[R] install package av on centos 7
I tried to install the av package on centos7 server with R 3.5.3 but am
receiving an error. I installed ffmpeg and ffmpeg-devel on centos 7 as below
and then followed the instructions of specifying LIB_DIR and INCLUDE_DIR for
installing av.
any help would be appreciated. Thanks
```
> install.packages('av', configure.vars =
> c("LIB_DIR=/usr/lib64","INCLUDE_DIR=/usr/include/ffmpeg"))
Installing package into '/home/dominik/R/x86_64-pc-linux-gnu-library/3.5'
(as 'lib' is unspecified)
% Total% Received % Xferd Average Speed TimeTime Time Current
Dload Upload Total SpentLeft Speed
0 00 00 0 100 720k 100 720k0 0 987k 0
--:--:-- --:--:-- --:--:-- 987k
No protocol specified
* installing *source* package 'av' ...
** package 'av' successfully unpacked and MD5 sums checked
Found INCLUDE_DIR and/or LIB_DIR!
Using PKG_CFLAGS=-I/usr/include/ffmpeg
Using PKG_LIBS=-L/usr/lib64 -lavfilter
** libs
rm -f av.so formats.o info.o init.o util.o video.o
gcc -std=gnu99 -I/opt/microsoft/ropen/3.5.3/lib64/R/include -DNDEBUG
-I/usr/include/ffmpeg -DU_STATIC_IMPLEMENTATION -fvisibility=hidden -fpic
-DU_STATIC_IMPLEMENTATION -O2 -g -c formats.c -o formats.o
gcc -std=gnu99 -I/opt/microsoft/ropen/3.5.3/lib64/R/include -DNDEBUG
-I/usr/include/ffmpeg -DU_STATIC_IMPLEMENTATION -fvisibility=hidden -fpic
-DU_STATIC_IMPLEMENTATION -O2 -g -c info.c -o info.o
info.c: In function 'get_video_info':
info.c:33:14: error: 'AVStream' has no member named 'codecpar'
if(stream->codecpar->codec_type != AVMEDIA_TYPE_VIDEO)
^
info.c:35:49: error: 'AVStream' has no member named 'codecpar'
AVCodec *codec = avcodec_find_decoder(stream->codecpar->codec_id);
^
info.c:40:55: error: 'AVStream' has no member named 'codecpar'
SET_VECTOR_ELT(streamdata, 0, Rf_ScalarReal(stream->codecpar->width));
^
info.c:41:55: error: 'AVStream' has no member named 'codecpar'
SET_VECTOR_ELT(streamdata, 1, Rf_ScalarReal(stream->codecpar->height));
^
info.c:45:94: error: 'AVStream' has no member named 'codecpar'
SET_VECTOR_ELT(streamdata, 5, safe_string(av_get_pix_fmt_name((enum
AVPixelFormat) stream->codecpar->format)));
^
info.c: In function 'get_audio_info':
info.c:64:14: error: 'AVStream' has no member named 'codecpar'
if(stream->codecpar->codec_type != AVMEDIA_TYPE_AUDIO)
^
info.c:66:49: error: 'AVStream' has no member named 'codecpar'
AVCodec *codec = avcodec_find_decoder(stream->codecpar->codec_id);
^
info.c:70:55: error: 'AVStream' has no member named 'codecpar'
SET_VECTOR_ELT(streamdata, 0, Rf_ScalarReal(stream->codecpar->channels));
^
info.c:71:55: error: 'AVStream' has no member named 'codecpar'
SET_VECTOR_ELT(streamdata, 1,
Rf_ScalarReal(stream->codecpar->sample_rate));
^
info.c:74:55: error: 'AVStream' has no member named 'codecpar'
SET_VECTOR_ELT(streamdata, 4, Rf_ScalarReal(stream->codecpar->bit_rate));
^
info.c:77:54: error: 'AVStream' has no member named 'codecpar'
av_get_channel_layout_string(layout, 1024, stream->codecpar->channels,
stream->codecpar->channel_layout);
^
info.c:77:82: error: 'AVStream' has no member named 'codecpar'
av_get_channel_layout_string(layout, 1024, stream->codecpar->channels,
stream->codecpar->channel_layout);
^
make: *** [info.o] Error 1
ERROR: compilation failed for package 'av'
* removing '/home/dominik/R/x86_64-pc-linux-gnu-library/3.5/av'
Warning in install.packages :
installation of package 'av' had non-zero exit status
The downloaded source packages are in
'/tmp/RtmpMDafXK/downloaded_packages'
```
ffmpeg was installed successfully:
```
sudo rpm --import http://li.nux.ro/download/nux/RPM-GPG-KEY-nux.ro
sudo rpm -Uvh
http://li.nux.ro/download/nux/dextop/el7/x86_64/nux-dextop-release-0-5.el7.nux.noarch.rpm
Retrieving
http://li.nux.ro/download/nux/dextop/el7/x86_64/nux-dextop-release-0-5.el7.nux.noarch.rpm
Preparing... # [100%]
Updating / installing...
