[R] Extracting elements from a nested list

Tue Oct 19 01:52:18 CEST 2010

Does this do what you want:

> x <- lapply(seq_along(MaxGrowth), function(.num){
+     AllPredictedValues[[.num]][[MaxGrowth[[.num]]]]
+ })
> x
[[1]]
  [1] 2 1 2 2 2 2 0 2 2 0 0 2 2 0 0 2 2 1 2 0 1 1 0 0 0 2 0 0 0 2 2 0
0 1 0 0 2 0 1 0 2 0 0 2 1 0 0 0 2 1 0 2 2
 [54] 2 2 0 2 0 1 0 2 0 1 0 0 0 1 0 0 2 2 2 2 2 2 1 0 2 0 2 0 0 0 0 2
0 2 1 2 2 0 2 0 0 0 0 1 2 2 1
Levels: 0 1 2

[[2]]
  [1] 0 1 0 1 0 1 0 0 1 0 0 1 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0
0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
 [54] 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1
0 0 0 0 0 0 0 0 0 0 0 0 1 0 1
Levels: 0 1 2

[[3]]
  [1] 2 0 2 0 2 2 0 2 0 0 0 2 2 1 0 2 2 0 2 0 1 1 0 0 0 0 0 0 0 2 2 0
0 1 0 0 2 0 1 0 2 0 0 2 0 1 1 0 2 1 0 2 2
 [54] 2 2 0 2 0 0 0 2 0 0 0 0 0 0 1 0 2 2 2 2 2 2 0 0 2 0 2 0 0 0 1 0
0 2 0 2 2 0 2 0 0 0 0 0 2 2 1
Levels: 0 1 2

