Re: [R] [FORGED] p-value=0 running log-rank test

[Rolf Turner]

This is not an R question.

Your question indicates that you really need to learn some statistics.

To answer the very last part:

> pchisq(430,3,lower=FALSE)
[1] 7.020486e-93

And if that is not 0 to all intents and purposes, then God help us all.

cheers,

Rolf Turner

On 19/05/17 12:48, Anne Karin da Mota Borges wrote:

Dear all,

I have a question concerning the p-value. When running log-rank test I get
a p-value = 0.
What is it mean? Can this be true? Why aren´t there decimal points? Is
there a way to find out the exact p-value?

Here is the output:


survdiff(Surv(tempo2,status)~tphist, data=base,rho=0)

Call:
survdiff(formula = Surv(tempo2, status) ~ tphist, data = base,
rho = 0)

   N Observed Expected (O-E)^2/E (O-E)^2/V
tphist=1 513   19   40.87911.710  9.00e+01
tphist=2  4943.892 0.003  3.27e-03
tphist=3  2391.68631.717  3.29e+01
tphist=4  18   150.543   385.172  3.91e+02

 Chisq= 430  on 3 degrees of freedom, p= 0

Thank you in advance.


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[R] p-value=0 running log-rank test

[Anne Karin da Mota Borges]

Dear all,

I have a question concerning the p-value. When running log-rank test I get
a p-value = 0.
What is it mean? Can this be true? Why aren´t there decimal points? Is
there a way to find out the exact p-value?

Here is the output:

> survdiff(Surv(tempo2,status)~tphist, data=base,rho=0)
Call:
survdiff(formula = Surv(tempo2, status) ~ tphist, data = base,
rho = 0)

   N Observed Expected (O-E)^2/E (O-E)^2/V
tphist=1 513   19   40.87911.710  9.00e+01
tphist=2  4943.892 0.003  3.27e-03
tphist=3  2391.68631.717  3.29e+01
tphist=4  18   150.543   385.172  3.91e+02

 Chisq= 430  on 3 degrees of freedom, p= 0

Thank you in advance.

Anne.

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Re: [R] Question about change the length of a string.

[Yen Lee]

Hi Jeff,

Thank you a lot!

Best,
Yen

-Original Message-
From: Jeff Newmiller [mailto:jdnew...@dcn.davis.ca.us] 
Sent: Thursday, May 18, 2017 11:11 PM
To: r-help@r-project.org; Yen Lee 
Subject: Re: [R] Question about change the length of a string.

http://stackoverflow.com/questions/2261079/how-to-trim-leading-and-trailing-whitespace-in-r
--
Sent from my phone. Please excuse my brevity.

On May 18, 2017 8:57:51 PM PDT, Yen Lee  wrote:
>Hello everyone,
>
> 
>
>I have a question and I need your precious kind help. 
>
> 
>
>I am working on matching two string. However, the length of the two 
>strings is different. For example, one is "example" (nchar=7), the 
>other one is
>"example   " (nchar=10). The R considers them as different strings but
>they
>will be the same if I could make the first string has 10 characters, 
>"example" followed by 3 spaces. However, I could not find a way to do 
>it. Is there anyone able to provide some suggestions?
>
> 
>
>I appreciate your precious time
>
> 
>
>Best,
>
>Yen
>
> 
>
>
>   [[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] Question about change the length of a string.

[Jeff Newmiller]

http://stackoverflow.com/questions/2261079/how-to-trim-leading-and-trailing-whitespace-in-r
-- 
Sent from my phone. Please excuse my brevity.

On May 18, 2017 8:57:51 PM PDT, Yen Lee  wrote:
>Hello everyone,
>
> 
>
>I have a question and I need your precious kind help. 
>
> 
>
>I am working on matching two string. However, the length of the two
>strings
>is different. For example, one is "example" (nchar=7), the other one is
>"example   " (nchar=10). The R considers them as different strings but
>they
>will be the same if I could make the first string has 10 characters,
>"example" followed by 3 spaces. However, I could not find a way to do
>it. Is
>there anyone able to provide some suggestions? 
>
> 
>
>I appreciate your precious time
>
> 
>
>Best,
>
>Yen 
>
> 
>
>
>   [[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] Question about change the length of a string.

[Yen Lee]

Hello everyone,

 

I have a question and I need your precious kind help. 

 

I am working on matching two string. However, the length of the two strings
is different. For example, one is "example" (nchar=7), the other one is
"example   " (nchar=10). The R considers them as different strings but they
will be the same if I could make the first string has 10 characters,
"example" followed by 3 spaces. However, I could not find a way to do it. Is
there anyone able to provide some suggestions? 

 

I appreciate your precious time

 

Best,

Yen 

 


[[alternative HTML version deleted]]

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[R-es] me parece interesante

[Javier Marcuzzi]

Estimados

Hay un sitio con recursos en varios lenguajes, entre estos R, puede ser bueno 
como fuente de recursos y hay unos datos que me parecieron interesantes de R 
(muy resumido como curiosidad).

https://libraries.io/languages/R

Javier Rubén Marcuzzi


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Re: [R] RCommander issue

[Jim Lemon]

Hi Thambu,
Try downloading the package from CRAN. For example, I just tried this
with the rgdal package as I need to use it.

The package (rgdal_1.2-7.zip) is saved on the local hard disk. Then
find out where your R executable is. For me it is:

E:\jim\R\R-3.3.3\bin\x64\R.exe

Now open a "Command Prompt" window. Change your directory to where the
package file is saved. Then using your path to the R executable and
the name of the package file you downloaded:

E:\jim\R\R-3.3.3\bin\x64\R CMD INSTALL rgdal_1.2-7.zip

This seems to work without invoking install.packages, which may be
failing to establish an internet connection for the download.

Jim


On Thu, May 18, 2017 at 9:11 PM, thambu david  wrote:
> thanks JIm, i do have internet connection and i was trying to set the CRAN
> mirror from R , before when i got the message
>
>
>
> On Thursday, 18 May 2017 3:20 AM, Jim Lemon  wrote:
>
>
> Hi Thambu,
> You must have an internet connection, otherwise you couldn't access
> CRAN. Have you tried using "install.packages" (see the help page) from
> within R? Your message isn't clear whether you have used that function
> or tried to download the package from a Web browser and then install
> it from the local copy of the package.
>
> There are a lot of things that can go wrong when accessing the
> internet and the error message does not seem to implicate R.
>
> Jim
>
> On Wed, May 17, 2017 at 10:27 PM, thambu david  wrote:
>> :-)
>> I did try with the right spelling with no luck
>> is there any simple ting i am making an error with?
>>
>>
>
>

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[R] S4 Question; show generic method

[stephen sefick]

Hello,

I changed the name of a function, and updated my R packages with
update.packages(ask=FALSE), and I have had a curious (to me) change in show
method. I am developing a package, and a generic show method I defined for
my S4 class is not autocalled when I type the object name - object_name;
however, the show(object_name) works as expected. I have included the
output from sessionInfo():

R version 3.3.2 (2016-10-31)
Platform: x86_64-redhat-linux-gnu (64-bit)
Running under: CentOS Linux 7 (Core)

locale:
 [1] LC_CTYPE=en_US.utf8   LC_NUMERIC=C
 [3] LC_TIME=en_US.utf8LC_COLLATE=en_US.utf8
 [5] LC_MONETARY=en_US.utf8LC_MESSAGES=en_US.utf8
 [7] LC_PAPER=en_US.utf8   LC_NAME=C
 [9] LC_ADDRESS=C  LC_TELEPHONE=C
[11] LC_MEASUREMENT=en_US.utf8 LC_IDENTIFICATION=C

attached base packages:
[1] stats graphics  grDevices utils datasets  methods   base

other attached packages:
[1] genotypeR_0.0.0.9000 ggplot2_2.2.1

loaded via a namespace (and not attached):
 [1] Rcpp_0.12.10 lattice_0.20-35  zoo_1.8-0MASS_7.3-47
 [5] grid_3.3.2   plyr_1.8.4   gtable_0.2.0 magrittr_1.5
 [9] scales_0.4.1 stringi_1.1.5rlang_0.1.1  reshape2_1.4.2
[13] lazyeval_0.2.0   doBy_4.5-15  Matrix_1.2-10tools_3.3.2
[17] stringr_1.2.0munsell_0.4.3colorspace_1.3-2 tibble_1.3.1

I appreciate any help with this problem.
kindest regards,

Stephen

-- 
Let's not spend our time and resources thinking about things that are so
little or so large that all they really do for us is puff us up and make us
feel like gods.  We are mammals, and have not exhausted the annoying little
problems of being mammals.

