Re: [R] why the length and width of a plot region produced by the dev.new() function cannot be correctly set?

[David Winsemius]

> On May 3, 2018, at 6:28 PM, sunyeping via R-help  wrote:
> 
> When I check the size of the plot region usingdev.size("in")a new plot region 
> is produced and in the Rconsole I get[1] 5.33 5.322917

Your test is all mangleed together. You failed in your duty to read the list 
info and the Posting guide . NO HTML!

> If I mean to produce a plot region with size setting 
> bydev.new(length=3,width=3)a plot region is produced, but the size is 
> [2.281250, 5.322917], as detected by the de.size function. If I 
> type:dev.new(length=10,width=10)I get a plot region of with the size of 
> [7.614583, 5.322917]. It seems that the width of the new plot region cannot 
> be set, and tt is always 5.322917. The length of the new plot region can be 
> set, but it is always smaller that the values I set.What do I miss? What is 
> the correct way of setting the dimension of the new plot region? I will be 
> grateful to any help.Best regards,

The size of the device is not the size of the plot region. You need to take 
into account the margins. See ?par


> 
>   [[alternative HTML version deleted]]
> 
> __
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
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> and provide commented, minimal, self-contained, reproducible code.

David Winsemius
Alameda, CA, USA

'Any technology distinguishable from magic is insufficiently advanced.'   
-Gehm's Corollary to Clarke's Third Law

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[R] why the length and width of a plot region produced by the dev.new() function cannot be correctly set?

[sunyeping via R-help]

When I check the size of the plot region usingdev.size("in")a new plot region 
is produced and in the Rconsole I get[1] 5.33 5.322917If I mean to produce 
a plot region with size setting bydev.new(length=3,width=3)a plot region is 
produced, but the size is [2.281250, 5.322917], as detected by the de.size 
function. If I type:dev.new(length=10,width=10)I get a plot region of with the 
size of [7.614583, 5.322917]. It seems that the width of the new plot region 
cannot be set, and tt is always 5.322917. The length of the new plot region can 
be set, but it is always smaller that the values I set.What do I miss? What is 
the correct way of setting the dimension of the new plot region? I will be 
grateful to any help.Best regards,

[[alternative HTML version deleted]]

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Re: [R] adding overall constraint in optim()

[Joshua Ulrich]

On Thu, May 3, 2018 at 2:03 PM, Michael Ashton
 wrote:
> Thanks Bert. But everyone on that forum wants to use finance tools rather 
> than general optimization stuff! And I am not optimizing a traditional 
> Markowitz mean-variance problem. Plus, smarter people here. :-)
>
I'm very confused by these statements.  Most of the "finance tools"
use general-purpose global and/or stochastic optimization packages
(e.g. rugarch uses nloptr and Rsolnp, PortfolioAnalytics uses DEoptim,
pso, GenSA).  And most (all?) of those optimization packages have ways
to specify box, equality, and nonlinear inequality constraints.

And I can't recall the last time someone emailed the list about
optimizing a traditional Markowitz mean-variance problem... maybe 10
years ago?

>> On May 3, 2018, at 3:01 PM, Bert Gunter  wrote:
>>
>> You can't -- at least as I  read the docs for ?optim (but I'm pretty
>> ignorant about this, so maybe there's a way to tweak it so you can).
>>
>> See here:   https://cran.r-project.org/web/views/Optimization.html
>> for other R optimization capabilities.
>>
>> Also,  given your credentials, the r-sig-finance list might be a
>> better place for you to post your query.
>>
>> Cheers,
>> Bert
>>
>>
>> Bert Gunter
>>
>> "The trouble with having an open mind is that people keep coming along
>> and sticking things into it."
>> -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
>>
>>
>> On Thu, May 3, 2018 at 10:52 AM, Michael Ashton
>>  wrote:
>>> Hi –
>>>
>>> This is giving me a headache. I’m trying to do a relatively simple 
>>> optimization – actually trying to approximate the output from the Excel 
>>> Solver function but at roughly 1000x the speed. 
>>>
>>> The optimization parameters look like this. The only trouble is that I want 
>>> to add a constraint that sum(wgt.vect)=1, and I can’t figure out how to do 
>>> that in optim.
>>>
>>> Mo.vect <- as.vector(tail(head(mo,i),1))
>>> wgt.vect <- as.vector(tail(head(moWeightsMax,i),1))
>>> cov.mat <- cov(tail(head(morets,i+12),12))
>>> opt.fun <- function(wgt.vect) -sum(Mo.vect %*% wgt.vect) / (t(wgt.vect) 
>>> %*% (cov.mat %*% wgt.vect))
>>>
>>> LowerBounds<-c(0.2,0.05,0.1,0,0,0)
>>> UpperBounds<-c(0.6,0.3,0.6,0.15,0.1,0.2)
>>>
>>> OptimSolution<-optim(wgt.vect, fn=opt.fun, 
>>> method="L-BFGS-B",lower=LowerBounds,upper=UpperBounds)
>>>
>>>
>>> Any thoughts are appreciated!
>>>
>>> Mike
>>>
>>> Michael Ashton, CFA
>>> Managing Principal
>>>
>>> Enduring Investments LLC
>>> W: 973.457.4602
>>> C: 551.655.8006
>>>
>>>
>>>[[alternative HTML version deleted]]
>>>
>>> __
>>> R-help@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.
> __
> R-help@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.



-- 
Joshua Ulrich  |  about.me/joshuaulrich
FOSS Trading  |  www.fosstrading.com
R/Finance 2018 | www.rinfinance.com

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Re: [R] Converting a list to a data frame

[William Dunlap via R-help]

If you require that the 'type' column be a factor with a level for each
element of the input list, then you need to do that after calling
dplyr::bind_rows(), just as with the base-R solutions.

