The issue of dropping unused factor levels when subsetting has come up before. Common solutions include using character vectors where possible by declaring
options(stringsAsFactors = FALSE)
Sometimes, though, ordered factors are necessary for plotting, in which case we can use convenience functions like droplevels to create a wrapper for subset:
subsetDrop <- function(...){droplevels(subset(...))}
I realize that subsetDrop mostly solves this problem, but there are some situations where subsetting via [ is more convenient (and less typing!).
My question is how much further, for the sake of convenience, can we push this to be the 'default' behavior of R by overriding [ for data frames to automatically drop factor levels. For instance, the Hmisc package contains dropUnusedLevels which overrides [.factor for subsetting a single factor (which is no longer necessary, since the default [.factor appears to have a drop argument for dropping unused levels). I'm looking for a similar solution that would allow me to subset data frames using [ but automatically dropping unused factor levels (and of course preserving order in the case of ordered factors).
I'd be really wary of changing the default behavior; you never know when another function you use depends on the usual default behavior. I'd instead write a similar function to your subsetDrop but for [, like
sel <- function(x, ...) droplevels(x[...])
Then
> d <- data.frame(a=factor(LETTERS[1:5]), b=factor(letters[1:5]))
> str(d[1:2,])
'data.frame': 2 obs. of 2 variables:
$ a: Factor w/ 5 levels "A","B","C","D",..: 1 2
$ b: Factor w/ 5 levels "a","b","c","d",..: 1 2
> str(sel(d,1:2,))
'data.frame': 2 obs. of 2 variables:
$ a: Factor w/ 2 levels "A","B": 1 2
$ b: Factor w/ 2 levels "a","b": 1 2
If you really want to change the default, you could do something like
foo <- `[.data.frame`
`[.data.frame` <- function(...) droplevels(foo(...))
but make sure you know how namespaces work as this will work for anything called from the global namespace but the version in the base namespace is unchanged. Which might be a good thing, but it's something you want to make sure you understand. After this change the output is as you'd like.
> str(d[1:2,])
'data.frame': 2 obs. of 2 variables:
$ a: Factor w/ 2 levels "A","B": 1 2
$ b: Factor w/ 2 levels "a","b": 1 2
you can do that work by overwriting the default value of the drop argument like this:
formals(`[.factor`)$drop <- TRUE
UPDATE
as for data.frame, you can do by:
`[.data.frame` <- function(...)droplevels(base::`[.data.frame`(...))
actually similar as #Aaron's one.
if you want to cancel this behavior, then:
rm(`[.data.frame`)
will do that.
> d <- data.frame(a=letters[1:10], b=LETTERS[1:10])
> str(d[1:5, ])
'data.frame': 5 obs. of 2 variables:
$ a: Factor w/ 10 levels "a","b","c","d",..: 1 2 3 4 5
$ b: Factor w/ 10 levels "A","B","C","D",..: 1 2 3 4 5
> `[.data.frame` <- function(...)droplevels(base::`[.data.frame`(...))
> str(d[1:5, ])
'data.frame': 5 obs. of 2 variables:
$ a: Factor w/ 5 levels "a","b","c","d",..: 1 2 3 4 5
$ b: Factor w/ 5 levels "A","B","C","D",..: 1 2 3 4 5
> rm(`[.data.frame`)
> str(d[1:5, ])
'data.frame': 5 obs. of 2 variables:
$ a: Factor w/ 10 levels "a","b","c","d",..: 1 2 3 4 5
$ b: Factor w/ 10 levels "A","B","C","D",..: 1 2 3 4 5
I think that changing the default is very dangerous, see my response here.
Most cases where people are concerned with dropping factor levels you either really don't need to (sumarizing something that you forced to have 1 value is silly) or there is a better way to accomplish what you are trying. The possible side effects from auto dropping is potentially worse than the couple of keystrokes saved. Also if you are doing reproducable research then you should not be depending on or even allowing the computer to change data without your specific say so.
Related
I have been tearing my hair out over this for the last hour, the following code was working perfectly a couple of hours ago, and now I have no idea why it doesn't anymore. I have searched for other questions regarding the undefined columns selected error, but I think I have corrected for all of the info in those answers. I am sure there is some tiny thing I have overlooked or accidently left in, but I can't see it!
I have a data frame with both factor and numeric variables, I want to subset so that I keep all of the factor variables, and remove numeric variables whose columns have a mean < 0.1.