1:nux-dextop-release-0-5.el7.nux # [100%]
sudo yum install ffmpeg ffmpeg-devel -y
```
Result:
```
(base) [dominik@cppc-server ffmpeg]$ ffmpeg
ffmpeg version 2.8.15 Copyright (c) 2000-2018 the FFmpeg developers
built with gcc 4.8.5 (GCC) 20150623 (Red Hat 4.8.5-36)
Re: [R] Having problems with the ifelse and negative numbers
or even simpler sqrt(abs(A)) On Mon, Dec 9, 2019 at 8:45 AM Eric Berger wrote: > > Hi Bob, > You wrote "the following error message" - > when in fact it is a Warning and not an error message. I think your > code does what you hoped it would do, in the sense it successfully > calculates the sqrt(abs(negativeNumber)), where appropriate. > > If you want to run the code without seeing this warning message you can run > > ifelse( A < 0, suppressWarnings(sqrt(-A)), A ) > > and you should be fine. > > HTH, > Eric > > On Mon, Dec 9, 2019 at 3:18 PM Kevin Thorpe wrote: > > > > The sqrt(-A) is evaluated for all A. The result returned is conditional on > > the first argument but the other two arguments are evaluated on the entire > > vector. > > > > Kevin > > > > -- > > Kevin E. Thorpe > > Head of Biostatistics, Applied Health Research Centre (AHRC) > > Li Ka Shing Knowledge Institute of St. Michael's > > Assistant Professor, Dalla Lana School of Public Health > > University of Toronto > > email: kevin.tho...@utoronto.ca Tel: 416.864.5776 Fax: 416.864.3016 > > > > > > On 2019-12-09, 7:58 AM, "R-help on behalf of rsherry8" > > wrote: > > > > Please consider the following two R statements: > > A = runif(20, min=-1,max=1) > > ifelse( A < 0, sqrt(-A), A ) > > > > The second statement produces the following error message: > > rt(-A) : NaNs produced > > > > I understand that you cannot take the square root of a negative number > > but I thought the condition A < 0 > > would take care of that issue. It appears not to be. > > > > What am I missing? > > > > Thanks, > > Bob > > > > __ > > 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. > > __ > 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.
Re: [R] Where is the SD in output of glm with Gaussian distribution
Dear Bert,
It's perhaps a bit pedantic to point it out, but the dispersion is estimated
from the Pearson statistic (sum of squared residuals or weighted squared
residuals) not from the residual deviance. You can see this in the code for
summary.glm().
Best,
John
-
John Fox, Professor Emeritus
McMaster University
Hamilton, Ontario, Canada
Web: http::/socserv.mcmaster.ca/jfox
> On Dec 9, 2019, at 10:45 AM, Bert Gunter wrote:
>
> In addition, as John's included output shows, only 1 parameter, the
> intercept, is fit. As he also said, the sd is estimated from the residual
> deviance -- it is not a model parameter.
>
> Suggest you spend some time with a glm tutorial/text.