> l <- list( A = data.frame(X=1:2, Y=11:12), B = data.frame(X=integer(),
Y=integer()), C = data.frame(X=3L, Y=13L) )
> str(d <- dplyr::bind_rows(l, .id = "Which") )
'data.frame':   3 obs. of  3 variables:
 $ Which: chr  "A" "A" "C"
 $ X: int  1 2 3
 $ Y: int  11 12 13
> d$Which <- factor(d$Which, levels=names(l))
> str(d)
'data.frame':   3 obs. of  3 variables:
 $ Which: Factor w/ 3 levels "A","B","C": 1 1 3
 $ X: int  1 2 3
 $ Y: int  11 12 13

Sometimes you need the names of the of the 0-row data.frames in the output
to make plots, etc., comparable across various samples of the data.


Bill Dunlap
TIBCO Software
wdunlap tibco.com

On Thu, May 3, 2018 at 10:28 AM, Hadley Wickham  wrote:

> On Wed, May 2, 2018 at 11:53 AM, Jeff Newmiller
>  wrote:
> > Another approach:
> >
> > 
> > library(tidyr)
> > L <- list( A = data.frame( x=1:2, y=3:4 )
> >  , B = data.frame( x=5:6, y=7:8 )
> >  )
> > D <- data.frame( Type = names( L )
> >, stringsAsFactors = FALSE
> >)
> > D$data <- L
> > unnest(D, data)
> > #>   Type x y
> > #> 1A 1 3
> > #> 2A 2 4
> > #> 3B 5 7
> > #> 4B 6 8
> > 
>
> I think a slightly more idiomatic tidyverse solution is dplyr::bind_rows()
>
> l <- list(
>   A = data.frame(x = 1:2, y = 3:4),
>   B = data.frame(x = 5:6, y = 7:8)
> )
>
> dplyr::bind_rows(l, .id = "type")
> #>   type x y
> #> 1A 1 3
> #> 2A 2 4
> #> 3B 5 7
> #> 4B 6 8
>
> This also has the advantage of returning a data frame when the inputs
> are data frames.
>
> Hadley
>
> --
> http://hadley.nz
>
> __
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> 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.
>

[[alternative HTML version deleted]]

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Re: [R] adding overall constraint in optim()

[John C. Nash]

This looks like what I call a sumscale problem i.e., some sort of simple
function of the parameters sums to a constant. I've done some work on
these, but don't have it with me just now. There are several approaches,
but they can be quite tricky. Will send some info in about a week or so
if you are still stuck and contact me offline.

JN
(you'll see my name on 3/5 of optim routines via ?optim)


On 2018-05-03 01:52 PM, Michael Ashton wrote:
> Hi –
> 
> This is giving me a headache. I’m trying to do a relatively simple 
> optimization – actually trying to approximate the output from the Excel 
> Solver function but at roughly 1000x the speed. 
> 
> The optimization parameters look like this. The only trouble is that I want 
> to add a constraint that sum(wgt.vect)=1, and I can’t figure out how to do 
> that in optim.
> 
>  Mo.vect <- as.vector(tail(head(mo,i),1))
>  wgt.vect <- as.vector(tail(head(moWeightsMax,i),1))
>  cov.mat <- cov(tail(head(morets,i+12),12))
>  opt.fun <- function(wgt.vect) -sum(Mo.vect %*% wgt.vect) / (t(wgt.vect) 
> %*% (cov.mat %*% wgt.vect))
> 
>  LowerBounds<-c(0.2,0.05,0.1,0,0,0)
>  UpperBounds<-c(0.6,0.3,0.6,0.15,0.1,0.2)
> 
>  OptimSolution<-optim(wgt.vect, fn=opt.fun, 
> method="L-BFGS-B",lower=LowerBounds,upper=UpperBounds)
> 
> 
> Any thoughts are appreciated!
> 
> Mike
> 
> Michael Ashton, CFA
> Managing Principal
> 
> Enduring Investments LLC
> W: 973.457.4602
> C: 551.655.8006
> 
> 
>   [[alternative HTML version deleted]]
> 
> __
> R-help@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.
>

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Re: [R] adding overall constraint in optim()

[David Winsemius]

> On May 3, 2018, at 10:52 AM, Michael Ashton 
>  wrote:
> 
> Hi –
> 
> This is giving me a headache. I’m trying to do a relatively simple 
> optimization – actually trying to approximate the output from the Excel 
> Solver function but at roughly 1000x the speed. 
> 
> The optimization parameters look like this. The only trouble is that I want 
> to add a constraint that sum(wgt.vect)=1, and I can’t figure out how to do 
> that in optim.
> 
> Mo.vect <- as.vector(tail(head(mo,i),1))
> wgt.vect <- as.vector(tail(head(moWeightsMax,i),1))
> cov.mat <- cov(tail(head(morets,i+12),12))
> opt.fun <- function(wgt.vect) -sum(Mo.vect %*% wgt.vect) / (t(wgt.vect) 
> %*% (cov.mat %*% wgt.vect))
> 
> LowerBounds<-c(0.2,0.05,0.1,0,0,0)
> UpperBounds<-c(0.6,0.3,0.6,0.15,0.1,0.2)
> 
> OptimSolution<-optim(wgt.vect, fn=opt.fun, 
> method="L-BFGS-B",lower=LowerBounds,upper=UpperBounds)
> 
> 
> Any thoughts are appreciated!

Have you reviewed the documentation for constrOptim {in pkg-stats}?

> 
> 
>   [[alternative HTML version deleted]]

I was surprised that emoji made it through. Nonetheless it is usually much 
safer to posting in plain-text. Didn't seem to be a problem in this case but 
code and data are often mangled.


