2) There were typos in the variable list that caused the loop to crash when they were encountered.
> 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" <
[hidden email]> 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" <
[hidden email]> 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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