>
> Bert
>
> On Mon, Dec 9, 2019 at 7:17 AM Marc Girondot via R-help <
> r-help@r-project.org> wrote:
>
>> Let do a simple glm:
>>
>>> y=rnorm(100)
>>> gnul <- glm(y ~ 1)
>>> gnul$coefficients
>> (Intercept)
>> 0.1399966
>>
>> The logLik shows the fit of two parameters (DF=2) (intercept) and sd
>>
>>> logLik(gnul)
>> 'log Lik.' -138.7902 (df=2)
>>
>> But where is the sd term in the glm object?
>>
>> If I do the same with optim, I can have its value
>>
>>> dnormx <- function(x, data) {1E9*-sum(dnorm(data, mean=x["mean"],
>> sd=x["sd"], log = TRUE))}
>>> parg <- c(mean=0, sd=1)
>>> o0 <- optim(par = parg, fn=dnormx, data=y, method="BFGS")
>>> o0$value/1E9
>> [1] 138.7902
>>> o0$par
>> meansd
>>
>> 0.1399966 0.9694405
>>
>> But I would like have the value in the glm.
>>
>> (and in the meantime, I don't understand why gnul$df.residual returned
>> 99... for me it should be 98=100 - number of observations) -1 (for mean)
>> - 1 (for sd); but it is statistical question... I have asked it in
>> crossvalidated [no answer still] !)
>>
>> Thanks
>>
>> Marc
>>
>> __
>> 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.
>>
>
> [[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.
Re: [R] how to create a txt file with parsed columns
Thanks for getting back to me, I resolved my problem with this:
library(reshape2)
c=dcast(a, rs ~ GENE)
d=merge(c,b,by="rs")
d[is.na(d)] <- 0
On Sun, Dec 8, 2019 at 11:03 PM Jim Lemon wrote:
>
> Hi Ana,
> Is this what you want?
>
> a<-read.table(text="GENErs BETA
> 1 ENSG0154803 rs2605134 0.0360182
> 2 ENSG0154803 rs7405677 0.0525463
> 3 ENSG0154803 rs7211573 0.0525531
> 4 ENSG0154803 rs2746026 0.0466392
> 5 ENSG0141030 rs2605134 0.0806140
> 6 ENSG0141030 rs7405677 0.0251654
> 7 ENSG0141030 rs7211573 0.0252775
> 8 ENSG0141030 rs2746026 0.0976396
> 9 ENSG0205309 rs2605134 0.0838975
> 10 ENSG0205309 rs7405677 -0.2148500
> 11 ENSG0205309 rs7211573 -0.2148170
> 12 ENSG0205309 rs2746026 0.1013920
> 13 ENSG0215030 rs2605134 0.1261050
> 14 ENSG0215030 rs7405677 0.0165236
> 15 ENSG0215030 rs7211573 0.0163509
> 16 ENSG0215030 rs2746026 0.1201180
> 17 ENSG0141026 rs2605134 0.0485897
> 18 ENSG0141026 rs7405677 -0.0929964
> 19 ENSG0141026 rs7211573 -0.0930321
> 20 ENSG0141026 rs2746026 0.0623033",
> header=TRUE,stringsAsFactors=FALSE)
> b<-read.table(text="rs GWAS
> 1 rs2605134 0.0315177
> 2 rs7405677 -0.0816389
> 3 rs7211573 -0.0797796
> 4 rs2746026 0.0199350
> 5 rs11658521 0.0728377
> 6 rs9914107 0.0720096
> 7 rs56964223 0.0723903",
> header=TRUE,stringsAsFactors=FALSE)
> ab<-merge(a,b,by="rs")
> library(prettyR)
> abc<-stretch_df(ab,idvar="rs",to.stretch=c("GENE","BETA"))
>
> Jiim
>
> On Mon, Dec 9, 2019 at 11:10 AM Ana Marija
> wrote:
> >
> > Hello,
> >
> > I have two data frames:
> >
> > head(a)
> > GENErs BETA
> > 1 ENSG0154803 rs2605134 0.0360182
> > 2 ENSG0154803 rs7405677 0.0525463
> > 3 ENSG0154803 rs7211573 0.0525531
> > 4 ENSG0154803 rs2746026 0.0466392
> > 5 ENSG0141030 rs2605134 0.0806140
> > 6 ENSG0141030 rs7405677 0.0251654
> > 7 ENSG0141030 rs7211573 0.0252775
> > 8 ENSG0141030 rs2746026 0.0976396