David Winsemius
Alameda, CA, USA

'Any technology distinguishable from magic is insufficiently advanced.'   
-Gehm's Corollary to Clarke's Third Law

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Re: [R] adding overall constraint in optim()

[Michael Ashton]

Thanks Bert. But everyone on that forum wants to use finance tools rather than 
general optimization stuff! And I am not optimizing a traditional Markowitz 
mean-variance problem. Plus, smarter people here. :-)

> On May 3, 2018, at 3:01 PM, Bert Gunter  wrote:
> 
> You can't -- at least as I  read the docs for ?optim (but I'm pretty
> ignorant about this, so maybe there's a way to tweak it so you can).
> 
> See here:   https://cran.r-project.org/web/views/Optimization.html
> for other R optimization capabilities.
> 
> Also,  given your credentials, the r-sig-finance list might be a
> better place for you to post your query.
> 
> Cheers,
> Bert
> 
> 
> Bert Gunter
> 
> "The trouble with having an open mind is that people keep coming along
> and sticking things into it."
> -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
> 
> 
> On Thu, May 3, 2018 at 10:52 AM, Michael Ashton
>  wrote:
>> Hi –
>> 
>> This is giving me a headache. I’m trying to do a relatively simple 
>> optimization – actually trying to approximate the output from the Excel 
>> Solver function but at roughly 1000x the speed. 
>> 
>> The optimization parameters look like this. The only trouble is that I want 
>> to add a constraint that sum(wgt.vect)=1, and I can’t figure out how to do 
>> that in optim.
>> 
>> Mo.vect <- as.vector(tail(head(mo,i),1))
>> wgt.vect <- as.vector(tail(head(moWeightsMax,i),1))
>> cov.mat <- cov(tail(head(morets,i+12),12))
>> opt.fun <- function(wgt.vect) -sum(Mo.vect %*% wgt.vect) / (t(wgt.vect) 
>> %*% (cov.mat %*% wgt.vect))
>> 
>> LowerBounds<-c(0.2,0.05,0.1,0,0,0)
>> UpperBounds<-c(0.6,0.3,0.6,0.15,0.1,0.2)
>> 
>> OptimSolution<-optim(wgt.vect, fn=opt.fun, 
>> method="L-BFGS-B",lower=LowerBounds,upper=UpperBounds)
>> 
>> 
>> Any thoughts are appreciated!
>> 
>> Mike
>> 
>> Michael Ashton, CFA
>> Managing Principal
>> 
>> Enduring Investments LLC
>> W: 973.457.4602
>> C: 551.655.8006
>> 
>> 
>>[[alternative HTML version deleted]]
>> 
>> __
>> R-help@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.
__
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Re: [R] adding overall constraint in optim()

[Bert Gunter]

You can't -- at least as I  read the docs for ?optim (but I'm pretty
ignorant about this, so maybe there's a way to tweak it so you can).

See here:   https://cran.r-project.org/web/views/Optimization.html
for other R optimization capabilities.

Also,  given your credentials, the r-sig-finance list might be a
better place for you to post your query.

Cheers,
Bert


Bert Gunter

"The trouble with having an open mind is that people keep coming along
and sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )


On Thu, May 3, 2018 at 10:52 AM, Michael Ashton
 wrote:
> Hi –
>
> This is giving me a headache. I’m trying to do a relatively simple 
> optimization – actually trying to approximate the output from the Excel 
> Solver function but at roughly 1000x the speed. 
>
> The optimization parameters look like this. The only trouble is that I want 
> to add a constraint that sum(wgt.vect)=1, and I can’t figure out how to do 
> that in optim.
>
>  Mo.vect <- as.vector(tail(head(mo,i),1))
>  wgt.vect <- as.vector(tail(head(moWeightsMax,i),1))
>  cov.mat <- cov(tail(head(morets,i+12),12))
>  opt.fun <- function(wgt.vect) -sum(Mo.vect %*% wgt.vect) / (t(wgt.vect) 
> %*% (cov.mat %*% wgt.vect))
>
>  LowerBounds<-c(0.2,0.05,0.1,0,0,0)
>  UpperBounds<-c(0.6,0.3,0.6,0.15,0.1,0.2)
>
>  OptimSolution<-optim(wgt.vect, fn=opt.fun, 
> method="L-BFGS-B",lower=LowerBounds,upper=UpperBounds)
>
>
> Any thoughts are appreciated!
>
> Mike
>
> Michael Ashton, CFA
> Managing Principal
>
> Enduring Investments LLC
> W: 973.457.4602
> C: 551.655.8006
>
>
> [[alternative HTML version deleted]]
>
> __
> R-help@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.

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Re: [R] MCMCglmm - metric of the estimates

[Bert Gunter]

The r-sig-mixed-models list would be a better place for this post if
you do not get an answer here.

Cheers,
Bert


Bert Gunter

"The trouble with having an open mind is that people keep coming along
and sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )


On Thu, May 3, 2018 at 10:56 AM, Petr Palíšek  wrote:
> Hi,
>
> my question is probably amateurish but I can't seem to find the answer
> anywhere.
>
> In what metric are the MCMCglmm package's posterior means for family =
> "categorical"?
>
> I suppose that they can't be odds ratios and probabilites as my numbers are
> outside their bounds. So I'm thinking – are they just basic regression
> coefficients conceptually equal to those obtained by lme4::glmer?
>
> Thank you!
>
> [[alternative HTML version deleted]]
>
> __
> R-help@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.

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[R] MCMCglmm - metric of the estimates

[Petr Palíšek]

Hi,

my question is probably amateurish but I can't seem to find the answer
anywhere.

In what metric are the MCMCglmm package's posterior means for family =
"categorical"?

I suppose that they can't be odds ratios and probabilites as my numbers are
outside their bounds. So I'm thinking – are they just basic regression
coefficients conceptually equal to those obtained by lme4::glmer?

Thank you!

