[R] apply question

2013-10-27 Thread Andras Farkas
Dear All,

please help with the following problem:

I have


t -seq(0,24,by=6)
a -600
g -0.05
b -a*exp(-g*t)
  
I would like to establish a vector called z (for example) based on b where the 
results are calculated as :

z -c(a-b[1],b[1]-b[2],b[2]-b[3],b[3]-b[4],b[4]-b[5])

so the results are:

[1]   0.0 155.50907 115.20395  85.34519  63.22527


as always your input is appreciated

Andras 
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Re: [R] apply question

2013-10-27 Thread Patrick Burns

Homework?  A hint is:

?diff

Pat


On 27/10/2013 11:37, Andras Farkas wrote:

Dear All,

please help with the following problem:

I have


t -seq(0,24,by=6)
a -600
g -0.05
b -a*exp(-g*t)

I would like to establish a vector called z (for example) based on b where the 
results are calculated as :

z -c(a-b[1],b[1]-b[2],b[2]-b[3],b[3]-b[4],b[4]-b[5])

so the results are:

[1]   0.0 155.50907 115.20395  85.34519  63.22527


as always your input is appreciated

Andras
[[alternative HTML version deleted]]



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Re: [R] apply question

2013-10-27 Thread Andras Farkas
not homework, but thanks for the hint!

Andras



On Sunday, October 27, 2013 7:51 AM, Patrick Burns pbu...@pburns.seanet.com 
wrote:
 
Homework?  A hint is:

?diff

Pat


On 27/10/2013 11:37, Andras Farkas wrote:
 Dear All,

 please help with the following problem:

 I have


 t -seq(0,24,by=6)
 a -600
 g -0.05
 b -a*exp(-g*t)

 I would like to establish a vector called z (for example) based on b where 
 the results are calculated as :

 z -c(a-b[1],b[1]-b[2],b[2]-b[3],b[3]-b[4],b[4]-b[5])

 so the results are:

 [1]   0.0 155.50907 115.20395  85.34519  63.22527


 as always your input is appreciated

 Andras
     [[alternative HTML version deleted]]



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twitter: @burnsstat @portfolioprobe
http://www.portfolioprobe.com/blog
http://www.burns-stat.com
(home of:
  'Impatient R'
  'The R Inferno'
  'Tao Te Programming')
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[R] apply question

2012-08-21 Thread Chet Seligman
This works, where zz is a dataframe:

for(i in 1:nrow(zz)) {
 zzz[i,1]-paste(zz[i,1],zz[i,2],sep=_)
 }

I would like to use apply to concatentate two columns of text along with
a separator.
How?

Chet

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Re: [R] apply question

2012-08-21 Thread Jeff Newmiller
What is wrong with utilizing the vectorized nature of paste?

 zz[[1]] - paste(zz[[1]],zz[[2]],sep=_)
---
Jeff NewmillerThe .   .  Go Live...
DCN:jdnew...@dcn.davis.ca.usBasics: ##.#.   ##.#.  Live Go...
  Live:   OO#.. Dead: OO#..  Playing
Research Engineer (Solar/BatteriesO.O#.   #.O#.  with
/Software/Embedded Controllers)   .OO#.   .OO#.  rocks...1k
--- 
Sent from my phone. Please excuse my brevity.

Chet Seligman chet.selig...@gmail.com wrote:

This works, where zz is a dataframe:

for(i in 1:nrow(zz)) {
 zzz[i,1]-paste(zz[i,1],zz[i,2],sep=_)
 }

I would like to use apply to concatentate two columns of text along
with
a separator.
How?

Chet

   [[alternative HTML version deleted]]

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Re: [R] apply question

2012-08-21 Thread R. Michael Weylandt
You don't need loops or apply with paste since it's vectorized:

x - data.frame(n = 1:5, l = letters[1:5], stringsAsFactors = FALSE)

paste(x[,1], x[,2], sep = _)

Cheers,
Michael

On Tue, Aug 21, 2012 at 10:57 AM, Chet Seligman chet.selig...@gmail.com wrote:
 This works, where zz is a dataframe:

 for(i in 1:nrow(zz)) {
  zzz[i,1]-paste(zz[i,1],zz[i,2],sep=_)
  }

 I would like to use apply to concatentate two columns of text along with
 a separator.
 How?

 Chet

 [[alternative HTML version deleted]]

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Re: [R] apply question

2012-08-21 Thread Bert Gunter
... or using the handy with()

with(x,paste(n,l,sep=_))

Slightly greater clarity and robustness, perhaps...

-- Bert

On Tue, Aug 21, 2012 at 9:24 AM, R. Michael Weylandt
michael.weyla...@gmail.com wrote:
 You don't need loops or apply with paste since it's vectorized:

 x - data.frame(n = 1:5, l = letters[1:5], stringsAsFactors = FALSE)

 paste(x[,1], x[,2], sep = _)

 Cheers,
 Michael

 On Tue, Aug 21, 2012 at 10:57 AM, Chet Seligman chet.selig...@gmail.com 
 wrote:
 This works, where zz is a dataframe:

 for(i in 1:nrow(zz)) {
  zzz[i,1]-paste(zz[i,1],zz[i,2],sep=_)
  }

 I would like to use apply to concatentate two columns of text along with
 a separator.
 How?