> > 9 ENSG0205309 rs2605134 0.0838975
> > 10 ENSG0205309 rs7405677 -0.2148500
> > 11 ENSG0205309 rs7211573 -0.2148170
> > 12 ENSG0205309 rs2746026 0.1013920
> > 13 ENSG0215030 rs2605134 0.1261050
> > 14 ENSG0215030 rs7405677 0.0165236
> > 15 ENSG0215030 rs7211573 0.0163509
> > 16 ENSG0215030 rs2746026 0.1201180
> > 17 ENSG0141026 rs2605134 0.0485897
> > 18 ENSG0141026 rs7405677 -0.0929964
> > 19 ENSG0141026 rs7211573 -0.0930321
> > 20 ENSG0141026 rs2746026 0.0623033
> >
> > head(b)
> > rs GWAS
> > 1 rs2605134 0.0315177
> > 2 rs7405677 -0.0816389
> > 3 rs7211573 -0.0797796
> > 4 rs2746026 0.0199350
> > 5 rs11658521 0.0728377
> > 6 rs9914107 0.0720096
> > 7 rs56964223 0.0723903
> >
> > Data frame a has:
> > > length(unique(a$GENE))
> > [1] 51
> > > dim(a)
> > [1] 287 3
> >
> > and the whole data frame b is shown
> >
> > I would like to create a txt file which would have rs match for each
> > ENSG from data frame b. If a particular ENSG does not have matching rs
> > from data frame b the value under it would be zero. So the txt file
> > would have 7 rows (for all those unique rs from data frame b) and 53
> > columns (for 51 ENSGs and one for unique rs and one for GWAS)
> >
> > So one row of that txt file would look like this.
> >
> > GENES ENSG0154803 ENSG0141030 ENSG0205309
> > ENSG0215030ENSG0141026 GWAS
> > rs2605134 0.0360182 0.0806140 0.0838975
> > 0.1261050 0.0485897 0.0315177
> > …
> >
> > Please advise,
> > Ana
> >
> > __
> > 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.
Re: [R] Where is the SD in output of glm with Gaussian distribution
Le 09/12/2019 à 16:45, Bert Gunter a écrit :
> In addition, as John's included output shows, only 1 parameter, the
> intercept, is fit. As he also said, the sd is estimated from the
> residual deviance -- it is not a model parameter.
>
> Suggest you spend some time with a glm tutorial/text.
I tried ! But I miss this point. I understand now this point. Thanks a
lot... big progress for me.
But still I don't understand why AIC calculation uses 2 parameters if
the SD is estimated from the residual deviance.
> y=rnorm(100)
> gnul <- glm(y ~ 1)
> logLik(gnul)
'log Lik.' -136.4343 (df=2)
> AIC(gnul)
[1] 276.8687
> -2*logLik(gnul)+2*2
'log Lik.' 276.8687 (df=2)
This is not intuitive when to count SD as a parameter (in AIC) or not in
df.resuidual !
>
> Bert
>
> On Mon, Dec 9, 2019 at 7:17 AM Marc Girondot via R-help
> mailto:r-help@r-project.org>> wrote:
>
> Let do a simple glm:
>
> > y=rnorm(100)
> > gnul <- glm(y ~ 1)
> > gnul$coefficients
> (Intercept)
> 0.1399966
>
> The logLik shows the fit of two parameters (DF=2) (intercept) and sd
>
> > logLik(gnul)
> 'log Lik.' -138.7902 (df=2)
>
> But where is the sd term in the glm object?
>
> If I do the same with optim, I can have its value
>
> > dnormx <- function(x, data) {1E9*-sum(dnorm(data, mean=x["mean"],
> sd=x["sd"], log = TRUE))}
> > parg <- c(mean=0, sd=1)
> > o0 <- optim(par = parg, fn=dnormx, data=y, method="BFGS")
> > o0$value/1E9
> [1] 138.7902
> > o0$par
> mean sd
>
> 0.1399966 0.9694405
>
> But I would like have the value in the glm.