>

On Mon, Oct 18, 2010 at 5:36 PM, Gregory Ryslik <rsaber at comcast.net> wrote:
> Hi,
>
> It seems that the files did not make it through the mailer. Perhaps it didn't like my extensions. I have now attached the files as .txt's as well as copied in the contents of each file:
>
>
>
>
> MaxGrowth.txt:
>
> list(4L, 3L, 4L)
>
> AllPredictedValues.txt
>
> list(list(structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
> 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
> 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
> 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
> 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
> 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
> 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = c("0", "1",
> "2"), class = "factor"), structure(c(3L, 1L, 3L, 3L, 3L, 3L,
> 3L, 3L, 3L, 1L, 3L, 3L, 3L, 1L, 3L, 3L, 3L, 1L, 3L, 1L, 1L, 1L,
> 1L, 1L, 1L, 3L, 1L, 1L, 1L, 3L, 3L, 1L, 3L, 1L, 3L, 3L, 3L, 3L,
> 1L, 3L, 3L, 3L, 1L, 3L, 1L, 1L, 1L, 1L, 3L, 1L, 1L, 3L, 3L, 3L,
> 3L, 3L, 3L, 3L, 1L, 3L, 3L, 3L, 1L, 3L, 3L, 3L, 1L, 1L, 3L, 3L,
> 3L, 3L, 3L, 3L, 3L, 1L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 1L, 3L, 1L,
> 3L, 1L, 3L, 3L, 1L, 3L, 3L, 1L, 3L, 3L, 1L, 3L, 3L, 1L), .Label = c("0",
> "1", "2"), class = "factor"), structure(c(3L, 2L, 3L, 3L, 3L,
> 3L, 3L, 3L, 3L, 1L, 3L, 3L, 3L, 1L, 3L, 3L, 3L, 2L, 3L, 1L, 2L,
> 2L, 1L, 1L, 1L, 3L, 1L, 1L, 1L, 3L, 3L, 1L, 3L, 2L, 3L, 3L, 3L,
> 3L, 2L, 3L, 3L, 3L, 1L, 3L, 2L, 1L, 1L, 1L, 3L, 2L, 1L, 3L, 3L,
> 3L, 3L, 3L, 3L, 3L, 2L, 3L, 3L, 3L, 2L, 3L, 3L, 3L, 2L, 1L, 3L,
> 3L, 3L, 3L, 3L, 3L, 3L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 1L, 3L,
> 1L, 3L, 2L, 3L, 3L, 1L, 3L, 3L, 1L, 3L, 3L, 2L, 3L, 3L, 2L), .Label = c("0",
> "1", "2"), class = "factor"), structure(c(3L, 2L, 3L, 3L, 3L,
> 3L, 1L, 3L, 3L, 1L, 1L, 3L, 3L, 1L, 1L, 3L, 3L, 2L, 3L, 1L, 2L,
> 2L, 1L, 1L, 1L, 3L, 1L, 1L, 1L, 3L, 3L, 1L, 1L, 2L, 1L, 1L, 3L,
> 1L, 2L, 1L, 3L, 1L, 1L, 3L, 2L, 1L, 1L, 1L, 3L, 2L, 1L, 3L, 3L,
> 3L, 3L, 1L, 3L, 1L, 2L, 1L, 3L, 1L, 2L, 1L, 1L, 1L, 2L, 1L, 1L,
> 3L, 3L, 3L, 3L, 3L, 3L, 2L, 1L, 3L, 1L, 3L, 1L, 1L, 1L, 1L, 3L,
> 1L, 3L, 2L, 3L, 3L, 1L, 3L, 1L, 1L, 1L, 1L, 2L, 3L, 3L, 2L), .Label = c("0",
> "1", "2"), class = "factor")), list(structure(c(2L, 2L, 2L, 2L,
> 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,
> 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,
> 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,
> 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,
> 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,
> 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L
> ), .Label = c("0", "1", "2"), class = "factor"), structure(c(2L,
> 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 2L, 2L, 2L, 1L, 2L, 2L, 2L,
> 1L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 1L, 1L, 1L, 2L, 2L, 1L, 2L,
> 2L, 2L, 2L, 2L, 2L, 1L, 2L, 2L, 2L, 1L, 2L, 1L, 1L, 1L, 1L, 2L,
> 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 2L, 2L, 2L, 1L, 2L, 2L,
> 2L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 2L, 2L, 2L, 2L, 2L,