-K. Mullis

"A big computer, a complex algorithm and a long time does not equal
science."

  -Robert Gentleman

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Re: [R] Extracting metadata information to corresponding dissimilarity matrix

[Rune Grønseth]

Brilliant, David, thank you so much!

Cheers, 

Rune 

> 16. mai 2017 kl. 18.44 skrev David L Carlson :
> 
> Fixing a typo in the original, adding a simplification, and using 
> dissimilarity instead of similarity:
> 
> set.seed(42)
> dta <- data.frame(ID=1:7, gender=sample(c("M", "F"), 7, replace=TRUE),
> age=sample.int(75, 7))
> dsim <- dist(dta$age) # distance, already lower triangular
> dsim
> 
> dta1 <- dta
> names(dta1) <- paste0(names(dta), "1") # generalizes to more than 3 columns
> dta2 <- dta
> names(dta2) <- paste0(names(dta), "2")
> 
> dta12 <- merge(dta2, dta1) # order is important
> dta12 <- dta12[dta12$ID1 < dta12$ID2, ] # get rid of duplicates
> 
> dta12 <- data.frame(dta12, dsim=as.vector(dsim)) # Typo was here
> dta12 <- dta12[, c("ID1", "ID2", "gender1", "gender2", "age1", "age2", 
> "dsim")]
> dta12
> 
> David C
> 
> 
> -Original Message-
> From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of David L 
> Carlson
> Sent: Tuesday, May 16, 2017 11:21 AM
> To: Rune Grønseth ; r-help@r-project.org
> Subject: Re: [R] Extracting metadata information to corresponding 
> dissimilarity matrix
> 
> I think this is what you are trying to do. I've created a data set with 7 
> rows and a similarity matrix based on age:
> 
> set.seed(42)
> dta <- data.frame(ID=1:7, gender=sample(c("M", "F"), 7, replace=TRUE),
> age=sample.int(75, 7))
> sim <- max(dist(dta$age)) - dist(dta$age) # already lower triangular
> sim
> 
> #1  2  3  4  5  6
> # 2 24   
> # 3 21 59
> # 4 40 46 43 
> # 5  0 38 41 22  
> # 6  7 45 48 29 55   
> # 7 55 31 28 47  7 14
> 
> # Now duplicate dta:
> dta1 <- dta
> names(dta1) <- c("ID1", "gender1", "age1")
> dta2 <- dta
> names(dta2) <- c("ID2", "gender2", "age2")
> 
> # Now merge and eliminate unneeded rows
> dta12 <- merge(dta2, dta1) # order is important
> dta12 <- dta12[dta12$ID1 < dta12$ID2, ]
> 
> # Finally combine the similarities with the combined data and rearrange
> # the variable names
> dta12 <- data.frame(dta12mod, sim=as.vector(sim))
> dta12 <- dta12[, c("ID1", "ID2", "gender1", "gender2", "age1", "age2", "sim")]
> dta12
> 
> #ID1 ID2 gender1 gender2 age1 age2 sim
> # 21   2   F   F   11   49  24
> # 31   3   F   M   11   52  21
> # 41   4   F   F   11   33  40
> # 51   5   F   F   11   73   0
> # 61   6   F   F   11   66   7
> # 71   7   F   F   11   18  55
> # 10   2   3   F   M   49   52  59
> # 11   2   4   F   F   49   33  46
> # 12   2   5   F   F   49   73  38
> # 13   2   6   F   F   49   66  45
> # 14   2   7   F   F   49   18  31
> # 18   3   4   M   F   52   33  43
> # 19   3   5   M   F   52   73  41
> # 20   3   6   M   F   52   66  48
> # 21   3   7   M   F   52   18  28
> # 26   4   5   F   F   33   73  22
> # 27   4   6   F   F   33   66  29
> # 28   4   7   F   F   33   18  47
> # 34   5   6   F   F   73   66  55
> # 35   5   7   F   F   73   18   7
> # 42   6   7   F   F   66   18  14
> 
> -
> David L Carlson
> Department of Anthropology
> Texas A University
> College Station, TX 77840-4352
> 
> -Original Message-
> From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Rune Grønseth
> Sent: Tuesday, May 16, 2017 4:31 AM
> To: r-help@r-project.org
> Subject: [R] Extracting metadata information to corresponding dissimilarity 
> matrix
> 
> Hi,
> I am R beginner. I've tried googling and reading, but this might be too 
> simple to be found in the documentation. 
> 
> I have a dissimilarity index (symmetric matrix) from which I have extracted 
> the unique values using the exodist package command "lower". There are 14 
> observations, so there are 91 unique comparisons.
> 
> After this I'd like to extract corresponding metadata from a separate data 
> frame (the 14 observations organized in rows identified by a 
> samplenumber-vector, and other variables as gender, age, et cetera). The aim 
> is to have a new data frame with 91 rows and metadata vectors giving me the 
> value of the dissimilarity index,  gender each of the two observations that 
> are compared by the dissimilarity metric. So if I'm looking for gender 
> differences, I need 5 vectors in the data frame: samplenumber1, 
> samplenumber2, gender1, gender2 and dissimilarity metric.
> 
> Does anyone have suggestions or experiences in reformatting data in this 
> manner? This is just a test-dataset. My full data-set is for more than 100 
> observations, so I need a more general code, if that is possible.
> 
> With great appreciation of any help.
> 
> Rune Grønseth 
> 
> ---
> 
> Rune Grønseth, MD, PhD, postdoctoral fellow
> Department of Thoracic Medicine
> Haukeland University Hospital
> N-5021 Bergen
> Norway
> 
>   [[alternative HTML version 

Re: [R] Randomly select one row by group from a matrix

[David L Carlson]

You can modify your original code to get what you want:

do.call(rbind, lapply(split(data.frame(test), test[,c("id")]), function(x) 
as.matrix(x[sample(nrow(x), 1), ])))
#   xcor ycor id
# 146  1
# 242  2

But Bert's way is simpler:

indx <- tapply(seq_along(test[, "id"]), test[, "id"], sample, size=1)
test[indx, ]
#  xcor ycor id
# [1,]46  1
# [2,]42  2

-
David L Carlson
Department of Anthropology
Texas A University
College Station, TX 77840-4352



-Original Message-
From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Bert Gunter
Sent: Thursday, May 18, 2017 12:39 PM
To: Ulrik Stervbo 
Cc: r-help@r-project.org; Marine Regis 
Subject: Re: [R] Randomly select one row by group from a matrix

If I understand corrrectly, this is easily accomplished in base R via
?tapply and indexing.

e.g.

set.seed(1234) ## for reproducibility
grp <- sample.int(5,size = 30,rep = TRUE) ## a grouping vector
## Could be just a column of your matrix or frame

indx <- tapply(seq_along(grp),grp, sample,size =1)
> indx  ## just to show you what you get
 1  2  3  4  5
19 15 10  6 14

