[R] Ifelse statement on a factor level data frame

[Kate Ignatius]

Apologies - you're right.  Missed it in the pdf.

K.

On Sun, Sep 28, 2014 at 10:22 AM, Bert Gunter <gunter.berton at gene.com> wrote:
> Inline.
>
> Bert Gunter
> Genentech Nonclinical Biostatistics
> (650) 467-7374
>
> "Data is not information. Information is not knowledge. And knowledge
> is certainly not wisdom."
> Clifford Stoll
>
>
>
>
> On Sun, Sep 28, 2014 at 6:38 AM, Kate Ignatius <kate.ignatius at gmail.com> wrote:
>> Strange that,
>>
>> I did put everything with as.character but all I got was the same...
>>
>> class of dbpmn[,2]) = factor
>> class of dbpmn[,21]  = factor
>> class of  dbpmn[,20] = data.frame
>>
>> This has to be a problem ???
>
> Indeed -- your failure to read documentation.
>
> I suggest you do your due diligence, read Pat Burns's link, and follow
> the advice given you by posting a reproducible example. More than
> likely the last will be unnecessary as you will figure it out in the
> course of doing what you should do.
>
> Cheers,
> Bert
>
>>
>> I can put reproducible output here but not sure if this going to of
>> help here. I think its all about factors and data frames and
>> characters...
>>
>> K.
>>
>> On Sun, Sep 28, 2014 at 1:15 AM, Jim Lemon <jim at bitwrit.com.au> wrote:
>>> On Sun, 28 Sep 2014 12:49:41 AM Kate Ignatius wrote:
>>>> Quick question:
>>>>
>>>> I am running the following code on some variables that are factors:
>>>>
>>>> dbpmn$IID1new <- ifelse(as.character(dbpmn[,2]) ==
>>>> as.character(dbpmn[,(21)]), dbpmn[,20], '')
>>>>
>>>> Instead of returning some value it gives me this:
>>>>
>>>> c(1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1))
>>>>
>>>> Playing around with the code, gives me some kind of variation to it.
>>>> Is there some way to get me what I want.  The variable that its
>>>> suppose to give back is a bunch of sampleIDs.
>>>>
>>> Hi Kate,
>>> If I create a little example:
>>>
>>> dbpmn<-data.frame(V1=factor(sample(LETTERS[1:4],20,TRUE)),
>>>   V2=factor(sample(LETTERS[1:4],20,TRUE)),
>>>   V3=factor(sample(LETTERS[1:4],20,TRUE)))
>>> dbpmn[4]<-
>>>  ifelse(as.character(dbpmn[,1]) == as.character(dbpmn[,(2)]),
>>>  dbpmn[,3],"")
>>> dbpmn
>>>    V1 V2 V3 V4
>>> 1   B  D  C
>>> 2   C  A  D
>>> 3   C  B  A
>>> 4   A  B  C
>>> 5   B  D  B
>>> 6   D  D  A  1
>>> 7   D  D  D  4
>>> 8   B  C  A
>>> 9   B  D  B
>>> 10  D  C  A
>>> 11  A  D  C
>>> 12  A  C  B
>>> 13  A  A  A  1
>>> 14  D  C  A
>>> 15  C  D  B
>>> 16  A  A  B  2
>>> 17  A  C  C
>>> 18  B  B  C  3
>>> 19  C  C  C  3
>>> 20  D  D  D  4
>>>
>>> I get what I expect, the numeric value of the third element in dbpmn
>>> where the first two elements are equal. I think what you want is:
>>>
>>> dbpmn[4]<-
>>>  ifelse(as.character(dbpmn[,1]) == as.character(dbpmn[,(2)]),
>>>  as.character(dbpmn[,3]),"")
>>> dbpmn
>>>    V1 V2 V3 V4
>>> 1   B  D  C
>>> 2   C  A  D
>>> 3   C  B  A
>>> 4   A  B  C
>>> 5   B  D  B
>>> 6   D  D  A  A
>>> 7   D  D  D  D
>>> 8   B  C  A
>>> 9   B  D  B
>>> 10  D  C  A
>>> 11  A  D  C
>>> 12  A  C  B
>>> 13  A  A  A  A
>>> 14  D  C  A
>>> 15  C  D  B
>>> 16  A  A  B  B
>>> 17  A  C  C
>>> 18  B  B  C  C
>>> 19  C  C  C  C
>>> 20  D  D  D  D
>>>
>>> Jim
>>>
>>
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