[[alternative HTML version deleted]]

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Re: [R] Converting a list to a data frame

[Kevin E. Thorpe]

On 05/03/2018 01:28 PM, Hadley Wickham wrote:

On Wed, May 2, 2018 at 11:53 AM, Jeff Newmiller
 wrote:

Another approach:


library(tidyr)
L <- list( A = data.frame( x=1:2, y=3:4 )
  , B = data.frame( x=5:6, y=7:8 )
  )
D <- data.frame( Type = names( L )
, stringsAsFactors = FALSE
)
D$data <- L
unnest(D, data)
#>   Type x y
#> 1A 1 3
#> 2A 2 4
#> 3B 5 7
#> 4B 6 8



I think a slightly more idiomatic tidyverse solution is dplyr::bind_rows()

l <- list(
   A = data.frame(x = 1:2, y = 3:4),
   B = data.frame(x = 5:6, y = 7:8)
)

dplyr::bind_rows(l, .id = "type")
#>   type x y
#> 1A 1 3
#> 2A 2 4
#> 3B 5 7
#> 4B 6 8

This also has the advantage of returning a data frame when the inputs
are data frames.

Hadley



I _clearly_ need to learn the dplyr package.

--
Kevin E. Thorpe
Head of Biostatistics,  Applied Health Research Centre (AHRC)
Li Ka Shing Knowledge Institute of St. Michael's Hospital
Assistant Professor, Dalla Lana School of Public Health
University of Toronto
email: kevin.tho...@utoronto.ca  Tel: 416.864.5776  Fax: 416.864.3016

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[R] adding overall constraint in optim()

[Michael Ashton]

Hi –

This is giving me a headache. I’m trying to do a relatively simple optimization 
– actually trying to approximate the output from the Excel Solver function but 
at roughly 1000x the speed. 

The optimization parameters look like this. The only trouble is that I want to 
add a constraint that sum(wgt.vect)=1, and I can’t figure out how to do that in 
optim.

 Mo.vect <- as.vector(tail(head(mo,i),1))
 wgt.vect <- as.vector(tail(head(moWeightsMax,i),1))
 cov.mat <- cov(tail(head(morets,i+12),12))
 opt.fun <- function(wgt.vect) -sum(Mo.vect %*% wgt.vect) / (t(wgt.vect) 
%*% (cov.mat %*% wgt.vect))

 LowerBounds<-c(0.2,0.05,0.1,0,0,0)
 UpperBounds<-c(0.6,0.3,0.6,0.15,0.1,0.2)

 OptimSolution<-optim(wgt.vect, fn=opt.fun, 
method="L-BFGS-B",lower=LowerBounds,upper=UpperBounds)


Any thoughts are appreciated!

Mike

Michael Ashton, CFA
Managing Principal

Enduring Investments LLC
W: 973.457.4602
C: 551.655.8006


[[alternative HTML version deleted]]

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Re: [R] Converting a list to a data frame

[Hadley Wickham]

On Wed, May 2, 2018 at 11:53 AM, Jeff Newmiller
 wrote:
> Another approach:
>
> 
> library(tidyr)
> L <- list( A = data.frame( x=1:2, y=3:4 )
>  , B = data.frame( x=5:6, y=7:8 )
>  )
> D <- data.frame( Type = names( L )
>, stringsAsFactors = FALSE
>)
> D$data <- L
> unnest(D, data)
> #>   Type x y
> #> 1A 1 3
> #> 2A 2 4
> #> 3B 5 7
> #> 4B 6 8
> 

I think a slightly more idiomatic tidyverse solution is dplyr::bind_rows()

l <- list(
  A = data.frame(x = 1:2, y = 3:4),
  B = data.frame(x = 5:6, y = 7:8)
)

dplyr::bind_rows(l, .id = "type")
#>   type x y
#> 1A 1 3
#> 2A 2 4
#> 3B 5 7
#> 4B 6 8

This also has the advantage of returning a data frame when the inputs
are data frames.

Hadley

-- 
http://hadley.nz

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Re: [R] Converting a list to a data frame

[David Winsemius]

> On May 2, 2018, at 2:43 PM, David L Carlson  wrote:
> 
> Typo: dat[[z]] should be x[[z]]:
> 
> x2 <- do.call(rbind, lapply(names(x), function(z) 
>  data.frame(type=z, x[[z]])))
> x2
>  type x y
> 1A 1 3
> 2A 2 4
> 3B 5 7
> 4B 6 8

Sometimes one want to "bind" list that have some columns in common and others 
that you want to include from one but leave as NA for the rows from the other. 
The rbind.list function from package-plyr is from the pre-tidy/tibble era of 
Hadley's efforts and provides that facility out of the box. It also seems to 
"work" with the do.call(rbind strategy for named lists above.

x <- list(A=mtcars[c("mpg", "wt")] , B=mtcars[c("wt", "cyl")])  #mtcars
x2 <- do.call(rbind.fill, lapply(names(x), function(z) 
 data.frame(type=z, x[[z]])))
str(x2)
#-
'data.frame':   64 obs. of  4 variables:
 type: Factor w/ 2 levels "A","B": 1 1 1 1 1 1 1 1 1 1 ...
 mpg : num  21 21 22.8 21.4 18.7 18.1 14.3 24.4 22.8 19.2 ...
 wt  : num  2.62 2.88 2.32 3.21 3.44 ...
 cyl : num  NA NA NA NA NA NA NA NA NA NA ...

-- 
David, the other.