## now just use indx to extract rowd of your matrix or data frame,d:

selected <- d[indx,] ## one row per group


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 18, 2017 at 8:45 AM, Ulrik Stervbo  wrote:
> Hi Marine,
>
> your manipulation of the matrix is quite convoluted, and it helps to expand
> a bit:
>
> test_lst <- split(test, test[,c("id")])
> test_lst$`1`
>
> after splitting, your matrix has gone back to be a plain vector, which
> makes the sampling fail.
>
> The reason is that, a matrix - behind the scenes - is a vector with a
> dimension and when splitting the matrix you lose the dimension information.
>
> Do you really need to work with a matrix? I prefer data.frames because I
> can mix different types. Also with data.frame you can use the functionality
> of the dplyr library, which also makes things more readable:
>
> library(dplyr)
>
> test_df <- data.frame(xcor = rnorm(8), ycor = rnorm(8), id = c(1, 2))
>
> grouped_test_df <- group_by(test_df, id)
> sample_n(grouped_test_df, 1)
>
> HTH
> Ulrik
>
>
>
> On Thu, 18 May 2017 at 17:18 Marine Regis  wrote:
>
>> Hello,
>> I would like to randomly select one row by group from a matrix. Here is an
>> example where there is one row by group. The code gives an error message:
>> test <- matrix(c(4,4, 6,2, 1,2), nrow = 2, ncol = 3, dimnames = list(NULL,
>> c("xcor", "ycor", "id")))
>> do.call(rbind, lapply(split(test, test[,c("id")]), function(x)
>> x[sample(nrow(x), 1), ]))
>>  Show Traceback
>>
>>  Rerun with Debug
>>
>> Error in sample.int(length(x), size, replace, prob) :
>>   invalid first argument
>>
>>
>> How can I modify the code so that it works when there are several rows or
>> one row for a given group?
>> Thanks very much for your time
>> Have a nice day
>> Marine
>>
>>
>> [[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.
>>
>
> [[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.

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Re: [R-es] PLOTEO DEL DATO MENSUAL DESDE EL ACUMULADO

[Carlos Ortega]

Hola,

En ese caso, lo que tienes que usar en la función "lag()", que está en el
paquete base, aunque trate sobre series temporales.
O puedes utilizar funciones equivalentes en los varios paquetes que tratan
series temporales (forecast, xts...).

Gracias,
Carlos Ortega
www.qualityexcellence.es

El 18 de mayo de 2017, 18:44, Javier Valdes Cantallopts (DGA) <
javier.val...@mop.gov.cl> escribió:

> Interesante.
>
> Cada uno de los valores de mi tabla tiene asociada una fecha.
>
> Se le podría asignar un  tipo de “break” ,para que le “diff” sea asociado
> a cada mes? Por ejemplo?
>
>
>
> [image: Descripción: FIRMA3]
>
>
>
> *De:* Carlos Ortega [mailto:c...@qualityexcellence.es]
> *Enviado el:* jueves, 18 de mayo de 2017 12:35
> *Para:* Javier Valdes Cantallopts (DGA)
> *CC:* r-help-es@r-project.org
> *Asunto:* Re: PLOTEO DEL DATO MENSUAL DESDE EL ACUMULADO
>
>
>
> Hola,
>
>
>
> Hay una función en "R" que permite hacer ese cálculo "diff()".
>
>
>
> #-
>
> > DF <- read.table(textConnection(val_cum), as.is = TRUE)
>
> > DF
>
> V1
>
> 11
>
> 25
>
> 38
>
> 4   10
>
> 5   20
>
> 6   30
>
> 7   50
>
> 8   80
>
> 9  120
>
> 10 125
>
> 11 200
>
> 12 250
>
> > diff(DF$V1, 1)
>
>  [1]  4  3  2 10 10 20 30 40  5 75 50
>
> #-
>
>
>
> Saludos,
>
> Carlos Ortega
>
> www.qualiytexcellence.es
>
>
>
> El 18 de mayo de 2017, 16:57, Javier Valdes Cantallopts (DGA) <
> javier.val...@mop.gov.cl> escribió:
>
> Hola estimados:
>
> He tratado de encontrar la mejor forma de automatizar un proceso de “datos
> mensuales de precipitación”.
>
> El tema es que tengo solo la *serie acumulada anual de Pp*. Por ende,
> busco una forma de plotear un gráfico, *extrayendo “los datos mensuales
> desde el acumulado”*
>
> Ejemplo;
>
> ene
>
> feb
>
> mar
>
> abr
>
> may
>
> jun
>
> jul
>
> ago
>
> sep
>
> oct
>
> nov
>
> dic
>
>
>
> 1
>
> 5
>
> 8
>
> 10
>
> 20
>
> 30
>
> 50
>
> 80
>
> 120
>
> 125
>
> 200
>
> 250
>
> Datos ACUMULADOS
>
> *1*
>
> *4*
>
> *3*
>
> *2*
>
> *10*
>
> *10*
>
> *20*
>
> *30*
>
> *40*
>
> *5*
>
> *75*
>
> *50*
>
> *DATOS A PLOTEAR*
>
>
>
> He buscado por  time series y dapply, pero encuentro que son un poco
> engorrosos sus métodos.
>
> Saludos a todos.
>
>
>
> [image: Descripción: FIRMA3]
>
>
>
>
> --
>
>
> CONFIDENCIALIDAD: La información contenida en este mensaje y/o en los
> archivos adjuntos es de carácter confidencial o privilegiada y está
> destinada al uso exclusivo del emisor y/o de la persona o entidad a quien
> va dirigida. Si usted no es el destinatario, cualquier almacenamiento,
> divulgación, distribución o copia de esta información está estrictamente
> prohibido y sancionado por la ley. Si recibió este mensaje por error, por
> favor infórmenos inmediatamente respondiendo este mismo mensaje y borre
> todos los archivos adjuntos. Gracias.
>
> CONFIDENTIAL NOTE: The information transmitted in this message and/or
> attachments is confidential and/or privileged and is intented only for use
> of the person or entity to whom it is addressed. If you are not the
> intended recipient, any retention, dissemination, distribution or copy of
> this information is strictly prohibited and sanctioned by law. If you
> received this message in error, please reply us this same message and
> delete this message and all attachments. Thank you.
>
>
>
>
>
> --
>
> Saludos,
> Carlos Ortega
> www.qualityexcellence.es
>
> --
>
> CONFIDENCIALIDAD: La información contenida en este mensaje y/o en los
> archivos adjuntos es de carácter confidencial o privilegiada y está
> destinada al uso exclusivo del emisor y/o de la persona o entidad a quien
> va dirigida. Si usted no es el destinatario, cualquier almacenamiento,
> divulgación, distribución o copia de esta información está estrictamente
> prohibido y sancionado por la ley. Si recibió este mensaje por error, por
> favor infórmenos inmediatamente respondiendo este mismo mensaje y borre
> todos los archivos adjuntos. Gracias.
>
> CONFIDENTIAL NOTE: The information transmitted in this message and/or
> attachments is confidential and/or privileged and is intented only for use
> of the person or entity to whom it is addressed. If you are not the
> intended recipient, any retention, dissemination, distribution or copy of
> this information is strictly prohibited and sanctioned by law. If you
> received this message in error, please reply us this same message and
> delete this message and all attachments. Thank you.
>



-- 
Saludos,
Carlos Ortega
www.qualityexcellence.es
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Re: [R-es] Crear GUIDES con R

[Carlos Ortega]

Si, busca/mira "Shiny"...