> 2L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 2L, 1L, 2L, 2L, 1L, 2L, 2L, 1L,
> 2L, 2L, 2L), .Label = c("0", "1", "2"), class = "factor"), structure(c(1L,
> 2L, 1L, 2L, 1L, 2L, 1L, 1L, 2L, 1L, 1L, 2L, 1L, 1L, 1L, 2L, 1L,
> 1L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
> 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
> 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
> 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
> 1L, 1L, 1L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
> 2L, 1L, 2L), .Label = c("0", "1", "2"), class = "factor")), list(
>    structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
>    1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
>    1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
>    1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
>    1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
>    1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
>    1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = c("0",
>    "1", "2"), class = "factor"), structure(c(3L, 2L, 3L, 2L,
>    3L, 3L, 3L, 3L, 2L, 2L, 3L, 3L, 3L, 2L, 3L, 3L, 3L, 2L, 3L,
>    2L, 2L, 2L, 2L, 2L, 2L, 3L, 2L, 2L, 2L, 3L, 3L, 2L, 3L, 2L,
>    3L, 3L, 3L, 3L, 2L, 3L, 3L, 3L, 2L, 3L, 2L, 2L, 2L, 2L, 3L,
>    2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 2L, 3L, 3L, 3L, 2L, 3L,
>    3L, 3L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 2L, 3L, 3L, 3L,
>    3L, 3L, 3L, 3L, 2L, 2L, 2L, 3L, 2L, 3L, 3L, 2L, 3L, 3L, 2L,
>    3L, 3L, 2L, 3L, 3L, 2L), .Label = c("0", "1", "2"), class = "factor"),
>    structure(c(3L, 2L, 3L, 2L, 3L, 3L, 1L, 3L, 2L, 2L, 1L, 3L,
>    3L, 2L, 1L, 3L, 3L, 2L, 3L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 2L,
>    2L, 2L, 3L, 3L, 2L, 1L, 2L, 1L, 1L, 3L, 1L, 2L, 1L, 3L, 1L,
>    2L, 3L, 2L, 2L, 2L, 2L, 3L, 2L, 2L, 3L, 3L, 3L, 3L, 1L, 3L,
>    1L, 2L, 1L, 3L, 1L, 2L, 1L, 1L, 1L, 2L, 2L, 1L, 3L, 3L, 3L,
>    3L, 3L, 3L, 2L, 1L, 3L, 1L, 3L, 1L, 1L, 1L, 2L, 2L, 2L, 3L,
>    2L, 3L, 3L, 2L, 3L, 1L, 2L, 1L, 1L, 2L, 3L, 3L, 2L), .Label = c("0",
>    "1", "2"), class = "factor"), structure(c(3L, 1L, 3L, 1L,
>    3L, 3L, 1L, 3L, 1L, 1L, 1L, 3L, 3L, 2L, 1L, 3L, 3L, 1L, 3L,
>    1L, 2L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 3L, 3L, 1L, 1L, 2L,
>    1L, 1L, 3L, 1L, 2L, 1L, 3L, 1L, 1L, 3L, 1L, 2L, 2L, 1L, 3L,
>    2L, 1L, 3L, 3L, 3L, 3L, 1L, 3L, 1L, 1L, 1L, 3L, 1L, 1L, 1L,
>    1L, 1L, 1L, 2L, 1L, 3L, 3L, 3L, 3L, 3L, 3L, 1L, 1L, 3L, 1L,
>    3L, 1L, 1L, 1L, 2L, 1L, 1L, 3L, 1L, 3L, 3L, 1L, 3L, 1L, 1L,
>    1L, 1L, 1L, 3L, 3L, 2L), .Label = c("0", "1", "2"), class = "factor")))
>
>
>
>
>
>
> On Oct 18, 2010, at 4:09 PM, jim holtman wrote:
>
>> files did not make it through the mailer.  How did you attach them?
>> try outputting the data using 'dput' and then attaching a '.txt' file,
>> or just pasting them in the email.
>>
>> On Mon, Oct 18, 2010 at 2:40 PM, Gregory Ryslik <rsaber at comcast.net> wrote:
>>> Hi Everyone,
>>>
>>> This is closer to what I need but this returns me a matrix where each element is a factor. Instead I would want a list of lists. The first entry of the list should equal the first column of the matrix that mapply makes, the second entry to the second column etc...