I found the following code on another question on stackoverflow, which slightly modified worked well on my test data (smaller sub-dataset I am using for testing before trying out code on a big 3GB object)
meanfunction01 <- function(x){
if(is.numeric(x)){
mean(x) > 0.1
} else {
TRUE}
}
#then apply function to data table
Zdata <- Data1[,sapply(Data1, meanfunction01)]
I swear I was using this a few hours ago, then when i came back to it and tried to use it again it stopped working and now just returns the following error:
Error in `[.data.frame`(Data1, , sapply(Data1, meanfunction01)) :
undefined columns selected
I was trying to modify the function so that it would loop over multiple objects (I have 54 objects I want to apply it to, and didn't want to type them all manually), but I don't think I edited the original function, and now it has stopped working.
A brief str() of my data:
> str(Data1[1:10])
'data.frame': 11 obs. of 10 variables:
$ Name : Factor w/ 11688 levels "GTEX-1117F-0226-SM-5GZZ7",..: 8186 8242 8262 8270 8343 8388 8403 8621 8689 8709 ...
$ SEX : Factor w/ 2 levels "Female","Male": 1 2 2 1 1 2 2 1 2 1 ...
$ AGE : Factor w/ 6 levels "20-29","30-39",..: 4 4 1 3 3 1 3 3 3 2 ...
$ CIRCUMSTANCES: Factor w/ 5 levels "0","1","2","3",..: 1 1 1 1 1 1 1 1 1 1 ...
$ Tissue.x : Factor w/ 53 levels "Adipose_Subcutaneous",..: 7 7 7 7 7 7 7 7 7 7 ...
$ ENSG00000223972.4 : num 0 0.0701 0.0339 0.1149 0.0549 ...
$ ENSG00000227232.4 : num 12.5 17.2 13.1 16 15.7 ...
$ ENSG00000243485.2 : num 0.0717 0 0.1508 0 0.061 ...
$ ENSG00000237613.2 : num 0 0.0654 0 0.0402 0.0768 ...
$ ENSG00000268020.2 : num 0 0.0421 0.0611 0 0 ...
So if your only issue is changing the class of the integer variables in your data.frame but you have many columns (>10000) you may want to consider converting your data.frame into a data.table. Your code would then look like this:
library(data.table)
Data1<-data.table(Data1) #or if you have your data in csv document just use fread instead of read.csv which will automatically give you a data.table.
Then you just need to find the integer columns using this:
which(sapply(Data1,is.integer))
Putting it altogether using the data.table commands:
Data1[,which(sapply(Data1,is.integer)):=lapply(.SD,as.numeric),.SDcols=which(sapply(Data1,is.integer))]
Note you don't need to assign the above line of code into anything since data.table uses pointers which makes it much faster than data.frame or tibbles objects. So running the above line will update your Data1 object efficiently. The classes of the other non-integer columns (i.e., factors) will remain unchanged.
Please update if you have further questions but this should answer your comment. Best of luck!
I have a dataset where there are many columns with numbers in quotes which indicates that a variable is a factor. (ex: "8").
read.table automatically converts them in numerical variables even if stringsAsFactor is set as true.
Suppose I cannot convert them manually with as.factor, how can I import this dataset with those variables coded directly as factor?
That's because of the quote option. Set quote="". Example:
t <- '"1" "3"
"2" "4"'
> str(read.table(text=t))
'data.frame': 2 obs. of 2 variables:
$ V1: int 1 2
$ V2: int 3 4
> str(read.table(text=t, quote=""))
'data.frame': 2 obs. of 2 variables:
$ V1: Factor w/ 2 levels "\"1\"","\"2\"": 1 2
$ V2: Factor w/ 2 levels "\"3\"","\"4\"": 1 2
I have a 100k row dataframe on which I want to compute a Cochran–Mantel–Haenszel test.
My variables are the educational level and a computed score factored in quantiles, and my grouping variable is the sex, and the code line looks like this :
mantelhaen.test(db$education, db$score.grouped, db$sex)
This code throws this error and warning :
Error in qr.default(a, tol = tol) : NA/NaN/Inf in foreign function call (arg 1)
In addition: Warning message: In ntot * rowsums : NAs produced by integer overflow
The error seems to be caused by my first variable, as on 7 variables tested I got the problem with only 2 of them, which seems to share no obvious common thing.
Missing values and factor levels don't seem to differ between variables which throws error and variable which doesn't. I tried with complete cases (with na.omit) and the problem persists.
What does trigger this error ? does it mean ?
How can I get rid of it ?
Interesting posts : R: NA/NaN/Inf in foreign function call (arg 1), What is integer overflow in R and how can it happen?