Saludos,
Carlos Ortega
WWE.qualityexcellence.es



El El jue, may. 18, 2017 a las 6:55 PM, WILMER CONTRERAS SEPULVEDA <
wilme...@ufps.edu.co> escribió:

> Hola Buenos días. Quisiera saber si existe un paquete en R para crear un
> GUIDE como se hace facilmente en Matlab.
>
> Muchas gracias.
>
> Wilmer Contreras S.
>
> [[alternative HTML version deleted]]
>
> ___
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> https://stat.ethz.ch/mailman/listinfo/r-help-es
>
-- 
Saludos,
Carlos Ortega
www.qualityexcellence.es

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Re: [R] Randomly select one row by group from a matrix

[Bert Gunter]

If I understand corrrectly, this is easily accomplished in base R via
?tapply and indexing.

e.g.

set.seed(1234) ## for reproducibility
grp <- sample.int(5,size = 30,rep = TRUE) ## a grouping vector
## Could be just a column of your matrix or frame

indx <- tapply(seq_along(grp),grp, sample,size =1)
> indx  ## just to show you what you get
 1  2  3  4  5
19 15 10  6 14

## now just use indx to extract rowd of your matrix or data frame,d:

selected <- d[indx,] ## one row per group


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 18, 2017 at 8:45 AM, Ulrik Stervbo  wrote:
> Hi Marine,
>
> your manipulation of the matrix is quite convoluted, and it helps to expand
> a bit:
>
> test_lst <- split(test, test[,c("id")])
> test_lst$`1`
>
> after splitting, your matrix has gone back to be a plain vector, which
> makes the sampling fail.
>
> The reason is that, a matrix - behind the scenes - is a vector with a
> dimension and when splitting the matrix you lose the dimension information.
>
> Do you really need to work with a matrix? I prefer data.frames because I
> can mix different types. Also with data.frame you can use the functionality
> of the dplyr library, which also makes things more readable:
>
> library(dplyr)
>
> test_df <- data.frame(xcor = rnorm(8), ycor = rnorm(8), id = c(1, 2))
>
> grouped_test_df <- group_by(test_df, id)
> sample_n(grouped_test_df, 1)
>
> HTH
> Ulrik
>
>
>
> On Thu, 18 May 2017 at 17:18 Marine Regis  wrote:
>
>> Hello,
>> I would like to randomly select one row by group from a matrix. Here is an
>> example where there is one row by group. The code gives an error message:
>> test <- matrix(c(4,4, 6,2, 1,2), nrow = 2, ncol = 3, dimnames = list(NULL,
>> c("xcor", "ycor", "id")))
>> do.call(rbind, lapply(split(test, test[,c("id")]), function(x)
>> x[sample(nrow(x), 1), ]))
>>  Show Traceback
>>
>>  Rerun with Debug
>>
>> Error in sample.int(length(x), size, replace, prob) :
>>   invalid first argument
>>
>>
>> How can I modify the code so that it works when there are several rows or
>> one row for a given group?
>> Thanks very much for your time
>> Have a nice day
>> Marine
>>
>>
>> [[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.
>>
>
> [[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-es] PLOTEO DEL DATO MENSUAL DESDE EL ACUMULADO

[Javier Marcuzzi]

Estimado Javier Valdes Cantallopts

Entonces sus datos son distintos, es una fecha mensual y a esa fecha tiene 
asociada la cantidad de precipitaciones mensuales correspondientes a ese mes 
más la suma de los meses anteriores, por lo cual en diciembre tiene el 
acumulado anual, que en sus datos son 250, y no es 250 en un año para diciembre 
y 50 para diciembre de otro año, y no 13 columnas como está en la tabla que 
envía donde hay doce meses más la columna de acumulados.

Para graficar una forma sencilla a partir de la ayuda de Carlos Ortega, es a 
los datos de diff utilizarlos en xyplot (lattice), o a partir de datos con 
fechas (¿pero usted tiene fechas reales?)

Javier Rubén Marcuzzi

De: Javier Valdes Cantallopts (DGA)
Enviado: jueves, 18 de mayo de 2017 13:44
Para: Carlos Ortega
CC: r-help-es@r-project.org
Asunto: Re: [R-es] PLOTEO DEL DATO MENSUAL DESDE EL ACUMULADO

Interesante.
Cada uno de los valores de mi tabla tiene asociada una fecha.
Se le podría asignar un  tipo de “break” ,para que le “diff” sea asociado a 
cada mes? Por ejemplo?



De: Carlos Ortega [mailto:c...@qualityexcellence.es] 
Enviado el: jueves, 18 de mayo de 2017 12:35
Para: Javier Valdes Cantallopts (DGA)
CC: r-help-es@r-project.org
Asunto: Re: PLOTEO DEL DATO MENSUAL DESDE EL ACUMULADO

Hola,

Hay una función en "R" que permite hacer ese cálculo "diff()".

#-
> DF <- read.table(textConnection(val_cum), as.is = TRUE)
> DF
    V1
1    1
2    5
3    8
4   10
5   20
6   30
7   50
8   80
9  120
10 125
11 200
12 250
> diff(DF$V1, 1)
 [1]  4  3  2 10 10 20 30 40  5 75 50
#-

Saludos,
Carlos Ortega
www.qualiytexcellence.es

El 18 de mayo de 2017, 16:57, Javier Valdes Cantallopts (DGA) 
 escribió:
Hola estimados:
He tratado de encontrar la mejor forma de automatizar un proceso de “datos 
mensuales de precipitación”.
El tema es que tengo solo la serie acumulada anual de Pp. Por ende, busco una 
forma de plotear un gráfico, extrayendo “los datos mensuales desde el acumulado”
Ejemplo;
ene
feb
mar
abr
may
jun
jul
ago
sep
oct
nov
dic
 
1
5
8
10
20
30
50
80
120
125
200
250
Datos ACUMULADOS
1
4
3
2
10
10
20
30
40
5
75
50
DATOS A PLOTEAR
 
He buscado por  time series y dapply, pero encuentro que son un poco engorrosos 
sus métodos.
Saludos a todos.
 

 



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-- 
Saludos,
Carlos Ortega
www.qualityexcellence.es



CONFIDENCIALIDAD: La información contenida en este mensaje y/o en los archivos 
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[R-es] Crear GUIDES con R

[WILMER CONTRERAS SEPULVEDA]

Hola Buenos días. Quisiera saber si existe un paquete en R para crear un
GUIDE como se hace facilmente en Matlab.

Muchas gracias.

Wilmer Contreras S.

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Re: [R-es] PLOTEO DEL DATO MENSUAL DESDE EL ACUMULADO

[Javier Valdes Cantallopts (DGA)]

Interesante.
Cada uno de los valores de mi tabla tiene asociada una fecha.
Se le podría asignar un  tipo de “break” ,para que le “diff” sea asociado a 
cada mes? Por ejemplo?

[Descripción: FIRMA3]

De: Carlos Ortega [mailto:c...@qualityexcellence.es]
Enviado el: jueves, 18 de mayo de 2017 12:35
Para: Javier Valdes Cantallopts (DGA)
CC: r-help-es@r-project.org
Asunto: Re: PLOTEO DEL DATO MENSUAL DESDE EL ACUMULADO

Hola,

Hay una función en "R" que permite hacer ese cálculo "diff()".

#-
> DF <- read.table(textConnection(val_cum), as.is = TRUE)
> DF
V1
11
25
38
4   10
5   20
6   30
7   50
8   80
9  120
10 125
11 200
12 250
> diff(DF$V1, 1)
 [1]  4  3  2 10 10 20 30 40  5 75 50
#-

Saludos,
Carlos Ortega
www.qualiytexcellence.es

El 18 de mayo de 2017, 16:57, Javier Valdes Cantallopts (DGA) 
> escribió:
Hola estimados:
He tratado de encontrar la mejor forma de automatizar un proceso de “datos 
mensuales de precipitación”.
El tema es que tengo solo la serie acumulada anual de Pp. Por ende, busco una 
forma de plotear un gráfico, extrayendo “los datos mensuales desde el acumulado”
Ejemplo;
ene

feb

mar

abr

may

jun

jul

ago

sep

oct

nov

dic



1

5

8

10

20

30

50

80

120

125

200

250

Datos ACUMULADOS

1

4

3

2

10

10

20

30

40

5

75

50

DATOS A PLOTEAR


He buscado por  time series y dapply, pero encuentro que son un poco engorrosos 
sus métodos.
Saludos a todos.

[Descripción: FIRMA3]




CONFIDENCIALIDAD: La información contenida en este mensaje y/o en los archivos 
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exclusivo del emisor y/o de la persona o entidad a quien va dirigida. Si usted 
no es el destinatario, cualquier almacenamiento, divulgación, distribución o 
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message in error, please reply us this same message and delete this message and 
all attachments. Thank you.