>>>
>>> I've attached the two files that have all.predicted.values and max.growth from dput to make for easy testing. Thanks again!
>>>
>>> Kind regards,
>>> Greg
>>>
>>> On Oct 18, 2010, at 1:33 PM, Erich Neuwirth wrote:
>>>
>>>> You probably need mapply since you have 2 list of arguments which you want to use "in sync"
>>>>
>>>> mapply(function(x1,x2)x1[[x2]],all.predicted.values,max.growth)
>>>>
>>>> might be what you want.
>>>>
>>>>
>>>>
>>>> On Oct 18, 2010, at 5:17 PM, Gregory Ryslik wrote:
>>>>
>>>>> Unfortunately, that gives me null everywhere. Here's the data I have for all.predicted.values and max.growth. Perhaps this will help. Thus I want all.predicted.values[[1]][[4]] then all.predicted.values[[2]][3]] and then all.predicted.values[[3]][[4]].
>>>>>
>>>>> I've attached what your statement outputs at the end.
>>>>>
>>>>> Thanks again!
>>>>>
>>>>> Browse[2]> max.growth
>>>>> [[1]]
>>>>> [1] 4
>>>>>
>>>>> [[2]]
>>>>> [1] 3
>>>>>
>>>>> [[3]]
>>>>> [1] 4
>>>>>
>>>>> Browse[2]> all.predicted.values
>>>>> [[1]]
>>>>> [[1]][[1]]
>>>>>  [1] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
>>>>> [55] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
>>>>> Levels: 0 1 2
>>>>>
>>>>> [[1]][[2]]
>>>>>  [1] 2 2 2 0 2 0 2 2 2 2 2 2 2 2 0 2 2 2 2 2 2 2 2 2 2 0 0 2 2 2 2 0 0 0 2 2 0 0 2 2 0 2 2 2 2 2 0 2 2 2 0 2 2 0
>>>>> [55] 0 0 2 0 2 0 0 0 0 2 2 2 2 0 2 2 2 0 2 2 0 0 2 2 2 2 2 2 2 0 0 0 2 0 2 2 2 2 0 2 2 2 0 2 0 0
>>>>> Levels: 0 1 2
>>>>>
>>>>> [[1]][[3]]
>>>>>  [1] 0 0 0 0 2 0 0 0 0 0 0 0 0 2 0 0 0 0 0 2 0 2 2 2 0 0 0 2 0 0 2 0 0 0 0 0 0 0 2 0 0 0 0 0 2 2 0 0 0 2 0 0 0 0
>>>>> [55] 0 0 2 0 2 0 0 0 0 2 2 0 0 0 0 0 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 2 0 0 0 0 0 0 2 0 0
>>>>> Levels: 0 1 2
>>>>>
>>>>> [[1]][[4]]
>>>>>  [1] 0 0 0 0 2 0 0 0 0 0 0 0 0 2 0 0 0 0 0 2 0 2 2 2 0 0 0 2 0 0 2 0 0 0 0 0 0 0 2 0 0 0 0 0 2 2 0 0 0 2 0 0 0 0
>>>>> [55] 0 0 2 0 2 0 0 0 0 2 2 0 0 0 0 0 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 2 0 0 0 0 0 0 2 0 0
>>>>> Levels: 0 1 2
>>>>>
>>>>>
>>>>> [[2]]
>>>>> [[2]][[1]]
>>>>>  [1] 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
>>>>> [55] 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
>>>>> Levels: 0 1 2
>>>>>
>>>>> [[2]][[2]]
>>>>>  [1] 2 2 2 2 1 2 2 2 2 2 1 2 2 2 2 1 2 1 2 2 2 2 2 2 2 2 2 2 1 2 2 2 2 2 1 2 2 2 1 2 2 1 1 2 2 2 2 2 2 2 2 1 2 2
>>>>> [55] 2 2 2 2 1 2 2 2 2 1 2 2 1 1 1 2 2 2 1 2 1 2 1 2 1 2 2 2 1 1 2 2 1 2 2 1 1 2 1 1 1 2 2 1 2 2
>>>>> Levels: 0 1 2
>>>>>
>>>>> [[2]][[3]]
>>>>>  [1] 2 2 2 0 1 2 2 2 2 2 1 2 2 2 0 1 2 1 2 2 2 2 2 2 2 0 0 2 1 2 2 2 0 0 1 2 0 0 1 2 0 1 1 2 2 2 0 2 2 2 0 1 2 2
>>>>> [55] 0 2 2 2 1 0 0 0 0 1 2 2 1 1 1 2 2 0 1 2 1 0 1 2 1 2 2 2 1 1 2 2 1 2 2 1 1 2 1 1 1 2 2 1 0 2
>>>>> Levels: 0 1 2
>>>>>
>>>>>
>>>>> [[3]]
>>>>> [[3]][[1]]
>>>>>  [1] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
>>>>> [55] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
>>>>> Levels: 0 1 2
>>>>>
>>>>> [[3]][[2]]
>>>>>  [1] 2 2 2 0 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 0 0 2 2 2 2 2 0 0 2 2 2 0 2 2 0 2 2 2 2 2 0 2 2 2 0 2 2 2