>
> (and in the meantime, I don't understand why gnul$df.residual
> returned
> 99... for me it should be 98=100 - number of observations) -1 (for
> mean)
> - 1 (for sd); but it is statistical question... I have asked it in
> crossvalidated [no answer still] !)
>
> Thanks
>
> Marc
>
> __
> R-help@r-project.org mailing list --
> To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Where is the SD in output of glm with Gaussian distribution
In addition, as John's included output shows, only 1 parameter, the
intercept, is fit. As he also said, the sd is estimated from the residual
deviance -- it is not a model parameter.
Suggest you spend some time with a glm tutorial/text.
Bert
On Mon, Dec 9, 2019 at 7:17 AM Marc Girondot via R-help <
r-help@r-project.org> wrote:
> Let do a simple glm:
>
> > y=rnorm(100)
> > gnul <- glm(y ~ 1)
> > gnul$coefficients
> (Intercept)
>0.1399966
>
> The logLik shows the fit of two parameters (DF=2) (intercept) and sd
>
> > logLik(gnul)
> 'log Lik.' -138.7902 (df=2)
>
> But where is the sd term in the glm object?
>
> If I do the same with optim, I can have its value
>
> > dnormx <- function(x, data) {1E9*-sum(dnorm(data, mean=x["mean"],
> sd=x["sd"], log = TRUE))}
> > parg <- c(mean=0, sd=1)
> > o0 <- optim(par = parg, fn=dnormx, data=y, method="BFGS")
> > o0$value/1E9
> [1] 138.7902
> > o0$par
> meansd
>
> 0.1399966 0.9694405
>
> But I would like have the value in the glm.
>
> (and in the meantime, I don't understand why gnul$df.residual returned
> 99... for me it should be 98=100 - number of observations) -1 (for mean)
> - 1 (for sd); but it is statistical question... I have asked it in
> crossvalidated [no answer still] !)
>
> Thanks
>
> Marc
>
> __
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Where is the SD in output of glm with Gaussian distribution
summary(gnul)
shows the std error of the coefficient estimate
On Mon, Dec 9, 2019 at 5:16 PM Marc Girondot via R-help
wrote:
>
> Let do a simple glm:
>
> > y=rnorm(100)
> > gnul <- glm(y ~ 1)
> > gnul$coefficients
> (Intercept)
>0.1399966
>
> The logLik shows the fit of two parameters (DF=2) (intercept) and sd
>
> > logLik(gnul)
> 'log Lik.' -138.7902 (df=2)
>
> But where is the sd term in the glm object?
>
> If I do the same with optim, I can have its value
>
> > dnormx <- function(x, data) {1E9*-sum(dnorm(data, mean=x["mean"],
> sd=x["sd"], log = TRUE))}
> > parg <- c(mean=0, sd=1)
> > o0 <- optim(par = parg, fn=dnormx, data=y, method="BFGS")
> > o0$value/1E9
> [1] 138.7902
> > o0$par
> meansd
>
> 0.1399966 0.9694405
>
> But I would like have the value in the glm.
>
> (and in the meantime, I don't understand why gnul$df.residual returned
> 99... for me it should be 98=100 - number of observations) -1 (for mean)
> - 1 (for sd); but it is statistical question... I have asked it in
> crossvalidated [no answer still] !)