--
Saludos,
Carlos Ortega
www.qualityexcellence.es



CONFIDENCIALIDAD: La información contenida en este mensaje y/o en los archivos 
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Re: [R-es] PLOTEO DEL DATO MENSUAL DESDE EL ACUMULADO

[Carlos Ortega]

Hola,

Hay una función en "R" que permite hacer ese cálculo "diff()".

#-
> DF <- read.table(textConnection(val_cum), as.is = TRUE)
> DF
V1
11
25
38
4   10
5   20
6   30
7   50
8   80
9  120
10 125
11 200
12 250
> diff(DF$V1, 1)
 [1]  4  3  2 10 10 20 30 40  5 75 50
#-

Saludos,
Carlos Ortega
www.qualiytexcellence.es

El 18 de mayo de 2017, 16:57, Javier Valdes Cantallopts (DGA) <
javier.val...@mop.gov.cl> escribió:

> Hola estimados:
>
> He tratado de encontrar la mejor forma de automatizar un proceso de “datos
> mensuales de precipitación”.
>
> El tema es que tengo solo la *serie acumulada anual de Pp*. Por ende,
> busco una forma de plotear un gráfico, *extrayendo “los datos mensuales
> desde el acumulado”*
>
> Ejemplo;
>
> ene
>
> feb
>
> mar
>
> abr
>
> may
>
> jun
>
> jul
>
> ago
>
> sep
>
> oct
>
> nov
>
> dic
>
>
>
> 1
>
> 5
>
> 8
>
> 10
>
> 20
>
> 30
>
> 50
>
> 80
>
> 120
>
> 125
>
> 200
>
> 250
>
> Datos ACUMULADOS
>
> *1*
>
> *4*
>
> *3*
>
> *2*
>
> *10*
>
> *10*
>
> *20*
>
> *30*
>
> *40*
>
> *5*
>
> *75*
>
> *50*
>
> *DATOS A PLOTEAR*
>
>
>
> He buscado por  time series y dapply, pero encuentro que son un poco
> engorrosos sus métodos.
>
> Saludos a todos.
>
>
>
> [image: Descripción: FIRMA3]
>
>
>
> --
>
> CONFIDENCIALIDAD: La información contenida en este mensaje y/o en los
> archivos adjuntos es de carácter confidencial o privilegiada y está
> destinada al uso exclusivo del emisor y/o de la persona o entidad a quien
> va dirigida. Si usted no es el destinatario, cualquier almacenamiento,
> divulgación, distribución o copia de esta información está estrictamente
> prohibido y sancionado por la ley. Si recibió este mensaje por error, por
> favor infórmenos inmediatamente respondiendo este mismo mensaje y borre
> todos los archivos adjuntos. Gracias.
>
> CONFIDENTIAL NOTE: The information transmitted in this message and/or
> attachments is confidential and/or privileged and is intented only for use
> of the person or entity to whom it is addressed. If you are not the
> intended recipient, any retention, dissemination, distribution or copy of
> this information is strictly prohibited and sanctioned by law. If you
> received this message in error, please reply us this same message and
> delete this message and all attachments. Thank you.
>



-- 
Saludos,
Carlos Ortega
www.qualityexcellence.es
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Re: [R] Randomly select one row by group from a matrix

[Ulrik Stervbo]

Hi Marine,

your manipulation of the matrix is quite convoluted, and it helps to expand
a bit:

test_lst <- split(test, test[,c("id")])
test_lst$`1`

after splitting, your matrix has gone back to be a plain vector, which
makes the sampling fail.

The reason is that, a matrix - behind the scenes - is a vector with a
dimension and when splitting the matrix you lose the dimension information.

Do you really need to work with a matrix? I prefer data.frames because I
can mix different types. Also with data.frame you can use the functionality
of the dplyr library, which also makes things more readable:

library(dplyr)

test_df <- data.frame(xcor = rnorm(8), ycor = rnorm(8), id = c(1, 2))

grouped_test_df <- group_by(test_df, id)
sample_n(grouped_test_df, 1)

HTH
Ulrik



On Thu, 18 May 2017 at 17:18 Marine Regis  wrote:

> Hello,
> I would like to randomly select one row by group from a matrix. Here is an
> example where there is one row by group. The code gives an error message:
> test <- matrix(c(4,4, 6,2, 1,2), nrow = 2, ncol = 3, dimnames = list(NULL,
> c("xcor", "ycor", "id")))
> do.call(rbind, lapply(split(test, test[,c("id")]), function(x)
> x[sample(nrow(x), 1), ]))
>  Show Traceback
>
>  Rerun with Debug
>
> Error in sample.int(length(x), size, replace, prob) :
>   invalid first argument
>
>
> How can I modify the code so that it works when there are several rows or
> one row for a given group?
> Thanks very much for your time
> Have a nice day
> Marine
>
>
> [[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.
>

[[alternative HTML version deleted]]

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Re: [R-es] PLOTEO DEL DATO MENSUAL DESDE EL ACUMULADO

[Javier Valdes Cantallopts (DGA)]

Hola
Solo extraer los montos correspondientes a cada mes. Solo tengo la serie 
acumulada anual.

[Descripción: FIRMA3]

De: Javier Marcuzzi [mailto:javier.ruben.marcu...@gmail.com]
Enviado el: jueves, 18 de mayo de 2017 11:14
Para: Javier Valdes Cantallopts (DGA); Carlos Ortega (c...@qualityexcellence.es)
CC: r-help-es@r-project.org
Asunto: RE: [R-es] PLOTEO DEL DATO MENSUAL DESDE EL ACUMULADO

Estimado Javier Valdes Cantallopts

Hay algo que no comprendo, usted dice que desea extraer los datos mensuales 
desde el acumulado.

¿Sería algo como, el año 2000 llovió 2300 ml? A partir de esto conocer las 
precipitaciones en enero, febrero, etc.

Ahora tengo mis dudas, si usted tiene los datos originales podría tener un 
modelo, y predict(….). ¿Pero, si hay un año o mes con sequía por efecto de la 
Niña o Niño?

Yo soy de Argentina y por mi zona geográfica (más o menos a la mitad de 
Argentina a la altura de Uruguay) hay efectos en las precipitaciones, desde 
sequía  a llover unos 600 ml en un día (una vez, pero recuerdo esos 60 cm de 
lluvia a 7 km de mi casa). Posiblemente se pueda en forma matemática, pero 
tengo mis dudas.

¿No tiene acceso a los registros mensuales o diarios para a partir de estos 
comenzar su trabajo? No soy meteorólogo, pero pienso que debería tener un 
modelo con los meses y una corrección con el efecto del Niño y Niña.

Javier Rubén Marcuzzi


De: Javier Valdes Cantallopts (DGA)
Enviado: jueves, 18 de mayo de 2017 11:57
Para: Carlos Ortega 
(c...@qualityexcellence.es)
CC: r-help-es@r-project.org
Asunto: [R-es] PLOTEO DEL DATO MENSUAL DESDE EL ACUMULADO

Hola estimados:
He tratado de encontrar la mejor forma de automatizar un proceso de “datos 
mensuales de precipitación”.
El tema es que tengo solo la serie acumulada anual de Pp. Por ende, busco una 
forma de plotear un gráfico, extrayendo “los datos mensuales desde el acumulado”
Ejemplo;
ene

feb

mar

abr

may

jun

jul

ago

sep

oct

nov

dic



1

5

8

10

20

30

50

80

120

125

200

250

Datos ACUMULADOS

1

4

3

2

10

10

20

30

40

5

75

50

DATOS A PLOTEAR


He buscado por  time series y dapply, pero encuentro que son un poco engorrosos 
sus métodos.
Saludos a todos.