> 
> David C
> 
> -Original Message-
> From: R-help  On Behalf Of David L Carlson
> Sent: Wednesday, May 2, 2018 3:51 PM
> To: William Dunlap ; Kevin E. Thorpe 
> 
> Cc: r-help mailing list 
> Subject: Re: [R] Converting a list to a data frame
> 
> Or add the type column first and then rbind:
> 
> x <- list(A=data.frame(x=1:2, y=3:4),B=data.frame(x=5:6,y=7:8))
> x2 <- do.call(rbind, lapply(names(x), function(z) 
>  data.frame(type=z, dat[[z]])))
> 
> 
> David L Carlson
> Department of Anthropology
> Texas A University
> College Station, TX 77843-4352
> 
> -Original Message-
> From: R-help  On Behalf Of William Dunlap via 
> R-help
> Sent: Wednesday, May 2, 2018 12:28 PM
> To: Kevin E. Thorpe 
> Cc: R Help Mailing List 
> Subject: Re: [R] Converting a list to a data frame
> 
>> x1 <- do.call(rbind, c(x, list(make.row.names=FALSE)))
>> x2 <- cbind(type=rep(names(x), vapply(x, nrow, 0)), x1)
>> str(x2)
> 'data.frame':   4 obs. of  3 variables:
> $ type: Factor w/ 2 levels "A","B": 1 1 2 2
> $ x   : int  1 2 5 6
> $ y   : int  3 4 7 8
> 
> 
> Bill Dunlap
> TIBCO Software
> wdunlap tibco.com
> 
> On Wed, May 2, 2018 at 10:11 AM, Kevin E. Thorpe 
> wrote:
> 
>> I suspect this is pretty easy, but I'm having trouble figuring it out.
>> Basically, I have a list of data frames such as the following example:
>> 
>> list(A=data.frame(x=1:2, y=3:4),B=data.frame(x=5:6,y=7:8))
>> 
>> I would like to turn this into  data frame where the list elements are 
>> essentially rbind'ed together and the element name becomes a new variable.
>> For example, I would like to turn the list above into a data frame 
>> that looks like this:
>> 
>> data.frame(type=c("A","A","B","B"),x=c(1:2,5:6),y=c(3:4,7:8))
>> 
>> Appreciate any pointers.
>> 
>> Kevin
>> 
>> --
>> Kevin E. Thorpe
>> Head of Biostatistics,  Applied Health Research Centre (AHRC) Li Ka 
>> Shing Knowledge Institute of St. Michael's Hospital Assistant 
>> Professor, Dalla Lana School of Public Health University of Toronto
>> email: kevin.tho...@utoronto.ca  Tel: 416.864.5776  Fax: 416.864.3016
>> 
>> __
>> R-help@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/posti 
>> ng-guide.html and provide commented, minimal, self-contained, 
>> reproducible code.
>> 
> 
>   [[alternative HTML version deleted]]
> 
> __
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> and provide commented, minimal, self-contained, reproducible code.
> 
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David Winsemius
Alameda, CA, USA

'Any technology distinguishable from magic is insufficiently advanced.'   
-Gehm's Corollary to Clarke's Third Law

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Re: [R] Calling the curve function with a character object converted into an expression

[Sebastien Bihorel]

Thanks,

I always get confused by expression evaluation, when and how to use call, 
do.call, eval, parse/deparse, and all that good stuff. I always have to read 
documentation 10 times and still does not want to stick in my brain.

- Original Message -
From: "Bert Gunter" 
To: "Sebastien Bihorel" 
Cc: "R-help" 
Sent: Thursday, May 3, 2018 2:28:59 AM
Subject: Re: [R] Calling the curve function with a character object converted 
into an expression

Typo: should be NULL not NUL of course

An alternative approach closer to your original attempt is to use
do.call() to explicitly evaluate the expr argument:

w <- "1 + x^2"
do.call(curve, list(expr = parse(text = w), ylab ="y"))

Cheers,
Bert

Bert Gunter

"The trouble with having an open mind is that people keep coming along
and sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )


On Wed, May 2, 2018 at 10:35 PM, Bert Gunter  wrote:
> Sebastian:
>
> This is somewhat arcane, perhaps even a bug (correction on this
> welcomed). The problem is that the "expr" argument to curve() must be
> an actual expression, not a call to parse that evaluates to an
> expression. If you look at the code of curve() you'll see why
> (substitute() does not evaluate  expr in the code). Another simple
> workaround other than sticking in the eval() call is this:
>
> myf <- function(x)NUL
> body(myf)<- parse(text = "{1+x^2}")   ## note the additional "{   }"
> for safety, though not necessary here
> ## this idiom will continue to work for any text.
>
> curve(myf, from = 0, to = 10)  ## myf is now the name of a function
> that executes the parsed text.
>
> *** I would appreciate any wiser R programmers correcting any
> misunderstanding or error in my explanation ***
>
> Cheers,
> Bert
>
> Bert Gunter
>
> "The trouble with having an open mind is that people keep coming along
> and sticking things into it."
> -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
>
>
> On Wed, May 2, 2018 at 8:11 PM, Sebastien Bihorel
>  wrote:
>>
>> Hi,
>>
>> Down a cascade of function calls, I want to use the curve function with an 
>> expression that is a variable. For various reason, this variable must be a 
>> character object and cannot be an expression as required by the curve 
>> function. How do I convert my variable into a expression that is accepted by 
>> curve?
>>
>> Thanks in advance for your help.
>>
>> ## The following attempts do not work
>>
>> myf <- '1+x^2'
>> curve(myf, from = 0, to = 10) # obviously !
>> curve(parse(text=myf), from = 0, to = 10) # not sure why this does not work
>>
>> ## This works but does not feel elegant:
>> eval(parse(text=sprintf('curve(%s, from = 0, to = 10)', myf)))
>>
>> __
>> R-help@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.

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Re: [R] The stages of standard function evaluation

[Duncan Murdoch]

First of all, your message is a little hard to read because you posted 
in HTML.  This list removes the HTML, and often mangles messages, so you 
should always post in plain text.  But in this case your message was 
still pretty readable.


On 02/05/2018 11:04 PM, Andrew Hoerner wrote:

Dear R Help folks --

I have been trying to put together a list of the steps or stages of R
function evaluation, with particular focus on those that have "standard" or
"nonstandard" forms. This is both for my own edification and also because I
am thinking of joining the world of R bloggers and have been trying to put
together some draft posting that might be useful. I seem to have an
affirmative genius for finding incorrect interpretations of R's evaluation
rules; I'm trying to make that an asset.

I am hoping that you can tell me:


1. Is this list complete, or are there additional stages I am missing?
2. Have I inserted one or more imaginary stages?
3. Are the terms I use below to name each stage appropriate, or are
there other terms more widely used or recognizable?
4. Is the order correct?