>>>>> [55] 0 2 2 2 2 2 0 0 2 2 2 2 2 2 2 2 2 0 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
>>>>> Levels: 0 1 2
>>>>>
>>>>> [[3]][[3]]
>>>>>  [1] 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 1 1 1 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0 1 0 0
>>>>> [55] 0 0 0 0 1 0 0 0 0 1 0 0 1 1 1 0 0 0 1 0 1 1 1 0 1 0 0 0 1 1 0 0 1 0 0 1 1 0 1 1 1 0 0 1 1 0
>>>>> Levels: 0 1 2
>>>>>
>>>>> [[3]][[4]]
>>>>>  [1] 2 2 2 0 1 0 2 2 0 2 1 2 2 0 0 1 1 1 1 0 2 0 0 0 2 0 0 0 1 2 0 0 0 0 1 2 0 0 1 2 0 1 1 2 0 0 0 2 2 0 0 1 2 0
>>>>> [55] 0 0 0 0 1 0 0 0 0 1 0 2 1 1 1 2 0 0 1 2 1 1 1 2 1 2 2 2 1 1 0 0 1 0 2 1 1 2 1 1 1 2 0 1 1 0
>>>>> Levels: 0 1 2
>>>>>
>>>>>
>>>>> Browse[2]>   predicted.values.for.max.growth<-diag(sapply(all.predicted.values,'[[','max.growth'))
>>>>> Browse[2]> predicted.values.for.max.growth
>>>>> [[1]]
>>>>> NULL
>>>>>
>>>>> [[2]]
>>>>> [1] 0
>>>>>
>>>>> [[3]]
>>>>> [1] 0
>>>>>
>>>>> [[4]]
>>>>> [1] 0
>>>>>
>>>>> [[5]]
>>>>> NULL
>>>>>
>>>>> [[6]]
>>>>> [1] 0
>>>>>
>>>>> [[7]]
>>>>> [1] 0
>>>>>
>>>>> [[8]]
>>>>> [1] 0
>>>>>
>>>>> [[9]]
>>>>> NULL
>>>>>
>>>>>
>>>>>
>>>>> On Oct 18, 2010, at 11:08 AM, Henrique Dallazuanna wrote:
>>>>>
>>>>>> Try this:
>>>>>>
>>>>>> diag(sapply(all.predicted.values, '[[', 'max.growth'))
>>>>>>
>>>>>>
>>>>>> On Mon, Oct 18, 2010 at 12:59 PM, Gregory Ryslik <rsaber at comcast.net> wrote:
>>>>>> Hi,
>>>>>>
>>>>>> I have a list of n items and the ith element has m_i elements within it.
>>>>>>
>>>>>> I want to do something like:
>>>>>>
>>>>>> predicted.values<- lapply(all.predicted.values,'[[',max.growth[[i]])
>>>>>>
>>>>>> Where max.growth[[i]] is the element I want to extract from each of the ith predicted elements. Thus, for example, I want to extract the max.growth[[1]] element from  all.predicted.values[[1]] (which is itself a list). Then I want to extract max.growth[[2]] element from all.predicted.values[[2]].
>>>>>>
>>>>>> I realize I can do this with a for loop but then if I can do this as one line that would be preferable.
>>>>>>
>>>>>> Thanks!
>>>>>>
>>>>>> Greg
>>>>>>       [[alternative HTML version deleted]]
>>>>>>
>>>>>> ______________________________________________
>>>>>> R-help at 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.
>>>>>>
>>>>>>
>>>>>>
>>>>>> --
>>>>>> Henrique Dallazuanna
>>>>>> Curitiba-Paraná-Brasil
>>>>>> 25° 25' 40" S 49° 16' 22" O
>>>>>
>>>>>
>>>>>      [[alternative HTML version deleted]]
>>>>>
>>>>> ______________________________________________
>>>>> R-help at 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.
>>>>
>>>> --
>>>> Erich Neuwirth
>>>> Didactic Center for Computer Science and Institute for Scientific Computing
>>>> University of Vienna
>>>>
>>>>
>>>>
>>>>
>>>
>>> ______________________________________________
>>> R-help at 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.
>>>
>>
>>
>>
>> --
>> Jim Holtman
>> Cincinnati, OH
>> +1 513 646 9390
>>
>> What is the problem that you are trying to solve?
>
>
>

-- 
Jim Holtman
Cincinnati, OH
+1 513 646 9390

What is the problem that you are trying to solve?