Of course, what you could do is Google dqrls and get the source and
documentation. That is because it is in the publically available linpack.
If it were not publically available that would not work. Theoretically,
all FORTRAN or C code in R should be publically available.
Dave
From:
<cbe...@tajo.ucsd.edu>
To:
<r-de...@stat.math.ethz.ch>
Date:
04/27/2012 06:28 AM
Subject:
Re: [Rd] How does .Fortran "dqrls" work?
Sent by:
r-devel-boun...@r-project.org
yangleicq <yanglei...@126.com> writes:
> Hi, all.
> I want to write some functions like glm() so i studied it.
> In glm.fit(), it calls a fortran subroutine named "dqrfit" to compute
least
> squares solutions
> to the system
> x * b = y
>
> To learn how "dqrfit" works, I just follow how glm() calls "dqrfit" by
my
> own example, my codes are given below:
>
>> qr <-
>>
matrix(c(4.17,5.58,5.18,6.11,4.50,4.61,5.17,4.53,5.33,5.14,2.3,1.7,1.3,1.7,1.7,1.6,1,1.7,1.7,1.7),ncol=2)
>> qr
> [,1] [,2]
> [1,] 4.17 2.3
> [2,] 5.58 1.7
> [3,] 5.18 1.3
> [4,] 6.11 1.7
> [5,] 4.50 1.7
> [6,] 4.61 1.6
> [7,] 5.17 1.0
> [8,] 4.53 1.7
> [9,] 5.33 1.7
> [10,] 5.14 1.7
>> n=10
>> p=2
>> y <- c(4.81,4.17,4.41,3.59,5.87,3.83,6.03,4.89,4.32,4.69)
>> ny=1L
>> tol=1e-07
>> coefficients=double(p)
>> residuals=double(n)
>> effects=double(n)
>> rank=integer(1L)
>> pivot=1:n
>> qraux=double(n)
>> work=double(2*n)
>>
>>
>>
>> fittt<-.Fortran("dqrls", qr =qr, n = n,
> + p = p, y = y, ny = ny, tol = tol,
> coefficients=coefficients,
> + residuals = residuals, effects = effects,
> + rank = rank, pivot = pivot, qraux = qraux,
> + work = work, PACKAGE = "base")
>>
>> fittt$coefficients
> [1] 0 0
You have the args for .Fortran wrong. Try:
> fargs <- structure(list("dqrls", qr = structure(c(1, 1, 1, 1, 1, 1, 1,
+ 1, 1, 1, 4.17, 5.58, 5.18, 6.11, 4.5, 4.61, 5.17, 4.53, 5.33,
+ 5.14, 2.3, 1.7, 1.3, 1.7, 1.7, 1.6, 1, 1.7, 1.7, 1.7), .Dim = c(10L,
+ 3L)), n = 10L, p = 3L, y = c(4.81, 4.17, 4.41, 3.59, 5.87, 3.83,
+ 6.03, 4.89, 4.32, 4.69), ny = 1L, tol = 1e-11, coefficients = c(0,
+ 0, 0), residuals = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0), effects = c(0,
+ 0, 0, 0, 0, 0, 0, 0, 0, 0), rank = 0L, pivot = 1:3, qraux = c(0,
+ 0, 0), work = c(0, 0, 0, 0, 0, 0), PACKAGE = "base"), .Names = c("",
+ "qr", "n", "p", "y", "ny", "tol", "coefficients", "residuals",
+ "effects", "rank", "pivot", "qraux", "work", "PACKAGE"))
> do.call(.Fortran,fargs)$coef
[1] 11.176571 -0.883272 -1.262772
>
TIP: It often helps to use something like
debug(function.calling.Fortran)
and then step thru the function till the call you want to study is
invoked. Then inspect the inputs one-by-one and tinker with them and
recall the function or save them via
dput( list(...) , file="fargs" )
so you can later invoke the function as above.
HTH,
Chuck
>
> but when i use lm() which also calls "dqrls" internally to fit this
model,
> it gives reasonable result.
>
>> lm(y~qr)
>
> Call:
> lm(formula = y ~ qr)
>
> Coefficients:
> (Intercept) qr1 qr2
> 11.1766 -0.8833 -1.2628
>
>
> when I change the coefficients to be c(1,1), the output from "dqrls",
> fittt$coefficients also equals to c(1,1). That means the
.Fortran("dqrls",
> qr=qr,n=n,p=p,...) did nothing to the coefficients! I don't know why, is
> there anything I did wrong or missed? How can I get the result from
"dqrls"
> as what lm() or glm() gets from "dqrls"?
>
> Thanks in advance. Best Regards.
>
>
>
> --
> View this message in context:
http://r.789695.n4.nabble.com/How-does-Fortran-dqrls-work-tp4588973p4588973.html
> Sent from the R devel mailing list archive at Nabble.com.
>
--
Charles C. Berry Dept of Family/Preventive
Medicine
cberry at ucsd edu UC
San Diego
http://famprevmed.ucsd.edu/faculty/cberry/ La Jolla, San Diego 92093-0901
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