I begin each name with “Standard,” to express my belief that each of these
things has a usual or default form, but also that (unless I am mistaken)
almost none of them exist only in a single form true of all R functions. (I
have marked with an asterisk a few evaluation steps that I think may always
be followed).

It is my ultimate goal (which I do not feel at all close to accomplishing)
to determine a way to mechanically test for “standardness” along each of
these dimensions, so that each function could be assigned a logical vector
showing the ways that it is and is not standard. One thing I think is
conceptually or procedurally difficult about this project is that I think
“standardness” should be determined by what a function does, rather than by
how it does it, so that a primitive function that takes unevaluated
arguments positionally could still have standard matching, scoping, etc.,
by internal emulation. A related goal is to identify which evaluation steps
most often use an alternative form, and perhaps determine if there is more
than one such alternative. Finally, an easier short-term goal is simply to
find instances of one or more function with standard and non-standard
evaluation for each evaluation step.

For the most part below I am treating the evaluation of closures as the
standard from which “nonstandard” is defined. However, I do not assume that
other kinds of functions are automatically nonstandard on any particular
dimension below. Most of this comes from the R Language Definition, but
there are numerous places where I am by no means certain that my
interpretation is correct. I have highlighted some of these below with a
“??”.

I look forward to learning from you.

Warmest regards,

J. Andrew Hoerner


** Standard function recognition:* recognizing some or all of a string code
as a function. (Part of code line parsing)

*Standard environment construction:* construction of the execution
environment, and of pointers to the calling and enclosing environments.

*Standard function identification:* Get the name of the function, if any


This may be mangling, but it's really hard to tell whether the 3 
paragraphs above are supposed to be steps, headings, or what.  Assuming 
they are steps, the first one is wrong.


The parser looks at a string and breaks it down into tokens and 
subexpressions, making what you later call an AST.  The first step in 
function evaluation is recognizing that something is a function call, 
not recognizing it as a function.  For example, "mean" is the name of a 
function and also an expression evaluating to a function, "mean(1:10)" 
is a function call.


Once you have a function call, the next step looks at the expression 
used to specify the function.  In "mean(1:10)", that expression is 
"mean", but it could be an arbitrary R expression.  If it is a name like 
"mean" (or a string), then R looks for an object of mode "function" of 
that name in the current evaluation frame, or its parent frames.  These 
are not "constructed"; the current evaluation frame is always known, and 
contains a pointer to its parent.  If the function is specified by a 
more complex expression (e.g. in "fn[[1]](1:10)", the expression is 
"fn[[1]]") then that expression is evaluated.  It needs to return a 
function object or an error will be generated.


So these work:

mean(1:10)
list(mean)[[1]](1:10)
"mean"(1:10)

and these don't:

list("mean")(1:10)
c("mean")(1:10)

So now we have the function.  Its name is irrelevant.


** Standard f**unction scoping*: Search the current environment and then up
the chain of enclosing environments until you find the first binding
environment, an environment where the name of the function is bound, i.e.
linked to a pointer to the function definition. The binding environment is
usually (but not always) the same as the defining environment (i.e. the
enclosing 

Re: [R] Converting a list to a data frame

[Kevin E. Thorpe]

On 05/03/2018 03:59 AM, Martin Maechler wrote:

David L Carlson 
 on Wed, 2 May 2018 21:43:52 + writes:



Typo: dat[[z]] should be x[[z]]:

x2 <- do.call(rbind, lapply(names(x), function(z)
 data.frame(type=z, x[[z]])))
x2
   type x y
1A 1 3
2A 2 4
3B 5 7
4B 6 8

David C


Before this thread gets carried away to data.table and
tibbleverse (and as nobody else has done so) :

Let me nominate this beautiful solution by
Bill Dunlap and David Carlson to win the prize  with a 10 out 10 grade:

Beautiful use of  do.call() and lapply(), two of the most
versatile and important functions from the base R toolbox.

Congratulations!

Martin Maechler
R Core Team


First of all, thank you to all who responded. I have learned (and will 
learn) about several new things. The data.table/tibbleverse is one of 
those places I have yet to tread but it never seems to be a "good" time 
to learn as generally when these tools would be helpful is when I am 
trying to get something done but don't have the time to recast the 
problem in a different paradigm.


Second, I echo Martin's comment. The combined solution of Bill Dunlap 
and David Carlson is beautifully elegant.


Thanks again,

Kevin




-Original Message-
From: R-help  On Behalf Of David L Carlson
Sent: Wednesday, May 2, 2018 3:51 PM
To: William Dunlap ; Kevin E. Thorpe 

Cc: r-help mailing list 
Subject: Re: [R] Converting a list to a data frame

Or add the type column first and then rbind:

x <- list(A=data.frame(x=1:2, y=3:4),B=data.frame(x=5:6,y=7:8))
x2 <- do.call(rbind, lapply(names(x), function(z)
   data.frame(type=z, dat[[z]])))


David L Carlson
Department of Anthropology
Texas A University
College Station, TX 77843-4352

-Original Message-
From: R-help  On Behalf Of William Dunlap via 
R-help
Sent: Wednesday, May 2, 2018 12:28 PM
To: Kevin E. Thorpe 
Cc: R Help Mailing List 
Subject: Re: [R] Converting a list to a data frame


x1 <- do.call(rbind, c(x, list(make.row.names=FALSE)))
x2 <- cbind(type=rep(names(x), vapply(x, nrow, 0)), x1)
str(x2)

'data.frame':   4 obs. of  3 variables:
  $ type: Factor w/ 2 levels "A","B": 1 1 2 2
  $ x   : int  1 2 5 6
  $ y   : int  3 4 7 8


Bill Dunlap
TIBCO Software
wdunlap tibco.com

On Wed, May 2, 2018 at 10:11 AM, Kevin E. Thorpe 
wrote:


I suspect this is pretty easy, but I'm having trouble figuring it out.
Basically, I have a list of data frames such as the following example:

list(A=data.frame(x=1:2, y=3:4),B=data.frame(x=5:6,y=7:8))

I would like to turn this into  data frame where the list elements are
essentially rbind'ed together and the element name becomes a new variable.
For example, I would like to turn the list above into a data frame
that looks like this:

data.frame(type=c("A","A","B","B"),x=c(1:2,5:6),y=c(3:4,7:8))

Appreciate any pointers.

