date:20170108

Re: [R] using if else function to complete a column in data frame

2017-01-08 Thread David Winsemius


> On Jan 8, 2017, at 9:21 PM, Cacique Samurai  wrote:
> 
> Hello all!
> 
> I´m trying to complete the "movimento" column in dataframe based in
> the values of "kmr" column in two sequential lines, as below:
> 
> data example (dput in the end of email):
> 
> IDkmr movimento
> 510.700 314.20NA
> 110.700 278.74NA
> 210.700 278.74NA
> 310.700 278.74NA
> 410.700 278.74NA
> 494 100.700 269.94NA
> 500 100.700 278.74NA
> 499 100.700 314.20NA
> 495 100.700 278.74NA
> 498 100.700 278.74NA
> 496 100.700 255.40NA
> 497 100.700 255.10NA
> 
> Once I have different IDs, I wrote this function:
> 
> move = function (x){
> 
>  for (j in x$ID){
> 
>for (i in 2:length(x$kmr)-1){
> 
>  if (x$kmr[i+1]  < x$kmr[i]) {
>x$movimento[i+1] <- "jusante"
>  } else if (x$kmr[i+1] > x$kmr[i]) {
>x$movimento[i+1] <- "montante"
>  } else {
>x$movimento[i+1] <- "parado"
>  }
> 
>}
> 
>  }
> 
>  return (x)
> }
> 
> Worked pretty well with just one ID, but with many IDs the function
> didn´t detach different IDs.
> 
> IDkmr movimento
> 510.700 314.20  
> 110.700 278.74   jusante
> 210.700 278.74parado
> 310.700 278.74parado
> 410.700 278.74parado
> 494 100.700 269.94   jusante <-- this should be 

The inner loop was not "recognizing" (or more accurately you were not causing 
the code to account for the fact) that you wanted this to be done within values 
of ID. Each pass of the outer loop was doing the same process inside the inner 
loop.


> 500 100.700 278.74  montante
> 499 100.700 314.20  montante
> 495 100.700 278.74   jusante
> 498 100.700 278.74parado
> 496 100.700 255.40   jusante
> 497 100.700 255.10   jusante
> 
> I also tried remove the first If condition and pass this function
> using lapply in the splitted original data-frame, but didn´t work as
> well.
> 
> Some onde can help?
> 
> Thanks in advanced,
> 
> Raoni
> 
> structure(list(ID = c("10.700", "10.700", "10.700", "10.700",
> "10.700", "100.700", "100.700", "100.700", "100.700", "100.700",
> "100.700", "100.700"), kmr = c(314.2, 278.74, 278.74, 278.74,
> 278.74, 269.94, 278.74, 314.2, 278.74, 278.74, 255.4, 255.1),
>movimento = c(NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA,
>NA)), .Names = c("ID", "kmr", "movimento"), row.names = c(5L,
> 1L, 2L, 3L, 4L, 494L, 500L, 499L, 495L, 498L, 496L, 497L), class = 
> "data.frame")
> 
> -- 
> Raoni Rosa Rodrigues
> Research Associate of Fish Transposition Center CTPeixes
> Universidade Federal de Minas Gerais - UFMG
> Brasil
> rodrigues.ra...@gmail.com
> 
> __
> 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

__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Re: [R] using if else function to complete a column in data frame

2017-01-08 Thread PIKAL Petr

Hi

Split option seems to me the most effective and ifelse is in this case not 
necessary.

You can use this function to estimate levels

fff <- function(x) sign(diff(x))
test.s <- split(test, test$ID)
for (i in 1:length(test.s)) test.s[[i]]$movimento[-1] <- fff(test.s[[i]][,2])
test <- do.call(rbind, test.s)
test[,3] <- factor(test[,3], levels=c(-1, 0, 1), labels=c("jusante", "parado", 
"montante"))
test

To avoid extension of row names you can use
library (plyr)
test<-ldply (test.s, data.frame)

instead of do.call.

I tried assigning factor within for cycle

fff<- function(x) factor(sign(diff(x)), levels=c(-1, 0, 1), labels=c("jusante", 
"parado", "montante"))
test.s <- split(test, test$ID)
for (i in 1:length(test.s)) test.s[[i]]$movimento[-1] <- fff(test.s[[i]][,2])
test <- do.call(rbind, test.s)
test

but assigned in this case is not factor but numeric vector, which seems to me 
strange. Maybe somebody could explain this behaviour