>
> Thanks
>
> Marc
>
> __
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Where is the SD in output of glm with Gaussian distribution
Dear Marc,
For your simple model, the standard deviation of y is the square-root of the
estimated dispersion parameter:
> set.seed(123)
> y <- rnorm(100)
> gnul <- glm(y ~ 1)
> summary(gnul)
Call:
glm(formula = y ~ 1)
Deviance Residuals:
Min1QMedian3Q Max
-2.39957 -0.58426 -0.02865 0.60141 2.09693
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 0.090410.091280.990.324
(Dispersion parameter for gaussian family taken to be 0.8332328)
Null deviance: 82.49 on 99 degrees of freedom
Residual deviance: 82.49 on 99 degrees of freedom
AIC: 268.54
Number of Fisher Scoring iterations: 2
> sqrt(0.8332328)
[1] 0.9128159
> mean(y)
[1] 0.09040591
> sd(y)
[1] 0.9128159
I hope this helps,
John
-
John Fox, Professor Emeritus
McMaster University
Hamilton, Ontario, Canada
Web: http::/socserv.mcmaster.ca/jfox
> On Dec 9, 2019, at 10:16 AM, Marc Girondot via R-help
> wrote:
>
> Let do a simple glm:
>
> > y=rnorm(100)
> > gnul <- glm(y ~ 1)
> > gnul$coefficients
> (Intercept)
> 0.1399966
>
> The logLik shows the fit of two parameters (DF=2) (intercept) and sd
>
> > logLik(gnul)
> 'log Lik.' -138.7902 (df=2)
>
> But where is the sd term in the glm object?
>
> If I do the same with optim, I can have its value
>
> > dnormx <- function(x, data) {1E9*-sum(dnorm(data, mean=x["mean"],
> > sd=x["sd"], log = TRUE))}
> > parg <- c(mean=0, sd=1)
> > o0 <- optim(par = parg, fn=dnormx, data=y, method="BFGS")
> > o0$value/1E9
> [1] 138.7902
> > o0$par
> meansd
>
> 0.1399966 0.9694405
>
> But I would like have the value in the glm.
>
> (and in the meantime, I don't understand why gnul$df.residual returned 99...
> for me it should be 98=100 - number of observations) -1 (for mean) - 1 (for
> sd); but it is statistical question... I have asked it in crossvalidated [no
> answer still] !)
>
> Thanks
>
> Marc
>
> __
> 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] A better scales::dollar() ?
You can get pretty close with label_number_si():
pounds <- scales::label_number_si(prefix = "£")
pounds(10 ^ (0:7))
#> [1] "£1""£10" "£100" "£1K" "£10K" "£100K" "£1M" "£10M"
Created on 2019-12-09 by the [reprex
package](https://reprex.tidyverse.org) (v0.3.0.9001)
Hadley
On Fri, Dec 6, 2019 at 5:51 AM POLWART, Calum (COUNTY DURHAM AND
DARLINGTON NHS FOUNDATION TRUST) via R-help
wrote:
>
> I'm writing a quite large document in Rmarkdown which has financial data in
> it. I format that data using scales::dollar() currently something like this:
>
> >
> > require (scales)
> > x = 10
> > cat (dollar (x, prefix ="£", big.mark=","))
>
> £100,000
>
> But actually, I'd quite like to get £100k out in that instance so I'd do:
>
> > cat (dollar (x/10^3, prefix ="£", suffix="k" ))
>
> £100k
>
> But x could be 100 or 10,000,000. I want some form of 'automatic' version
> that might give me something like:
>
> >
> > require (scales)
> > y = 0:7
> > x = 1^y
> > dollar(x, prefix="£")
> [1] "£1" "£10" "£100""£1,000" "£10,000"
> "£100,000""£1,000,000" "£10,000,000"
>
> But what I want is more like:
>
> £1.00 £10.00 £100 £1k £10k £100k £1m £10m
>
> I'm sure I can write something as a little function, but before I do - is
> there something already out there?
>
> I have a similar need to format milligrams through to kilograms.
>
>
>
>
> This message may contain confidential information. If ...{{dropped:27}}
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[R] Where is the SD in output of glm with Gaussian distribution
Let do a simple glm:
> y=rnorm(100)
> gnul <- glm(y ~ 1)
> gnul$coefficients
(Intercept)
0.1399966
The logLik shows the fit of two parameters (DF=2) (intercept) and sd
> logLik(gnul)
'log Lik.' -138.7902 (df=2)
But where is the sd term in the glm object?
If I do the same with optim, I can have its value
> dnormx <- function(x, data) {1E9*-sum(dnorm(data, mean=x["mean"],
sd=x["sd"], log = TRUE))}
> parg <- c(mean=0, sd=1)
> o0 <- optim(par = parg, fn=dnormx, data=y, method="BFGS")
> o0$value/1E9
[1] 138.7902
> o0$par
mean sd
0.1399966 0.9694405
But I would like have the value in the glm.