Kevin




--
Kevin E. Thorpe
Head of Biostatistics,  Applied Health Research Centre (AHRC)
Li Ka Shing Knowledge Institute of St. Michael's Hospital
Assistant Professor, Dalla Lana School of Public Health
University of Toronto
email: kevin.tho...@utoronto.ca  Tel: 416.864.5776  Fax: 416.864.3016

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[R] GA/SWARM Hyperparameter (HP) Optimisation for Classification based Machine Learning

[Alexander Gracian via R-help]

Hi,
I believe that Caret uses a  grid-serach approach. I was wondering if:
1 There are more efficient implementations for HP tuning for classification 
algos (eg XGboost, CatBoost, SVM, RF etc), using say GM/SWARM approaches, akin 
to Google's approach AutoML for Image related Net problems?
2 This one is most probably wishful thinking, but is anyone looking at GM/SWARM 
at HP tuning across models (ensemble models). eg the best set of HP for 
combined XGBoost + SVM, which accounts for the correlation/interaction of the 
prediction assumptions.
BestAlex
[[alternative HTML version deleted]]

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Re: [ESS] ess-insert-function-outline

[Patrick Connolly]

G'day Alex,

On Wed, 02-May-2018 at 09:19AM -0500, Alex Branham wrote:

[...]

|> 
|> It looks like ess-insert-function-outline was retired to the old/
|> directory, so it isn't defined in newer ESS versions. 

What could be the thinking behind that?  The outline file is still
there in the etc/ directory, so I never would have imagined dumping
ess-insert-function-outline into the old/ directory.

|> I'm guessing you updated ESS while tinkering with .emacs/.Rprofile?

Good guess.

|> 
|> I never used it myself, but it looks like it's supposed to insert a
|> function skeleton? If that's the case, there are a few replacements you
|> can use, including the builtin skeleton.el or yasnippet, available on
|> ELPA.

How do I use that?  There's /usr/share/emacs/24.3/lisp/skeleton.elc
that I could copy to the lisp/ directory or make a link from there to
it.  Would that make sense?

If not, I'll just stick with an older version that works fine.
There's nothing new that makes it worth forgoing the function
skeleton.

Thanks for your help.

-- 
~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.   
   ___Patrick Connolly   
 {~._.~}   Great minds discuss ideas
 _( Y )_ Average minds discuss events 
(:_~*~_:)  Small minds discuss people  
 (_)-(_)  . Eleanor Roosevelt
  
~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.

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Re: [R] Package for Molecular Properties

[Spencer Graves]

library(sos)


(mp <- findFn('{molecular properties}'))


  ** found 7 matches in 4 packages and opened two web pages in my 
default browser with (a) the 7 matches and (b) the 4 packages. The first 
function was something for amino acids, like you suggested.  Two others 
returned compound and substance information from PubChem.



  Does this help?
  Spencer


On 2018-05-02 19:17, bbb_...@verizon.net wrote:

All

Is there a package or library that will, given a nucleotide sequence

1. calculate the extinction coefficient at 260 nm for (Beer-Lambert's law)
2. calculate molecular weight
3. return it's complementary sequence

I was able to find several packages that can do similar calculations for an 
amino acid sequence for proteins but none for nucleic acids.

Any pointers, etc. would be great.

thank you in advance
Indr

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Re: [R] Converting a list to a data frame

[Martin Maechler]

> David L Carlson 
> on Wed, 2 May 2018 21:43:52 + writes:

> Typo: dat[[z]] should be x[[z]]:
> 
> x2 <- do.call(rbind, lapply(names(x), function(z) 
> data.frame(type=z, x[[z]])))
> x2
>   type x y
> 1A 1 3
> 2A 2 4
> 3B 5 7
> 4B 6 8
> 
> David C

Before this thread gets carried away to data.table and
tibbleverse (and as nobody else has done so) :

Let me nominate this beautiful solution by
Bill Dunlap and David Carlson to win the prize  with a 10 out 10 grade:

Beautiful use of  do.call() and lapply(), two of the most
versatile and important functions from the base R toolbox.

Congratulations!

Martin Maechler
R Core Team


> -Original Message-
> From: R-help  On Behalf Of David L Carlson
> Sent: Wednesday, May 2, 2018 3:51 PM
> To: William Dunlap ; Kevin E. Thorpe 
> 
> Cc: r-help mailing list 
> Subject: Re: [R] Converting a list to a data frame
> 
> Or add the type column first and then rbind:
> 
> x <- list(A=data.frame(x=1:2, y=3:4),B=data.frame(x=5:6,y=7:8))
> x2 <- do.call(rbind, lapply(names(x), function(z) 
>   data.frame(type=z, dat[[z]])))
> 
> 
> David L Carlson
> Department of Anthropology
> Texas A University
> College Station, TX 77843-4352
> 
> -Original Message-
> From: R-help  On Behalf Of William Dunlap via 
> R-help
> Sent: Wednesday, May 2, 2018 12:28 PM
> To: Kevin E. Thorpe 
> Cc: R Help Mailing List 
> Subject: Re: [R] Converting a list to a data frame
> 
> > x1 <- do.call(rbind, c(x, list(make.row.names=FALSE)))
> > x2 <- cbind(type=rep(names(x), vapply(x, nrow, 0)), x1)
> > str(x2)
> 'data.frame':   4 obs. of  3 variables:
>  $ type: Factor w/ 2 levels "A","B": 1 1 2 2
>  $ x   : int  1 2 5 6
>  $ y   : int  3 4 7 8
> 
> 
> Bill Dunlap
> TIBCO Software
> wdunlap tibco.com
> 
> On Wed, May 2, 2018 at 10:11 AM, Kevin E. Thorpe 
> wrote:
> 
> > I suspect this is pretty easy, but I'm having trouble figuring it out.
> > Basically, I have a list of data frames such as the following example:
> >
> > list(A=data.frame(x=1:2, y=3:4),B=data.frame(x=5:6,y=7:8))
> >
> > I would like to turn this into  data frame where the list elements are 
> > essentially rbind'ed together and the element name becomes a new variable.
> > For example, I would like to turn the list above into a data frame 
> > that looks like this:
> >
> > data.frame(type=c("A","A","B","B"),x=c(1:2,5:6),y=c(3:4,7:8))
> >
> > Appreciate any pointers.
> >
> > Kevin
> >
> > --
> > Kevin E. Thorpe
> > Head of Biostatistics,  Applied Health Research Centre (AHRC) Li Ka 
> > Shing Knowledge Institute of St. Michael's Hospital Assistant 
> > Professor, Dalla Lana School of Public Health University of Toronto
> > email: kevin.tho...@utoronto.ca  Tel: 416.864.5776  Fax: 416.864.3016
> >
> __
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[R] Package for Molecular Properties

