[R] subset columns from list with variable substitution

[Jon Ween, MD]

Thanks Don

Please forgive my poor mail-liost etiquette.

I had a couple of errors:

1) the counter logic in the loop was "i in name list", my typo in the post.
2) There were typos in the variable list that caused the loop to crash when they were encountered.

Thanks for your help, really appreciated.

Best

Jon

On 2012-05-25, at 1:43 PM, MacQueen, Don wrote:

> The select argument to subset() is supposed to name the columns you want
> to keep.
> So is the syntax I gave, table[,list1], and it is the correct way when
> list1 is a character vector (which it is).
> 
> Your error message says that at least one of the values in list1 is not
> the name of a column in your data frame.
> 
> 
> So, compare
>  names(table)
> with
>   list1
> (it looks like your data frame doesn't really have the column names you
> think it has)
> 
> If you just type
>  i %in% namelist
> by itself, what do you get? Is it what you are trying to loop over?
> Heck, what do you get from
>  print(i)
> ?
> 
> If you are trying to loop over the names in namelist, you MUST use
>  for (i in namelist)
> the expression  i %in% namelist
> will give you some combination of logical values TRUE and/or FALSE,
> depending on the value of i at the time when you use the expression.
> 
> So if it "doesn't work", it's for some other reason. Also, just saying it
> doesn't work isn't enough information. Please read and follow the posting
> guide (mentioned at the end of every R-help email)
> 
> You are using i as a loop index, not only in
> 
> If I do your i %in% namelist in a fresh R session, I get:
>> i %in% namelist
>  Error in match(x, table, nomatch = 0L) : object 'i' not found
> So you must already have i defined. This is probably confusing the issue.
> 
> 
> 
> 
> Also, since you are assigning a value to table2 inside the loop, the value
> when the loop is done will be whatever it was the last time through the
> loop.
> 
> By the way, table() is a built in R function, so 'table' not a good choice
> for other uses.
> 
> Just for understanding R terminology, a list in R has a special structure.
> You don't have any lists in your example. Your objects list1, list2,
> namelist, group1, and group2 are all character vectors.
> 
> The help page for subset() gives examples of how to use the select
> argument, and specifying a character vector is not one of the ways to use
> it.
> 
> 
> But given how you define list1, the correct "display" of it is
> 
>> list1
>  [1] "a" "b" "c" "d"
> 
> This appears to be what you want, so that's not the explanation.
> 
> Try
>  length(list1)
>  str(list1)
> to learn a little more.
> 
> 
> 
> -- 
> Don MacQueen
> 
> Lawrence Livermore National Laboratory
> 7000 East Ave., L-627
> Livermore, CA 94550
> 925-423-1062
> 
> 
> 
> 
> 
> On 5/25/12 12:32 PM, "jween" <jween at rockwoodclinic.com> wrote:
> 
>> Thanks Don
>> 
>> but 
>> 
>> table[,list1]
>> 
>> did not work either:
>> 
>> Error in `[.data.frame`(table, , list1) : undefined columns selected.
>> 
>> I'm guessing my list (list1) is not structured right? Displaying it has no
>> commas, so the whole list may be taken as a single variable rather than a
>> sequence of variables? I've tried various ways of reformatting (c(),
>> as.list(), etc), but no go.
>> 
>> also
>> 
>> "i in namelist" does not work while
>> "i %in% namelist" does. I don't really reference "i" in any function, only
>> using it in the conditional.
>> 
>> 
>> Any other suggestions?
>> 
>> Thanks
>> 
>> Jon
>> 
>> On 5/25/12 11:09 AM, "jween" <jween at rockwoodclinic.com> wrote:
>> 
>>> Hi there, I would like to use a list variable to select columns in a
>>> subset
>>> from a parent table:
>>> 
>>> I have a data frame "table" with column headers a,b,c,d,e,x,y,z
>>> 
>>> and list variables
>>> 
>>> list1=c("a","b","c","d")
>>> list2=c("a","b","x",y","z")
>>> namelist=c("peter","paul","mary","jane")
>>> group1=c("peter","paul")
>>> group2=c("mary","jane")
>>> 
>>> I would like to subset "table" based on the list variable in a for loop:
>>> 
>>> for (i %in% namelist){
>>>    if (i %in% group1){table2<-subset(table, select=list1)}
>>>    else {{table2<-subset(table, select=list2)}
>>> }
>>> 
>>> the "select=list1" syntax does not work. What would be the correct way
>>> to do
>>> this?
>>> 
>>> Many Thanks
>>> 
>>> Jon
>>> 
>>> 
> 
> 
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