Cheers
Petr

> -Original Message-
> From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Cacique
> Samurai
> Sent: Monday, January 9, 2017 6:22 AM
> To: R help 
> Subject: [R] using if else function to complete a column in data frame
>
> Hello all!
>
> I´m trying to complete the "movimento" column in dataframe based in the
> values of "kmr" column in two sequential lines, as below:
>
> data example (dput in the end of email):
>
>  IDkmr movimento
> 510.700 314.20NA
> 110.700 278.74NA
> 210.700 278.74NA
> 310.700 278.74NA
> 410.700 278.74NA
> 494 100.700 269.94NA
> 500 100.700 278.74NA
> 499 100.700 314.20NA
> 495 100.700 278.74NA
> 498 100.700 278.74NA
> 496 100.700 255.40NA
> 497 100.700 255.10NA
>
> Once I have different IDs, I wrote this function:
>
> move = function (x){
>
>   for (j in x$ID){
>
> for (i in 2:length(x$kmr)-1){
>
>   if (x$kmr[i+1]  < x$kmr[i]) {
> x$movimento[i+1] <- "jusante"
>   } else if (x$kmr[i+1] > x$kmr[i]) {
> x$movimento[i+1] <- "montante"
>   } else {
> x$movimento[i+1] <- "parado"
>   }
>
> }
>
>   }
>
>   return (x)
> }
>
> Worked pretty well with just one ID, but with many IDs the function didn´t
> detach different IDs.
>
>  IDkmr movimento
> 510.700 314.20  
> 110.700 278.74   jusante
> 210.700 278.74parado
> 310.700 278.74parado
> 410.700 278.74parado
> 494 100.700 269.94   jusante <-- this should be 
> 500 100.700 278.74  montante
> 499 100.700 314.20  montante
> 495 100.700 278.74   jusante
> 498 100.700 278.74parado
> 496 100.700 255.40   jusante
> 497 100.700 255.10   jusante
>
> I also tried remove the first If condition and pass this function using 
> lapply in
> the splitted original data-frame, but didn´t work as well.
>
> Some onde can help?
>
> Thanks in advanced,
>
> Raoni
>
> structure(list(ID = c("10.700", "10.700", "10.700", "10.700", "10.700",
> "100.700", "100.700", "100.700", "100.700", "100.700", "100.700", "100.700"),
> kmr = c(314.2, 278.74, 278.74, 278.74, 278.74, 269.94, 278.74, 314.2, 278.74,
> 278.74, 255.4, 255.1),
> movimento = c(NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA,
> NA)), .Names = c("ID", "kmr", "movimento"), row.names = c(5L, 1L, 2L, 3L,
> 4L, 494L, 500L, 499L, 495L, 498L, 496L, 497L), class = "data.frame")
>
> --
> Raoni Rosa Rodrigues
> Research Associate of Fish Transposition Center CTPeixes Universidade
> Federal de Minas Gerais - UFMG Brasil rodrigues.ra...@gmail.com
>
> __
> 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] R install/libraries

2017-01-08 Thread Chris

I'd appreciate a tip on (re-)installing libraries for R on my windows machine 
I have a windows (Windows 7)  computer and a recent version of R installed 
(3.3.2).  I discovered that when I install R libraries they install to the "my 
documents" folder rather than to "program files"This seems to also create 
problems installing the dependent libraries when a library installs.
I'm presuming that R libraries are not installing to "program files" due to 
some permission/administrator privilege problems.
I have tried manually copying files from my documents to program files which 
didn't seem to completely fix the problem.
I also installed "r studio" but that hasn't resolved the problem of libraries 
not finding other libraries it depends on.A specific example, typically I use 
Frank Harrel's HMISC, however in my current installation it doesn't find 
"openssl". 
installation tips appreciated.
 Chris Barker, Ph.D.
Adjunct Associate Professor of Biostatistics - UIC-SPH
and
President and Owner
Statistical Planning and Analysis Services, Inc.
www.barkerstats.com
415 609 7473 
skype: barkerstats


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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

[R] gridExtra-arrangeGrob

2017-01-08 Thread Felipe Carrillo via R-help

 Hi;The code below used to work on my older version of gridExtra but doesn't 
work with the new version. Could someonegive me a hint on how to translate this 
code to the new version of gridExtra code? Thank you beforehand.
p1 <- ggplot(iris,aes(Sepal.Length,  Petal.Length, colour=Species)) +
geom_point() + theme_bw() + theme(legend.position='top')

 grid.arrange(p1, arrangeGrob(p1,p1,p1, heights=c(0.33, .33,.33), ncol=1), 
ncol=2)
 #Create 2 columns with different width using the 'widths' argument in the 
grid.arrange call
  grid.arrange(p1, arrangeGrob(p1,p1,p1, heights=c(0.33, .40,.27), ncol=1), 
ncol=2,widths=c(1.25,0.75))
p <- rectGrob()  
 grid.arrange(p, arrangeGrob(p,p,p, heights=c(0.33, .33,.33), ncol=1), ncol=2)

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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

[R] using if else function to complete a column in data frame

2017-01-08 Thread Cacique Samurai

Hello all!