[bbb_aaa]

All

Is there a package or library that will, given a nucleotide sequence

1. calculate the extinction coefficient at 260 nm for (Beer-Lambert's law)
2. calculate molecular weight 
3. return it's complementary sequence 

I was able to find several packages that can do similar calculations for an 
amino acid sequence for proteins but none for nucleic acids.

Any pointers, etc. would be great.

thank you in advance
Indr

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Re: [R] Converting a list to a data frame

[Tóth Dénes]

On 05/03/2018 07:58 AM, Jeff Newmiller wrote:

This is very nice to learn about, Denis, but it seems only fair to point out 
that the result of rbindlist is not a data frame.  You can convert it to a data 
frame easily, but the copy and indexing semantics of data tables are quite 
different than data tables, which could be a real headache for someone not 
prepared for those differences. (To learn more, read the data tables vignette.)

Tibbles  (as produced by unnest in my previous response) are not data tables 
either, but they behave much more like data frames than data tables do.

On May 2, 2018 1:30:37 PM MDT, "Tóth Dénes"  wrote:



On 05/02/2018 07:11 PM, Kevin E. Thorpe wrote:

I suspect this is pretty easy, but I'm having trouble figuring it

out.

Basically, I have a list of data frames such as the following

example:


list(A=data.frame(x=1:2, y=3:4),B=data.frame(x=5:6,y=7:8))

I would like to turn this into  data frame where the list elements

are

essentially rbind'ed together and the element name becomes a new
variable. For example, I would like to turn the list above into a

data

frame that looks like this:

data.frame(type=c("A","A","B","B"),x=c(1:2,5:6),y=c(3:4,7:8))

Appreciate any pointers.


Hi Kevin,

data.table::rbindlist does exactly what you want in a very efficient
way:

library(data.table)
dat <- list(A=data.frame(x=1:2, y=3:4),B=data.frame(x=5:6,y=7:8))
rbindlist(dat, idcol = "type")


In response to Jeff's note, this a solution which results in a 
data.frame instead of a data.table:


library(data.table)
dat <- list(A=data.frame(x=1:2, y=3:4),B=data.frame(x=5:6,y=7:8))
dat <- rbindlist(dat, idcol = "type")

# traditional way
dat_df <- as.data.frame(dat)

# no copy (assignment not needed, memory-efficient)
setDF(dat)

Further amendments would be needed to transform the 'type' variable to a 
factor, if it is required.




Regards,
Denes




Kevin



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Re: [R] Calling the curve function with a character object converted into an expression

[Bert Gunter]

Typo: should be NULL not NUL of course

An alternative approach closer to your original attempt is to use
do.call() to explicitly evaluate the expr argument:

w <- "1 + x^2"
do.call(curve, list(expr = parse(text = w), ylab ="y"))

Cheers,
Bert

Bert Gunter

"The trouble with having an open mind is that people keep coming along
and sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )


On Wed, May 2, 2018 at 10:35 PM, Bert Gunter  wrote:
> Sebastian:
>
> This is somewhat arcane, perhaps even a bug (correction on this
> welcomed). The problem is that the "expr" argument to curve() must be
> an actual expression, not a call to parse that evaluates to an
> expression. If you look at the code of curve() you'll see why
> (substitute() does not evaluate  expr in the code). Another simple
> workaround other than sticking in the eval() call is this:
>
> myf <- function(x)NUL
> body(myf)<- parse(text = "{1+x^2}")   ## note the additional "{   }"
> for safety, though not necessary here
> ## this idiom will continue to work for any text.
>
> curve(myf, from = 0, to = 10)  ## myf is now the name of a function
> that executes the parsed text.
>
> *** I would appreciate any wiser R programmers correcting any
> misunderstanding or error in my explanation ***
>
> Cheers,
> Bert
>
> Bert Gunter
>
> "The trouble with having an open mind is that people keep coming along
> and sticking things into it."
> -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
>
>
> On Wed, May 2, 2018 at 8:11 PM, Sebastien Bihorel
>  wrote:
>>
>> Hi,
>>
>> Down a cascade of function calls, I want to use the curve function with an 
>> expression that is a variable. For various reason, this variable must be a 
>> character object and cannot be an expression as required by the curve 
>> function. How do I convert my variable into a expression that is accepted by 
>> curve?
>>
>> Thanks in advance for your help.
>>
>> ## The following attempts do not work
>>
>> myf <- '1+x^2'
>> curve(myf, from = 0, to = 10) # obviously !
>> curve(parse(text=myf), from = 0, to = 10) # not sure why this does not work
>>
>> ## This works but does not feel elegant:
>> eval(parse(text=sprintf('curve(%s, from = 0, to = 10)', myf)))
>>
>> __
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>> and provide commented, minimal, self-contained, reproducible code.

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