Dear friends - I'm calculating buffer capacities by different methods and need very high precision and package Rmpfr is working beautifully. However, I have not been able to find out how to keep precision when finding correlations. library(Rmpfr) KA <- mpfr(10^-4.6, 128) x <- rnorm(100)*KA y <- rnorm(100)*x cor(x,y) # "x" must be numeric cor(as.numeric(x),as.numeric(y))# 0.2918954 In my concrete application I get cor = 1 for cor(as.numeric(dff$BB),as.numeric(BBVS)) even though I have str(summary((dff$BB)-(BBVS))) Class 'summaryMpfr' [package "Rmpfr"] of length 6 and precision 128 4.61351010833e-8 7.33418976521e-7 1.31009046563e-5 3.76407022709e-5 5.72386764888e-5 ... I am on windows 10 R version 3.6.1 Best wishes Troels Ring, Aalborg, Denmark This email has been scanned by BullGuard antivirus protection. For more info visit www.bullguard.com <http://www.bullguard.com/tracking.aspx?affiliate=bullguard&buyaffiliate=smt p&url=/> [[alternative HTML version deleted]]
Hello,
Why not write a function COR? Not one as general purpose as stats::cor
but a simple one, to compute the sample Pearson correlation only.
library(Rmpfr)
COR <- function(x, y){
precBits <- getPrec(x)[1]
n <- mpfr(length(x), precBits = precBits)
x.bar <- mean(x)
y.bar <- mean(y)
numer <- sum(x*y) - n*x.bar*y.bar
denom <- sqrt(sum(x*x) - n*x.bar*x.bar) * sqrt(sum(y*y) - n*y.bar*y.bar)
numer/denom
}
set.seed(2020)
KA <- mpfr(10^-4.6, 128)
x <- rnorm(100)*KA
y <- rnorm(100)*x
cor(as.numeric(x), as.numeric(y)) # -0.1874986
#[1] -0.1874986
COR(x, y)
#1 'mpfr' number of precision 128 bits
#[1] -0.1874985950531874160800643775644747505073
Hope this helps,
Rui Barradas
?s 10:42 de 12/07/20, tring at gvdnet.dk escreveu:> Dear friends - I'm calculating buffer capacities by different methods
and
> need very high precision and package Rmpfr is working beautifully. However,
> I have not been able to find out how to keep precision when finding
> correlations.
>
> library(Rmpfr)
>
> KA <- mpfr(10^-4.6, 128)
>
> x <- rnorm(100)*KA
>
> y <- rnorm(100)*x
>
> cor(x,y) # "x" must be numeric
>
> cor(as.numeric(x),as.numeric(y))# 0.2918954
>
>
>
> In my concrete application I get cor = 1 for
> cor(as.numeric(dff$BB),as.numeric(BBVS)) even though I have
>
>
>
> str(summary((dff$BB)-(BBVS)))
> Class 'summaryMpfr' [package "Rmpfr"] of length 6 and
precision 128
> 4.61351010833e-8 7.33418976521e-7 1.31009046563e-5 3.76407022709e-5
> 5.72386764888e-5 ...
>
>
>
> I am on windows 10
>
> R version 3.6.1
>
> Best wishes
> Troels Ring,
> Aalborg, Denmark
>
>
>
>
>
>
> This email has been scanned by BullGuard antivirus protection.
> For more info visit www.bullguard.com
>
<http://www.bullguard.com/tracking.aspx?affiliate=bullguard&buyaffiliate=smt
> p&url=/>
>
> [[alternative HTML version deleted]]
>
> ______________________________________________
> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
Thanks a lot - solved the issue!
BW
Troels
-----Oprindelig meddelelse-----
Fra: Rui Barradas <ruipbarradas at sapo.pt>
Sendt: 12. juli 2020 12:59
Til: tring at gvdnet.dk; r-help mailing list <r-help at r-project.org>
Emne: Re: [R] Rmpfr correlation
Hello,
Why not write a function COR? Not one as general purpose as stats::cor but a
simple one, to compute the sample Pearson correlation only.
library(Rmpfr)
COR <- function(x, y){
precBits <- getPrec(x)[1]
n <- mpfr(length(x), precBits = precBits)
x.bar <- mean(x)
y.bar <- mean(y)
numer <- sum(x*y) - n*x.bar*y.bar
denom <- sqrt(sum(x*x) - n*x.bar*x.bar) * sqrt(sum(y*y) - n*y.bar*y.bar)
numer/denom
}
set.seed(2020)
KA <- mpfr(10^-4.6, 128)
x <- rnorm(100)*KA
y <- rnorm(100)*x
cor(as.numeric(x), as.numeric(y)) # -0.1874986
#[1] -0.1874986
COR(x, y)
#1 'mpfr' number of precision 128 bits
#[1] -0.1874985950531874160800643775644747505073
Hope this helps,
Rui Barradas
?s 10:42 de 12/07/20, tring at gvdnet.dk escreveu:> Dear friends - I'm calculating buffer capacities by different methods
and
> need very high precision and package Rmpfr is working beautifully. However,
> I have not been able to find out how to keep precision when finding
> correlations.
>
> library(Rmpfr)
>
> KA <- mpfr(10^-4.6, 128)
>
> x <- rnorm(100)*KA
>
> y <- rnorm(100)*x
>
> cor(x,y) # "x" must be numeric
>
> cor(as.numeric(x),as.numeric(y))# 0.2918954
>
>
>
> In my concrete application I get cor = 1 for
> cor(as.numeric(dff$BB),as.numeric(BBVS)) even though I have
>
>
>
> str(summary((dff$BB)-(BBVS)))
> Class 'summaryMpfr' [package "Rmpfr"] of length 6 and
precision 128
> 4.61351010833e-8 7.33418976521e-7 1.31009046563e-5 3.76407022709e-5
> 5.72386764888e-5 ...
>
>
>
> I am on windows 10
>
> R version 3.6.1
>
> Best wishes
> Troels Ring,
> Aalborg, Denmark
>
>
>
>
>
>
> This email has been scanned by BullGuard antivirus protection.
> For more info visit www.bullguard.com
>
<http://www.bullguard.com/tracking.aspx?affiliate=bullguard&buyaffiliate=smt
> p&url=/>
>
> [[alternative HTML version deleted]]
>
> ______________________________________________
> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
This email has been scanned by BullGuard antivirus protection.
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