 Chet

 [[alternative HTML version deleted]]

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Internal Contact Info:
Phone: 467-7374
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[R] apply question

2011-09-17 Thread clsnyder
Why does one method work, and the other not?

attach(iris)
head(iris)

mean(Sepal.Length)
[1] 5.84

apply(iris, Sepal.Length, mean)
Error in if (d2 == 0L) { : missing value where TRUE/FALSE needed

I seem to get this error frequently with the apply functions but other
manual methods work normally on the same data. 

TIA

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Re: [R] apply question

2011-09-17 Thread Jorge I Velez
Hi clsnyder,

You forgot the MARGIN argument in you function call.  Take a look at ?apply

apply(iris, 2, length)
apply(iris, 2, mean)

HTH,
Jorge
*

*

On Sat, Sep 17, 2011 at 6:42 PM, clsnyder  wrote:

 Why does one method work, and the other not?

 attach(iris)
 head(iris)

 mean(Sepal.Length)
 [1] 5.84

 apply(iris, Sepal.Length, mean)
 Error in if (d2 == 0L) { : missing value where TRUE/FALSE needed

 I seem to get this error frequently with the apply functions but other
 manual methods work normally on the same data.

 TIA

 --
 View this message in context:
 http://r.789695.n4.nabble.com/apply-question-tp3821122p3821122.html
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Re: [R] apply question

2010-05-03 Thread Petr PIKAL
Hi

r-help-boun...@r-project.org napsal dne 30.04.2010 23:11:54:

 Hello David,
 
 On Apr 30, 2010, at 11:00 PM, David Winsemius wrote:
  Note: Loops may be just as fast or faster than apply calls.
  
 How come!? is this true also for other similar functions: lapply, tapply 
and sapply?
 
 Then the only advantage of these above is only syntactic sugar? 

Yes and no. If your loop(s) is poorly written it can be much slower then 
*apply operator but usually not from loop point of view but from the code 
itself. 

Regards
Petr

 
  
  indices - replicate(B,sample(1:N, size=N, replace=TRUE))
  theta_star - apply(indices,2,statistic,data)
  
  Why not:
  
  theta_star - apply(indices,2, function(x) statistic(data, x) )
  
  May end up as a data class that needs further work, depending on what 
 statistic return.
  
 Nice! thank you!
 
 Best regards,
 Giovanni
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[R] apply question

2010-04-30 Thread Giovanni Azua
Hello,

I have a bootstrap implementation loop that I would like to replace by a faster 
batch operator.

The loop is as follows:

for (b in 1:B) {   
  indices - sample(1:N, size=N, replace=TRUE) # sample n elements with 
replacement 
  theta_star[b,] = statistic(data,indices) # execute statistic callback 
function
}

I would like to replace this by something like (does not work):

indices - replicate(B,sample(1:N, size=N, replace=TRUE))
theta_star - apply(indices,2,statistic,data)

This, however, does not work because the statistic function receives indices as 
second parameter and not as first parameter. While I could change the signature 
of my statistic function I would like to know how to achieve this without 
having to do so.

TIA,
Best regards,
Giovanni
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Re: [R] apply question

2010-04-30 Thread David Winsemius


On Apr 30, 2010, at 4:52 PM, Giovanni Azua wrote:


Hello,

I have a bootstrap implementation loop that I would like to replace  
by a faster batch operator.


The loop is as follows:

for (b in 1:B) {
 indices - sample(1:N, size=N, replace=TRUE) # sample n elements  
with replacement
 theta_star[b,] = statistic(data,indices) # execute statistic  
callback function

}

I would like to replace this by something like (does not work):


Note: Loops may be just as fast or faster than apply calls.



indices - replicate(B,sample(1:N, size=N, replace=TRUE))
theta_star - apply(indices,2,statistic,data)


Why not:

theta_star - apply(indices,2, function(x) statistic(data, x) )

May end up as a data class that needs further work, depending on what  
statistic return.




This, however, does not work because the statistic function receives  
indices as second parameter and not as first parameter. While I  
could change the signature of my statistic function I would like to  
know how to achieve this without having to do so.


TIA,
Best regards,
Giovanni
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David Winsemius, MD
West Hartford, CT

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Re: [R] apply question

2010-04-30 Thread Giovanni Azua
Hello David,

On Apr 30, 2010, at 11:00 PM, David Winsemius wrote:
 Note: Loops may be just as fast or faster than apply calls.
 
How come!? is this true also for other similar functions: lapply, tapply and 
sapply?

Then the only advantage of these above is only syntactic sugar? 

 
 indices - replicate(B,sample(1:N, size=N, replace=TRUE))
 theta_star - apply(indices,2,statistic,data)
 
 Why not:
 
 theta_star - apply(indices,2, function(x) statistic(data, x) )
 
 May end up as a data class that needs further work, depending on what 
 statistic return.
 
Nice! thank you!

Best regards,
Giovanni
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