[Descripción: FIRMA3]




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[R] Randomly select one row by group from a matrix

[Marine Regis]

Hello,
I would like to randomly select one row by group from a matrix. Here is an 
example where there is one row by group. The code gives an error message:
test <- matrix(c(4,4, 6,2, 1,2), nrow = 2, ncol = 3, dimnames = list(NULL, 
c("xcor", "ycor", "id")))
do.call(rbind, lapply(split(test, test[,c("id")]), function(x) 
x[sample(nrow(x), 1), ]))
 Show Traceback

 Rerun with Debug

Error in sample.int(length(x), size, replace, prob) :
  invalid first argument


How can I modify the code so that it works when there are several rows or one 
row for a given group?
Thanks very much for your time
Have a nice day
Marine


[[alternative HTML version deleted]]

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Re: [R-es] PLOTEO DEL DATO MENSUAL DESDE EL ACUMULADO

[Javier Marcuzzi]

Estimado Javier Valdes Cantallopts

Hay algo que no comprendo, usted dice que desea extraer los datos mensuales 
desde el acumulado.

¿Sería algo como, el año 2000 llovió 2300 ml? A partir de esto conocer las 
precipitaciones en enero, febrero, etc.

Ahora tengo mis dudas, si usted tiene los datos originales podría tener un 
modelo, y predict(….). ¿Pero, si hay un año o mes con sequía por efecto de la 
Niña o Niño?

Yo soy de Argentina y por mi zona geográfica (más o menos a la mitad de 
Argentina a la altura de Uruguay) hay efectos en las precipitaciones, desde 
sequía  a llover unos 600 ml en un día (una vez, pero recuerdo esos 60 cm de 
lluvia a 7 km de mi casa). Posiblemente se pueda en forma matemática, pero 
tengo mis dudas.

¿No tiene acceso a los registros mensuales o diarios para a partir de estos 
comenzar su trabajo? No soy meteorólogo, pero pienso que debería tener un 
modelo con los meses y una corrección con el efecto del Niño y Niña.

Javier Rubén Marcuzzi


De: Javier Valdes Cantallopts (DGA)
Enviado: jueves, 18 de mayo de 2017 11:57
Para: Carlos Ortega (c...@qualityexcellence.es)
CC: r-help-es@r-project.org
Asunto: [R-es] PLOTEO DEL DATO MENSUAL DESDE EL ACUMULADO

Hola estimados:
He tratado de encontrar la mejor forma de automatizar un proceso de “datos 
mensuales de precipitación”.
El tema es que tengo solo la serie acumulada anual de Pp. Por ende, busco una 
forma de plotear un gráfico, extrayendo “los datos mensuales desde el acumulado”
Ejemplo;
ene
feb
mar
abr
may
jun
jul
ago
sep
oct
nov
dic
 
1
5
8
10
20
30
50
80
120
125
200
250
Datos ACUMULADOS
1
4
3
2
10
10
20
30
40
5
75
50
DATOS A PLOTEAR
He buscado por  time series y dapply, pero encuentro que son un poco engorrosos 
sus métodos.
Saludos a todos.






CONFIDENCIALIDAD: La información contenida en este mensaje y/o en los archivos 
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exclusivo del emisor y/o de la persona o entidad a quien va dirigida. Si usted 
no es el destinatario, cualquier almacenamiento, divulgación, distribución o 
copia de esta información está estrictamente prohibido y sancionado por la ley. 
Si recibió este mensaje por error, por favor infórmenos inmediatamente 
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CONFIDENTIAL NOTE: The information transmitted in this message and/or 
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[R-es] PLOTEO DEL DATO MENSUAL DESDE EL ACUMULADO

[Javier Valdes Cantallopts (DGA)]

Hola estimados:
He tratado de encontrar la mejor forma de automatizar un proceso de "datos 
mensuales de precipitación".
El tema es que tengo solo la serie acumulada anual de Pp. Por ende, busco una 
forma de plotear un gráfico, extrayendo "los datos mensuales desde el acumulado"
Ejemplo;
ene

feb

mar

abr

may

jun

jul

ago

sep

oct

nov

dic



1

5

8

10

20

30

50

80

120

125

200

250

Datos ACUMULADOS

1

4

3

2

10

10

20

30

40

5

75

50

DATOS A PLOTEAR


He buscado por  time series y dapply, pero encuentro que son un poco engorrosos 
sus métodos.
Saludos a todos.

[Descripción: FIRMA3]




CONFIDENCIALIDAD: La información contenida en este mensaje y/o en los archivos 
adjuntos es de carácter confidencial o privilegiada y está destinada al uso 
exclusivo del emisor y/o de la persona o entidad a quien va dirigida. Si usted 
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copia de esta información está estrictamente prohibido y sancionado por la ley. 
Si recibió este mensaje por error, por favor infórmenos inmediatamente 
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Re: [R] violin plot help

[Ismail SEZEN]

> On 16 May 2017, at 17:06, Abdelrahman, Omar (RER) 
>  wrote:
> 
> I am trying to produce multiple violin plots by 3 categorical variables, each 
> violin representing 1 year worth of data. The variables are:
> 
> Watershed (7 levels: county canals)
> 
> Geography (5 levels: west; central; east; mouth; bay)
> 
> Parameter (8 levels: water quality chemical parameters)
> 
> Year (25 levels: 1992-2017)
> 
> I want to produce 1 plot for each Parameter-Watershed subdivided into 
> Geography with a violin for each year. I used facets with the following code 
> (not by year):

Hello Omer,
If you want violin plots for different categories, I would suggest lattice 
solution. For start;

https://www.r-bloggers.com/violin-and-boxplots-with-lattice-and-r/

Search terms: lattice, bwplot, panel.violin will help you find more examples.

> 
> ggplot () +
> 
> facet_grid (PARAMETER ~Wshed, scales="free_y") +
> 
> geom_violin (data=merged, aes(x=Geo, y=RESULT))
> 
> 
> 
> I do not want facets, they crowd the information so it is unreadable. I just 
> started with R this week and have not been able to figure out the foreach 
> protocol, or any other loop protocol. I tried to subset the data to do it 
> iteratively with the following code:
> 
> 
> 
> subdf<-subset (merged, Wshed = "AC")
> 
> 
> 
> but got an error: Error: unexpected input in "subdf=subset (merged, Wshed == 
> ""
> 
> Any help would be greatly appreciated.
> 
> Thanks,
> 
> Omar Abdelrahman, Biologist II
> Miami-Dade County, Department of Regulatory and Economic Resources
> Division of Environmental Resources Management (DERM)
> Overtown Transit Village
> 701 NW 1st Court, 5th Floor
> Miami, FL 33136-3912
> (305) 372-6872
> abd...@miamidade.gov
> www.miamidade.gov/environment
> 

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Re: [R] violin plot help

[Abdelrahman, Omar (RER)]

Many thanks Jeff! I knew it would require a loop approach, so I will now 
explore that with the code you suggested.

-Original Message-
From: Jeff Newmiller [mailto:jdnew...@dcn.davis.ca.us] 
Sent: Wednesday, May 17, 2017 5:19 PM
To: Abdelrahman, Omar (RER) ; R-help 

Subject: RE: [R] violin plot help

Your request is outside of the scope of ggplot2. There are a variety of ways to 
achieve your ends, but they all involve loops of one sort or another... e.g.

wsheds <- unique( merged$Wshed )
for ( w in wsheds ) {
print( ggplot( data=subset( merged, w == Wsheds ), ... ) ) }

I have become quite used to embedding my R code into rmarkdown files which 
allows me to mix multiple graphs, tables and text commentary in one place. 
RStudio makes this about as easy as it can be,  though support for that IDE is 
off topic here. 
--
Sent from my phone. Please excuse my brevity.

