Dear Matt,
You're absolutely right. The reason why my code gives different results is
because it calculates the standard deviation for each row/column in all the
arrays rather than for every cell. My bad.
You can easily get the results you posted running
>apply(p,1:2,sd,na.rm=TRUE)
Here is another option which gives the same results you posted :-)
Unfortunately there is not timing improvement :( (Note that I ran both
sd1 and sd2 functions using pp instead of p)
# -------------
# System times
# -------------
> pp <- array(c(1:5, rep(NA, times = 3)), dim = c(1000, 1000, 60))
> system.time(apply(pp, 1:2, sd1))
# user system elapsed
# 93.16 0.23 94.87
> system.time(apply(pp, 1:2, sd2))
# user system elapsed
# 66.51 0.30 67.84
# ----------
# Functions
# ----------
sd1<-function(x) sd(x,na.rm=TRUE)
sd2<- function(i){
if(sum(!is.na(i))==0) temp.sd <- NA
else temp.sd <- sd(i, na.rm = TRUE)
temp.sd
}
Regards,
Jorge
On Wed, Feb 25, 2009 at 3:23 PM, Matt Oliver <moli...@udel.edu> wrote:
> Hi Jorge, this does not seem to return the same thing as
>
> p <- array(c(1:5, rep(NA, times = 3)), dim = c(5, 5, 3))
> sd_fun <- function(i){
> if(sum(!is.na(i))==0){
> temp.sd <- NA
> }else{
> temp.sd <- sd(i, na.rm = TRUE)
> }
> return(temp.sd)
> }
>
> apply(p, 1:2, sd_fun)
>
> Am I missing something basic here?
>
>
>
> On Wed, Feb 25, 2009 at 11:47 AM, Jorge Ivan Velez <
> jorgeivanve...@gmail.com> wrote:
>
>>
>> Hi Mark,
>> There is a typo in the first way. My apologies. It should be:
>> # First
>> res<-apply(p,3,function(X)
>> c(scols=apply(X,2,sd,na.rm=TRUE),srows=apply(X,1,sd,na.rm=TRUE))
>> )
>> res
>>
>> HTH,
>>
>> Jorge
>>
>>
>> On Wed, Feb 25, 2009 at 11:42 AM, Jorge Ivan Velez <
>> jorgeivanve...@gmail.com> wrote:
>>
>>>
>>> Dear Matt,
>>>
>>> Here are two ways:
>>>
>>> # Data
>>> p <- array(c(1:5, rep(NA, times = 3)), dim = c(5, 5, 3))
>>>
>>> # First
>>> res<-apply(p,3,function(X)
>>> c(scols=apply(X,2,sd,na.rm=TRUE),srows=apply(X,2,sd,na.rm=TRUE))
>>> )
>>> res
>>>
>>> # Second
>>> res2<-apply(p,3,function(X)
>>>
>>> list(scols=apply(X,2,sd,na.rm=TRUE),srows=apply(X,1,sd,na.rm=TRUE))
>>> )
>>>
>>> lapply(res2,function(x) do.call(rbind,x))
>>>
>>> HTH,
>>>
>>> Jorge
>>>
>>>
>>> On Wed, Feb 25, 2009 at 10:53 AM, Matt Oliver <moli...@udel.edu> wrote:
>>>
>>>> Dear help, suppose I have this array and want to compute sd aross rows
>>>> and
>>>> columns.
>>>>
>>>> p <- array(c(1:5, rep(NA, times = 3)), dim = c(5, 5, 3))
>>>>
>>>> apply(p, 1:2, sd) fails because sd requires at least 2 numbers to
>>>> compute sd
>>>>
>>>> apply(p, 1:2, sd, na.rm = TRUE) fails for the same reason
>>>>
>>>> I crafted my own function that does what I want
>>>>
>>>> sd_fun <- function(i){
>>>> if(sum(!is.na(i))==0){
>>>> temp.sd <- NA
>>>> }else{
>>>> temp.sd <- sd(i, na.rm = TRUE)
>>>> }
>>>> return(temp.sd)
>>>> }
>>>>
>>>>
>>>> apply(p, 1:2, sd_fun)
>>>>
>>>> This does what I want, but when I scale up to large arrays like
>>>>
>>>> pp <- array(c(1:5, rep(NA, times = 3)), dim = c(1000, 1000, 60))
>>>>
>>>> the apply function takes a long time to run.
>>>>
>>>> Is there a faster, more efficient way to do this?
>>>>
>>>> Thanks in advance
>>>>
>>>> Matt
>>>>
>>>>
>>>> --
>>>> Matthew J. Oliver
>>>> Assistant Professor
>>>> College of Marine and Earth Studies
>>>> University of Delaware
>>>> 700 Pilottown Rd.
>>>> Lewes, DE, 19958
>>>> 302-645-4079
>>>> http://www.ocean.udel.edu/people/profile.aspx?moliver
>>>>
>>>> [[alternative HTML version deleted]]
>>>>
>>>> ______________________________________________
>>>> R-help@r-project.org mailing list
>>>> 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.
>>>>
>>>
>>>
>>
>
>
> --
> Matthew J. Oliver
> Assistant Professor
> College of Marine and Earth Studies
> University of Delaware
> 700 Pilottown Rd.
> Lewes, DE, 19958
> 302-645-4079
> http://www.ocean.udel.edu/people/profile.aspx?moliver
>
>
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