位都將用於指數,所以我們的2^(11-1)-1=1023最大的指數,但我們希望的是溢出最大指數所以double.max.exp是1024

# Maximum number of representations 
# double.base^double.max.exp 
base <- mpfr(2, 2048) 
max.exp <- mpfr(1024, 2048) 

# This is where the big part of the 1.79... comes from 
base^max.exp 

## 1 'mpfr' number of precision 2048 bits 
## [1] 179769313486231590772930519078902473361797697894230657273430081157732675805500963132708477322407536021120113879871393357658789768814416622492847430639474124377767893424865485276302219601246094119453082952085005768838150682342462881473913110540827237163350510684586298239947245938479716304835356329624224137216 

# Smallest definitive unit. 
# Find the largest negative integer... 
neg.ulp.digits <- -64; while((1 - 2^neg.ulp.digits) == 1) 
    neg.ulp.digits <<- neg.ulp.digits + 1 

neg.ulp.digits 

## [1] -53 

# It makes a real small number... 
neg.eps <- base^neg.ulp.digits 

neg.eps 

## 1 'mpfr' number of precision 2048 bits 
## [1] 1.11022302462515654042363166809082031250000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000e-16 

# Largest difinitive floating point number less than 1 
# times the number of representations 
xmax <- (1-neg.eps) * base^max.exp 

xmax 

## 1 'mpfr' number of precision 2048 bits 
## [1] 179769313486231570814527423731704356798070567525844996598917476803157260780028538760589558632766878171540458953514382464234321326889464182768467546703537516986049910576551282076245490090389328944075868508455133942304583236903222948165808559332123348274797826204144723168738177180919299881250404026184124858368 

identical(asNumeric(xmax), .Machine$double.xmax) 

## [1] TRUE