On May 17, 2017 12:59:36 PM PDT, "Abdelrahman, Omar (RER)" 
 wrote:
>Thank you, curly quotes got me! I was able to subset the data and 
>produce the violin plot. Now, is there a way to generate multiple plots 
>separately (no facets)? With so many levels of each variable, I am 
>trying to avoid doing it iteratively. Neither ggplot2 books nor web 
>searches have yielded anything (so far).
>Also I want a violin for each year within Geo. I did try to specify 
>year with the following:
>ggplot () +
>facet_grid (PARAMETER ~Wshed~year, scales="free_y") + geom_violin 
>(data=subdf, aes(x=Geo, y=RESULT, fill=Geo))
>
>which yielded
>-Error in combine_vars(data, params$plot_env, cols, drop = params$drop)
>: 
>  At least one layer must contain all variables used for faceting
>
>Also tried:
>ggplot () +
>facet_grid (PARAMETER ~Wshed, scales="free_y") + geom_violin 
>(data=subdf, aes(x=Geo~year, y=RESULT, fill=Geo))
>
>Do I need to specify "year(date)"; I loaded lubridate?
>
>
>-Original Message-
>From: Jeff Newmiller [mailto:jdnew...@dcn.davis.ca.us]
>Sent: Wednesday, May 17, 2017 10:05 AM
>To: Abdelrahman, Omar (RER) 
>Cc: R-help 
>Subject: RE: [R] violin plot help
>
>Here is an example that works... a reproducible example always includes 
>code AND enough sample data to exercise the code:
>
>
>dta <- read.table( text=
>"STATIONGeo Wshed   DATEPARAMETER  
>  RESULT
>BB36Bay C-100   1/10/2013   'Phosphorus, Total
>(TP)'  0.004
>BB36Bay C-100   1/10/2013   'Chlorophyll-A'
>  0.2
>BB52Bay C-100   1/10/2013   'Phosphorus, Total
>(TP)'  0.003
>BB52Bay C-100   1/10/2013   'Chlorophyll-A'
>  0.39
>CD01A   Mouth   C-100   1/10/2013   'Phosphorus, Total
>(TP)'  0.017
>CD01A   Mouth   C-100   1/10/2013   'Chlorophyll-A'
>  0.64
>CD02EastC-100   1/10/2013   'Phosphorus, Total
>(TP)'  0.01
>CD05Central C-100   1/10/2013   'Phosphorus, Total
>(TP)'  0.005
>CD06Central C-100   1/10/2013   'Phosphorus, Total
>(TP)'  0.01
>CD09Central C-100   1/10/2013   'Phosphorus, Total
>(TP)'  0.007
>BB36Bay C-100   2/7/2013'Chlorophyll-A'
>  0.18
>BB36Bay C-100   2/7/2013'Phosphorus, Total
>(TP)'  0.002
>BB52Bay C-100   2/7/2013'Phosphorus, Total
>(TP)'  0.002
>BB52Bay C-100   2/7/2013'Chlorophyll-A'
>  0.31
>CD01A   Mouth   C-100   2/7/2013'Phosphorus, Total
>(TP)'  0.004
>CD01A   Mouth   C-100   2/7/2013'Chlorophyll-A'
>  0.4
>CD02EastC-100   2/7/2013'Phosphorus, Total
>(TP)'  0.011
>CD05Central C-100   2/7/2013'Phosphorus, Total
>(TP)'  0.007
>CD06Central C-100   2/7/2013'Phosphorus, Total
>(TP)'  0.015
>CD09Central C-100   2/7/2013'Phosphorus, Total
>(TP)'  0.008
>CD01A   Mouth   C-100   3/7/2013'Phosphorus, Total
>(TP)'  0.007
>", header=TRUE)
># prints result to console without assigning it to a new variable 
>subset( dta, Geo == "East" ) 
>
>Note that [1] and [2] suggest the use of the dput function to help 
>create R code that creates the object just as you have it before the 
>troublesome line of code:
>
>
>dta <- structure(list(STATION = structure(c(1L, 1L, 2L, 2L, 3L, 3L, 4L, 
>5L, 6L, 7L, 1L, 1L, 2L, 2L, 3L, 3L, 4L, 5L, 6L, 7L, 3L) , .Label = 
>c("BB36", "BB52", "CD01A", "CD02", "CD05", "CD06", "CD09")
>  , class = "factor"),
> Geo = structure(c(1L, 1L, 1L, 1L, 4L, 4L, 3L, 2L, 2L, 2L,
> 1L, 1L, 1L, 1L, 4L, 4L, 3L, 2L, 2L, 2L, 4L), .Label = c("Bay",
> "Central", "East", "Mouth"), class = "factor"),
> Wshed = structure(c(1L,
> 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
> 

Re: [R] Help on reducing multiple loops

[John Kane via R-help]

Data? It's difficult to do anything without some test data.See How to make a 
great R reproducible example? or http://adv-r.had.co.nz/Reproducibility.html 
 with particular reference to the use of dput() as the best way to provide 
sample data.

  
|  
|   
|   
|   ||

   |

  |
|  
|   |  
How to make a great R reproducible example?
 When discussing performance with colleagues, teaching, sending a bug report or 
searching for guidance on mailing...  |   |

  |

  |

 
 

On Wednesday, May 17, 2017 6:10 PM, Sumanta Basak  
wrote:
 

 Hi All,

I've a data-set on product sub-product matrix on which I'm doing multiple
calculation, but unfortunately using nested loops, the programme is taking
long time to execute. Can anyone help me how to get rid of the following
jungle? Any direction would be helpful.

GA <- "India"
verticle <- "Prod1"

prod_data <- readRDS(paste0("/Prod_ladder_",GA,"_",verticle,".rds"))
setDF(prod_data)

Final_data <-
subset(prod_data[,c("P_KEY","Active_Prod_Id","Active_Prod_Nm")],!duplicated(prod_data[,c("P_KEY","Active_Prod_Id")]))

proximity_prod_mapping <- readRDS("Proximity_prod_mapping.rds")
dst_prod <- subset(prod_data[,c("P_KEY")],!duplicated(prod_data$P_KEY))


output_data <- c()
data_merge_final <- c()

system.time({
  for(i in 1 : length(dst_prod)){

    prod_data <- subset(prod_data,prod_data$P_KEY == dst_prod[i]) #
Subsetting data at prod level
    dst_prod <-
subset(prod_data[,c("Active_Prod_Id")],!duplicated(prod_data$Active_Prod_Id))
# Finding distinct prods of active prodloyee

    for(j in 1 : length(dst_prod)){
      # Subsetting data at prod level for active prod
      # Fetiching data for Anchor prod
      prod_data1 <-
subset(prod_data[,c("P_KEY","Active_Prod_Id","Active_Prod_Nm","Start_Date_1","End_Date_1")],prod_data$Active_Prod_Id
== dst_prod[j])
      prod_data1$Anchor_prod <- 1
      anc_max_End_Date_1 <- as.Date(max(prod_data1$End_Date_1),origin =
"1970-01-01")
      anc_prod_count <- sum(prod_data1$Anchor_prod)

