[R] Antwort: RE: Antwort: Re: Factors and Alternatives (SOLVED)

Tue May 9 15:37:42 CEST 2017

Hi David,
Hi Bob,

many thanks for your help.

Your solution - just to use all levels instead of just the one's found in 
the data - helped.

The original code looked like this:

-- cut --

c_v10_val_labs <- c(
  "1 = sehr gut",
  "2", "3", "4", "5",
  "6 = sehr schlecht"
)

# where c_v10_val_labs is handed over to my function as "val_labs".

  ds_results$value <- factor(ds_results$value,
                             levels = sort(unique(ds_results$value)),  # 
old code
                             labels = sort(unique(val_labs)))

-- cut --

If I write instead

-- cut --

  ds_results$value <- factor(ds_results$value,
                             levels = seq_along(val_labs),  # new code 1st 
version
                             labels = sort(unique(val_labs)))

-- cut --

Your solution builds a factor with all factor levels even if a value for 
factor is not present (not NA, but does just not occur in the data, i.e. 
not stated by any respondent).

In Zumel's book "Practical Data Science with R" (
https://www.amazon.de/Practical-Data-Science-Nina-Zumel/dp/1617291560), 
Shelter Island: Manning, 2014, p. 23-24, Listing 2-5, a mapping using 
subscripts is described:

-- cut --

mapping <- list(
'A40'='car (new)',
'A41'='car (used)',
'A42'='furniture/equipment',
'A43'='radio/television',
'A44'='domestic appliances',
...
)

for(i in 1:(dim(d))[2]) {
if(class(d[,i])=='character') {
d[,i] <- as.factor(as.character(mapping[d[,i]]))
}
}

-- cut -

Simple stated this would mean:

-- cut --

val_labs <- list(
  "1" = "1 = sehr gut",
  "2" = "2",
  "3" = "3",
  "4" = "4",
  "5" = "5",
  "6" = "6 = sehr schlecht"
)

set.seed(12345)
answers = c(sample(1:5, 10, replace = TRUE))

test <- factor(unlist(val_labs[answers]))

# or just

val_labs <- c(
  "1 = sehr gut",
  "2",
  "3",
  "4",
  "5",
  "6 = sehr schlecht"
)

set.seed(12345)
answers = c(sample(1:5, 10, replace = TRUE))

test <- val_labs[answers]

-- cut --

Adapting this to my code would give:

-- cut --

  ds_results$value <- factor(ds_results$value,
                             levels = sort(unique(ds_results$value)),
                             labels = 
val_labs[sort(unique(ds_results$value))])  # new code 2nd version

-- cut --

This results in a factor just as long as the vector of unique resulting 
values.

Both solutions work. Which version is best depends on the overall process 
and the purpose of the code. I document all this for use by readers who 
refer later to the list archives.

Using your version and running my code reveals that ggplot runs into 
difficulties cause the legend lacks values and the sequence and coloring 
of the legend is wrong. But that's another story.

Many thanks again for your help.

Kind regards

Georg

Von:    David L Carlson <dcarlson at tamu.edu>
An:     "G.Maubach at weinwolf.de" <G.Maubach at weinwolf.de>, "Bob O'Hara" 
<rni.boh at gmail.com>, 
Kopie:  r-help <r-help at r-project.org>
Datum:  09.05.2017 14:37
Betreff:        RE: [R] Antwort: Re:  Factors and Alternatives

I'm not sure I understand your question, but you can easily include all 
possible answers when you create the factor by using the levels= argument 
as Bob pointed out. Here is an example of values that range from 1 to 6, 
but value 3 is not represented. Notice that a factor level 3 is created 
even though it does not appear in the data:

> set.seed(42)
> x <- sample.int(6, 10, replace=TRUE)
> table(x)
x
1 2 4 5 6 
1 1 3 3 2 
> y <- factor(x, levels=1:6)
> y
 [1] 6 6 2 5 4 4 5 1 4 5
Levels: 1 2 3 4 5 6

-------------------------------------
David L Carlson
Department of Anthropology
Texas A&M University
College Station, TX 77840-4352

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 13:58
Betreff:        Re: Re: [R] Factors and Alternatives

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

-----Original Message-----
From: R-help [mailto:r-help-bounces at r-project.org] On Behalf Of 
G.Maubach at weinwolf.de
Sent: Tuesday, May 9, 2017 6:37 AM
To: Bob O'Hara <rni.boh at gmail.com>
Cc: r-help <r-help at r-project.org>
Subject: [R] Antwort: Re: Factors and Alternatives

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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-- 
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

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