[R] Factors and Alternatives

[Bob O'Hara]

For the problem you state, would it be enough to explicitly define your levels?

fac <- rep(c("a", "b", "d"), each=4)
fac.f <- factor(fac, levels=c("a", "b", "c", "d"))
table(fac.f)

# but be warned...
fac.f2 <- factor(fac.f)
table(fac.f2)

This has the advantage that the code explicitly documents what the
possible values are, so if something goes wrong down-stream, you know
it is a real problem (well, unless you have some type conversions
screwing things up). You might also want to do some defensive
programming, and put some checks in the code, to make sure your
factors have the right number of levels.

Bob

On 9 May 2017 at 13:36,  <G.Maubach at weinwolf.de> wrote:
> Hi Bob,
>
> many thanks for your reply.
>
> I have read the documentation. In my current project I use "item
> batteries" for dimensions of touchpoints which are rated by our customers.
> I wrote functions to analyse them. If I create a factor before filtering
> and analysing I lose the original values of the variable. If I use the
> original variable for filtering and analysis I might happen that for some
> dimensions values were not selected. This means they are not NA but none
> of the respondents chose "4" for instance on a scale from 1 to 6. That
> means that creating a factor from the analysed data with the complete
> scale (1:6) fails due the different vector length (amount of remaining
> unique values in the analysis vs values in the scale). As I have a
> function doing the analysis I am looking for a way to make my function
> robust to such circumstances and be able to use it to analyse all "item
> batteries". Thus my question. I believe my findings are not odd. Maybe
> there is a way dealing with that kind of problems in R and I am eager to
> learn how it can be solved using R.
>
> What would you suggest?
>
> Kind regards
>
> Georg
>
>
>
>
> Von:    "Bob O'Hara" <rni.boh at gmail.com>
> An:     G.Maubach at weinwolf.de,
> Kopie:  r-help <r-help at r-project.org>
> Datum:  09.05.2017 12:26
> Betreff:        Re: [R] Factors and Alternatives
>
>
>
> That's easy! First
>> str(test3)
>  Factor w/ 2 levels "WITHOUT Contact",..: 2 2 2 2 1 1 1 1 1 1
>
> tells you that the internal values are 1 and 2, and the labels are
> "WITHOUT Contact" and "WITH Contact". If you read the help page for
> factor() you'll see this:
>
> levels: an optional vector of the values (as character strings) that
>           ‘x’ might have taken.  The default is the unique set of
>           values taken by ‘as.character(x)’, sorted into increasing
>           order _of ‘x’_.  Note that this set can be specified as
>           smaller than ‘sort(unique(x))’.
>
>   labels: _either_ an optional character vector of (unique) labels for
>           the levels (in the same order as ‘levels’ after removing
>           those in ‘exclude’), _or_ a character string of length 1.
>
> So, when you create test3 you say that test can take values 0 and 1,
> and these should be labelled as "WITHOUT Contact" and "WITH Contact".
> So R internally codes "1" as 1 and "0" as 2 (internally R codes
> factors as integers, which can be both useful and dangerous), and then
> gives them labels "WITHOUT Contact" and "WITH Contact". It now doesn't
> care that they were 1 and 0, because you've told it to change the
> labels.
>
> If you want to filter by the original values, then don't change the
> labels (or at least not until after you've filtered by the original
> labels), or convert the filter to the new labels. You're asking for a
> data structure with two sets of labels, which sounds odd in general.
>
> Bob
>
> On 9 May 2017 at 12:12,  <G.Maubach at weinwolf.de> wrote:
>> Hi All,
>>
>> I am using factors in a study for the social sciences.
>>
>> I discovered the following:
>>
>> -- cut --
>>
>> library(dplyr)
>>
>> test1 <- c(rep(1, 4), rep(0, 6))
>> d_test1 <- data.frame(test)
>>
>> test2 <- factor(test1)
>> d_test2 <- data.frame(test2)
>>
>> test3 <- factor(test1,
>>                 levels = c(0, 1),
>>                 labels = c("WITHOUT Contact", "WITH Contact"))
>> d_test3 <- data.frame(test3)
>>
>> d_test1 %>% filter(test1 == 0)  # works OK
>> d_test2 %>% filter(test2 == 0)  # works OK
>> d_test3 %>% filter(test3 == 0)  # does not work, why?
>>
>> myf <- function(ds) {
>>   print(levels(ds$test3))
>>   print(labels(ds$test3))
>>   print(as.numeric(ds$test3))
>>   print(as.character(ds$test3))
>> }
>>
>> # This showsthat it is not possible to access the original
>> # values which were the basis to build the factor:
>> myf(d_test3)
>>
>> -- cut --
>>
>> Why is it not possible to use a factor with labels for filtering with
> the
>> original values?
>> Is there a data structure that works like a factor but gives also access
>> to the original values?
>>
>> Kind regards
>>
>> Georg
>>
>> ______________________________________________
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>> and provide commented, minimal, self-contained, reproducible code.
>
>
>
> --
> Bob O'Hara
> NOTE NEW ADDRESS!!!
> Institutt for matematiske fag
> NTNU
> 7491 Trondheim
> Norway
>
> Mobile: +49 1515 888 5440
> Journal of Negative Results - EEB: www.jnr-eeb.org
>
>



-- 
Bob O'Hara
NOTE NEW ADDRESS!!!
Institutt for matematiske fag
NTNU
7491 Trondheim
Norway

Mobile: +49 1515 888 5440
Journal of Negative Results - EEB: www.jnr-eeb.org