(and in the meantime, I don't understand why gnul$df.residual returned
99... for me it should be 98=100 - number of observations) -1 (for mean)
- 1 (for sd); but it is statistical question... I have asked it in
crossvalidated [no answer still] !)
Thanks
Marc
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Re: [R] Having problems with the ifelse and negative numbers
Hi Bob, You wrote "the following error message" - when in fact it is a Warning and not an error message. I think your code does what you hoped it would do, in the sense it successfully calculates the sqrt(abs(negativeNumber)), where appropriate. If you want to run the code without seeing this warning message you can run ifelse( A < 0, suppressWarnings(sqrt(-A)), A ) and you should be fine. HTH, Eric On Mon, Dec 9, 2019 at 3:18 PM Kevin Thorpe wrote: > > The sqrt(-A) is evaluated for all A. The result returned is conditional on > the first argument but the other two arguments are evaluated on the entire > vector. > > Kevin > > -- > Kevin E. Thorpe > Head of Biostatistics, Applied Health Research Centre (AHRC) > Li Ka Shing Knowledge Institute of St. Michael's > Assistant Professor, Dalla Lana School of Public Health > University of Toronto > email: kevin.tho...@utoronto.ca Tel: 416.864.5776 Fax: 416.864.3016 > > > On 2019-12-09, 7:58 AM, "R-help on behalf of rsherry8" > wrote: > > Please consider the following two R statements: > A = runif(20, min=-1,max=1) > ifelse( A < 0, sqrt(-A), A ) > > The second statement produces the following error message: > rt(-A) : NaNs produced > > I understand that you cannot take the square root of a negative number > but I thought the condition A < 0 > would take care of that issue. It appears not to be. > > What am I missing? > > Thanks, > Bob > > __ > 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. __ 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.
Re: [R] Having problems with the ifelse and negative numbers
The sqrt(-A) is evaluated for all A. The result returned is conditional on the first argument but the other two arguments are evaluated on the entire vector. Kevin -- Kevin E. Thorpe Head of Biostatistics, Applied Health Research Centre (AHRC) Li Ka Shing Knowledge Institute of St. Michael's Assistant Professor, Dalla Lana School of Public Health University of Toronto email: kevin.tho...@utoronto.ca Tel: 416.864.5776 Fax: 416.864.3016 On 2019-12-09, 7:58 AM, "R-help on behalf of rsherry8" wrote: Please consider the following two R statements: A = runif(20, min=-1,max=1) ifelse( A < 0, sqrt(-A), A ) The second statement produces the following error message: rt(-A) : NaNs produced I understand that you cannot take the square root of a negative number but I thought the condition A < 0 would take care of that issue. It appears not to be. What am I missing? Thanks, Bob __ 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.
[R] Having problems with the ifelse and negative numbers
Please consider the following two R statements: A = runif(20, min=-1,max=1) ifelse( A < 0, sqrt(-A), A ) The second statement produces the following error message: rt(-A) : NaNs produced I understand that you cannot take the square root of a negative number but I thought the condition A < 0 would take care of that issue. It appears not to be. What am I missing? Thanks, Bob __ 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] [R-pkgs] concurve v 2.3.0 - Comparing Functions, Bootstrapping, and Exporting Tables
Pleased to announce that the next version (2.3.0) of our package “concurve” is out on CRAN (https://cran.r-project.org/package=concurve). In addition to plotting confidence (consonance) curves, it can now plot two functions next to one another to see the amount of overlap. It can also plot likelihood and deviance functions along with consonance densities and distributions. And it can utilize bootstrapping (the BCa and t method) to approximate the curves, distributions, and densities. And to top it off, it can export tables of relevant statistics in various formats. We welcome any feedback and bug reports ( https://github.com/zadchow/concurve/issues) Best, Zad R. Chow Andrew D. Vigotsky [[alternative HTML version deleted]] ___ R-packages mailing list r-packa...@r-project.org https://stat.ethz.ch/mailman/listinfo/r-packages __ 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.