I´m trying to complete the "movimento" column in dataframe based in
the values of "kmr" column in two sequential lines, as below:

data example (dput in the end of email):

 IDkmr movimento
510.700 314.20NA
110.700 278.74NA
210.700 278.74NA
310.700 278.74NA
410.700 278.74NA
494 100.700 269.94NA
500 100.700 278.74NA
499 100.700 314.20NA
495 100.700 278.74NA
498 100.700 278.74NA
496 100.700 255.40NA
497 100.700 255.10NA

Once I have different IDs, I wrote this function:

move = function (x){

  for (j in x$ID){

for (i in 2:length(x$kmr)-1){

  if (x$kmr[i+1]  < x$kmr[i]) {
x$movimento[i+1] <- "jusante"
  } else if (x$kmr[i+1] > x$kmr[i]) {
x$movimento[i+1] <- "montante"
  } else {
x$movimento[i+1] <- "parado"
  }

}

  }

  return (x)
}

Worked pretty well with just one ID, but with many IDs the function
didn´t detach different IDs.

 IDkmr movimento
510.700 314.20  
110.700 278.74   jusante
210.700 278.74parado
310.700 278.74parado
410.700 278.74parado
494 100.700 269.94   jusante <-- this should be 
500 100.700 278.74  montante
499 100.700 314.20  montante
495 100.700 278.74   jusante
498 100.700 278.74parado
496 100.700 255.40   jusante
497 100.700 255.10   jusante

I also tried remove the first If condition and pass this function
using lapply in the splitted original data-frame, but didn´t work as
well.

Some onde can help?

Thanks in advanced,

Raoni

structure(list(ID = c("10.700", "10.700", "10.700", "10.700",
"10.700", "100.700", "100.700", "100.700", "100.700", "100.700",
"100.700", "100.700"), kmr = c(314.2, 278.74, 278.74, 278.74,
278.74, 269.94, 278.74, 314.2, 278.74, 278.74, 255.4, 255.1),
movimento = c(NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA,
NA)), .Names = c("ID", "kmr", "movimento"), row.names = c(5L,
1L, 2L, 3L, 4L, 494L, 500L, 499L, 495L, 498L, 496L, 497L), class = "data.frame")

-- 
Raoni Rosa Rodrigues
Research Associate of Fish Transposition Center CTPeixes
Universidade Federal de Minas Gerais - UFMG
Brasil
rodrigues.ra...@gmail.com

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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Re: [R] Linear optimization with quadratic constraints

2017-01-08 Thread ProfJCNash

Small example code to set up the problem?

JN

On 2017-01-07 06:26 AM, Preetam Pal wrote:
> Hi Guys,
> Any help with this,please?
> Regards,
> Preetam
> 
> On Thu, Jan 5, 2017 at 4:09 AM, Preetam Pal  wrote:
> 
>> Hello guys,
>>
>> The context is ordinary multivariate regression with k (>1) regressors,
>> i.e. *Y = XB + Error*, where
>> Y = n X 1 vector of predicted variable,
>> X = n X (k + 1) matrix of regressor variables(including ones in the first
>> column)
>> B = (k+1) vector of coefficients, including intercept.
>>
>> Say, I have already estimated B as B_hat = (X'X)^(-1) X'Y.
>>
>> I have to solve the following program:
>>
>> *minimize f(B) = LB*   ( L is a fixed vector 1 X (k+1)   )
>> such that:
>> *[(B-B_hat)' * X'X * (B-B_hat) ] / [ ( Y - XB_hat)' (Y - XB_hat) ] *  is
>> less than a given value *c*.
>>
>> Note that this is a linear optimization program *with respect to B* with
>> quadratic constraints.
>>
>> I don't understand how we can solve this optimization - I was going
>> through some online resources, each of which involve manually computing
>> gradients of the objective as well as constraint functions - which I want
>> to avoid (at least manually doing this).
>>
>>
>> Can you please help with solving this optimization problem? The inputs
>> would be:
>>
>>- X and Y
>>- B_hat
>>- L
>>- c
>>
>>
>> Please let me know if any further information is required - the set-up is
>> pretty general.
>>
>> Regards,
>> Preetam
>>
> 
> 
>

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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Re: [R] using if else function to complete a column in data frame

Re: [R] using if else function to complete a column in data frame

[R] R install/libraries

[R] gridExtra-arrangeGrob

[R] using if else function to complete a column in data frame

Re: [R] Linear optimization with quadratic constraints

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