      # Fetiching data for Proximate prod
      prox_prod_data <-
subset(proximity_prod_mapping[,c("Proximate_prod_ID")],proximity_prod_mapping$Anchor_prod_ID
== dst_prod[j])
      prod_data2 <-
subset(prod_data[,c("P_KEY","Active_Prod_Id","Active_Prod_Nm","Start_Date_1","End_Date_1")],prod_data$Active_Prod_Id
%in% c(prox_prod_data))
      prox_sill_count <- 0
      if(nrow(prod_data2) > 0){
        prod_data2$Proximity_prod <- 1
        prox_max_End_Date_1 <- as.Date(max(prod_data2$End_Date_1),origin =
"1970-01-01")
        prox_sill_count <- sum(prod_data2$Proximity_prod)
      }
      # library(plyr)
      prod_data <-rbind.fill(prod_data1,prod_data2)
      prod_data$exclude <- 0
      prod_data$Anchor_Active_Prod_Id <- dst_prod[j]

      prod_data$Start_Date_1 <- as.Date(prod_data$Start_Date_1,origin =
"1970-01-01")
      prod_data$End_Date_1 <- as.Date(prod_data$End_Date_1,origin =
"1970-01-01")

      if(prox_sill_count > 0){
        if(nrow(prod_data) > 1){
          # Trimming end date of proximity prods where end data of
proximity prod is greater that Anchor prod
          if((prox_max_End_Date_1 - anc_max_End_Date_1) > 0){
            prod_data$End_Date_1 <- ifelse(prod_data$Proximity_prod == 1 &
(prod_data$End_Date_1 - anc_max_End_Date_1) > 0,
anc_max_End_Date_1,prod_data$End_Date_1)
            prod_data$End_Date_1 <- as.Date(prod_data$End_Date_1,origin =
"1970-01-01")
          }
          prod_data$exclude <- ifelse(prod_data$Proximity_prod == 1 &
(as.Date(prod_data$Start_Date_1,origin = "1970-01-01") -
anc_max_End_Date_1) > 0,1,0)
          prod_data <- subset(prod_data,prod_data$exclude == 0)

          prod_data <-
arrange(prod_data,prod_data$Anchor_prod,desc(prod_data$End_Date_1),prod_data$Start_Date_1)

          prod_data$Anchor_prod <- ifelse(is.na
(prod_data$Anchor_prod),0,prod_data$Anchor_prod)
          prod_data$Proximity_prod <- ifelse(is.na
(prod_data$Proximity_prod),0,prod_data$Proximity_prod)
          prod_data$new_rec <- 0

          tot_loop <- nrow(prod_data)
          k=1
          # Looping to map start date and end date of each row with other
rows
          while(k <= tot_loop){
            excl_flag <- prod_data[k,c("exclude")]
            if(excl_flag == 0){
              st_dt1 <- as.Date(prod_data[k,c("Start_Date_1")])
              end_dt1 <- as.Date(prod_data[k,c("End_Date_1")])
              prod_flag1 <- prod_data[k,c("Anchor_prod")]

              if(k != nrow(prod_data)){
                tot_row <- nrow(prod_data)

                for(m in 1 : (tot_row -k)){
                  l = k+m
                  if(l != k){
                    st_dt2 <- as.Date(prod_data[l,c("Start_Date_1")])
                    end_dt2 <- as.Date(prod_data[l,c("End_Date_1")])
                    prod_flag2 <- prod_data[l,c("Anchor_prod")]

                    flag_excl <- prod_data[l,c("exclude")]
                    if(flag_excl ==0){
                      rec_check <- prod_data[l,c("new_rec")]
          

Re: [R] converting each column of a data frame into a matrix with n rows

[Davide Piffer]

Thanks David! It worked!

On 17 May 2017 at 23:39, David Winsemius  wrote:
>
>> On May 17, 2017, at 1:01 PM, Davide Piffer  wrote:
>>
>> Thanks! This gets closer to the solution but a small problem remains.
>> I get 2 rows and only one column, whereas I need a 2x2 matrix (like a
>> contingency table for Fisher's exact test).Also another issue is it
>> repeats the first number of the column, instead of using all 4.
>> For example, first vector of df is=c(564,3825,125, 377
>>
>> I get:
>>
>> [,1]
>> [1,] 564.3112
>> [2,] 564.3112
>>
>>
>> But I should get
>>
>>   [,1]   [.2]
>> [1,]564   125
>> [2,] 3825  377
>
> ( y <- data.frame(matrix(1:40, 4, 10)) )
> ( y.mat <- lapply(y, matrix, nrow=2))
>
>> head(y.mat, 3)
> $X1
>  [,1] [,2]
> [1,]13
> [2,]24
>
> $X2
>  [,1] [,2]
> [1,]57
> [2,]68
>
> $X3
>  [,1] [,2]
> [1,]9   11
> [2,]   10   12
>
>>
>> On 17 May 2017 at 22:35, David L Carlson  wrote:
>>> Not really enough info here since you don't specify much about the data 
>>> frame or how the results should be provided, but maybe something like this:
>>>
>>> y <- data.frame(matrix(1:100, 10, 10))
>>> y.mat <- lapply(y, matrix, nrow=2)
>>> str(y.mat)
>>> List of 10
>>> $ X1 : int [1:2, 1:5] 1 2 3 4 5 6 7 8 9 10
>>> $ X2 : int [1:2, 1:5] 11 12 13 14 15 16 17 18 19 20
>>> $ X3 : int [1:2, 1:5] 21 22 23 24 25 26 27 28 29 30
>>> $ X4 : int [1:2, 1:5] 31 32 33 34 35 36 37 38 39 40
>>> $ X5 : int [1:2, 1:5] 41 42 43 44 45 46 47 48 49 50
>>> $ X6 : int [1:2, 1:5] 51 52 53 54 55 56 57 58 59 60
>>> $ X7 : int [1:2, 1:5] 61 62 63 64 65 66 67 68 69 70
>>> $ X8 : int [1:2, 1:5] 71 72 73 74 75 76 77 78 79 80
>>> $ X9 : int [1:2, 1:5] 81 82 83 84 85 86 87 88 89 90
>>> $ X10: int [1:2, 1:5] 91 92 93 94 95 96 97 98 99 100
>>>
>>>
>>> -
>>> David L Carlson
>>> Department of Anthropology
>>> Texas A University
>>> College Station, TX 77840-4352
>>>
>>>
>>> -Original Message-
>>> From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Davide 
>>> Piffer
>>> Sent: Wednesday, May 17, 2017 2:18 PM
>>> To: r-help@r-project.org
>>> Subject: [R] converting each column of a data frame into a matrix with n 
>>> rows
>>>
>>> I need to convert each vector of a dataframe into a matrix with 2 rows
>>> and 2 columns (i.e. contingency table).
>>> Note I don't want to convert the entire df into a matrix! I want to
>>> apply a function that converts each 4 elements vector of a df into a 2
>>> x 2 matrix.
>>>
>>> I wrote something like this, but it will not work:
>>>
>>> f_matrix=function(x){ matrix (x)
>>>  nrow=2}
>>> matrix_y=apply(y,2,function(x) f_matrix (x))
>>>
>>> __
>>> 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.
>
> David Winsemius
> Alameda, CA, USA
>

__
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-es] 9 Jornadas R Granada. C4P

[José Luis Cañadas]

Hola.
Os paso información sobre las próximas jornadas de usuarios de R en
Granada.


La Asociación R-Hispano tiene el placer de anunciar la celebración de las
IX Jornadas de Usuarios de R. En esta ocasión, será la Universidad de
Granada , en la ETS de Ingenierías Informática y de Telecomunicación
, quien acoja las Jornadas los días 16 y 17 de
noviembre de 2017.

El objetivo de las Jornadas es proporcionar un punto de encuentro a la
diversidad de usuarios de R por lo que, independientemente de la filiación
o del área de trabajo o investigación, todo usuario de R es bienvenido.

Como en años anteriores, la presentación de trabajos en las Jornadas puede
ser en forma de *Presentación oral, *en distintas sesiones paralelas, que
reflejen el amplio abanico de análisis de datos mediante R, *Póster, en una
sesión específica,* o como un *Taller* que permita profundizar en aspectos
interesantes y menos conocidos de R.

El envío de trabajos ha de hacerse a través de enlace a easychair
habilitado en la web del congreso en http://r-es.org/9jornadasR/
 . En esta edición de las Jornadas también se
otorgará un premio al mejor trabajo presentado por un investigador joven
(ver bases en la web de las Jornadas).

Finalmente, queremos agradecer la colaboración de nuestros patrocinadores,
sin los que las Jornadas no serían posible. Si tú o tu institución estáis
interesados en patrocinar las Jornadas puedes consultar cómo hacerlo en
http://r-es.org/9jornadasR/


Las fechas importantes son:

Envío de comunicaciones (presentación/póster/taller): del 15/04/2017 al
15/06/2017

Decisión sobre envíos: 01/07/2017
Registro: Hasta completar aforo  En https://www.meetup.com/es-ES/
Granada-Geek/events/239294739/

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