[R] Subset and order at the same time?

[jim holtman]

Just write a function so that you have a "one-liner" in your script.
It will probably be a lot simpler than trying to type some convoluted
one-liner.

On Sat, May 12, 2012 at 2:58 PM, Noah Silverman <noahsilverman at ucla.edu> wrote:
> Bernard,
>
> Thanks, but I can't take that shortcut.
>
> The data is an xts object, and I may not want to order every group.  So, I need a way to just order one group at a time.
>
> Thoughts?
>
> --
> Noah Silverman
> UCLA Department of Statistics
> 8208 Math Sciences Building
> Los Angeles, CA 90095
>
> On May 12, 2012, at 11:38 AM, Berend Hasselman wrote:
>
>>
>> On 12-05-2012, at 20:04, Noah Silverman wrote:
>>
>>> Is there a way to order data and subset it at the same time??
>>>
>>>
>>>
>>> I want to sort all the members of group A by their values in column 3. (I'll then do the same for each subsequent group.)  This could be done in a loop building up another vector, but I like to avoid loops in R.
>>> -------------------------------------
>>> a <- temp[temp$group=="A",]
>>> a <- a[order(a[,3]),]
>>> temp[temp$group=="A",] <- a
>>> ------------------------------------------
>>>
>>> Iid like to do this in a single step for each group.  However, I can't figure out how to order and subset at the same time.
>>>
>>> This *does not work* but should illustrate what I'm trying to do
>>>
>>> temp[temp$group=="A",] <- temp[ temp$group=="A" & order(temp[temp$group=="A",3]) , ]
>>>
>>>
>>> Suggestions?
>>
>> set.seed(413)
>>
>> temp <- data.frame(group=rep(c("A","B","C"), rep=5), tt=1:15,val=round(runif(15),2), stringsAsFactors=FALSE)
>> idx <- order(temp$group,temp$val)
>> # or   idx <- order(temp[,1],temp[,3])
>> idx
>> z2 <- temp[idx,]
>> rownames(z2) <- NULL
>> z2
>>
>> Possible NA's etc. not taken into account.
>>
>> Berend
>
>
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>
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-- 
Jim Holtman
Data Munger Guru

What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.