ADDENDUM : Here is the result of str (failures are education and imc.cl):
str(db[c("education","score.grouped","sex", ...)])
'data.frame': 104382 obs. of 7 variables:
$ age.cl: Ord.factor w/ 5 levels "<30 ans"<"30-40 ans"<..: 5 2 1 1 3 4 2 3 4 4 ...
..- attr(*, "label")= chr "age"
$ emploi2 : Factor w/ 8 levels "Agriculteurs exploitants",..: 3 5 6 8 8 8 8 3 3 3 ...
..- attr(*, "label")= chr "CSP"
$ tabac : Factor w/ 4 levels "ancien fumeur",..: 4 1 4 4 3 4 4 1 4 4 ...
..- attr(*, "label")= chr "tabac"
$ situ_mari2 : Factor w/ 3 levels "Vit seul","Divorsé, séparé ou veuf",..: 3 2 1 1 1 3 1 3 2 3 ...
..- attr(*, "label")= chr "marriage"
$ education : Factor w/ 3 levels "Universitaire",..: 1 1 1 1 3 1 1 1 1 1 ...
$ revenu.cl : Factor w/ 4 levels "<1800 euros/uc",..: 3 4 2 NA 4 1 1 4 4 1 ...
$ imc.cl : Ord.factor w/ 6 levels "Maigre"<"Normal"<..: 2 2 1 2 3 1 3 2 2 3 ...
..- attr(*, "label")= chr "IMC"
EDIT : by diving inside the function, the error and warning are caused by a call to qr.solve. I don't understand anything about this but I'll try to dive deeper
EDIT2 : inside qr.solve, the error is thrown by a Fortran call to .F_dqrdc2. This is so much beyond my level my nose is starting to bleed.
EDIT3 : I tried to head my data to find out which line is in cause :
db2 = db %>% head(99787) #fails at 99788
db2 = db %>% tail(99698) #fails at 99699
mantelhaen.test(db2$education, db2$score.grouped, db2$sex)
This gives me not much information but maybe it could give you.
I was able to replicate the problem by making the data set bigger.
set.seed(101); n <- 500000
db <- data.frame(education=
factor(sample(1:3,replace=TRUE,size=n)),
score=
factor(sample(1:5,replace=TRUE,size=n)),
sex=
sample(c("M","F"),replace=TRUE,size=n))
After this, mantelhaen.test(db$education, db$score, db$sex) gives the reported error.
Thankfully, the real problem is not within the guts of the QR decomposition code: rather, it occurs when setting up a matrix prior to QR decomposition. There are two computations, ntot*colsums and ntot*rowsums, that overflow R's capacity for integer computation. There's a relatively easy way to work around this by creating a modified version of the function:
copy the source code: dump("mantelhaen.test",file="my_mh.R")
edit the source code
l. 1: modify name of function to my_mantelhaen.test (to avoid confusion)
lines 199 and 200: change ntot to as.numeric(ntot), converting the integer to double precision before the overflow happens
source("my_mh.R") to read in the new function
Now
my_mantelhaen.test(db$education, db$score, db$sex)
should work.
You should definitely test the new function against the old function for cases where it works to make sure you get the same answer.
Now posted to the R bug list, we'll see what happens ...
update 11 May 2018: this is fixed in the development version of R (3.6 to be).
I want to convert a factor data frame, to a factor matrix
But when I try the below code, the type of the matrix is still string
mydata=data.frame(f1=c("yes","yes","no","no"),f2=c("yes","no","no","yes"))
mydata[1:ncol(mydata)]=lapply(mydata[1:ncol(mydata)],factor)
mymatrix=as.matrix(mydata)
#this line didn't help (the matrix still string)
mymatrix=apply(mymatrix,FUN =factor,MARGIN = 2)
Maybe this will get you what you need?
mymatrix = matrix(mydata, ncol = 2)
str(mymatrix)
gives you
List of 2
$ : Factor w/ 2 levels "no","yes": 2 2 1 1
$ : Factor w/ 2 levels "no","yes": 2 1 1 2
- attr(*, "dim")= int [1:2] 1 2
You would need to explain a bit more what you want to do to get more precise help.
How does sort work, that is using what method to sort column in
data.frame (barley$site, barley$year, barley$variety)
as following
library(lattice)
barley <- barley[order(barley$site, barley$year, barley$variety), ]
You probably want:
barley[order(as.character(barley$site), as.numeric(barley$year), as.character(barley$variety)),]
As you have it you are ordering by the underlying levels of the data.frame, which leads to really odd stuff. Look at the structure of the data frame:
'data.frame': 120 obs. of 4 variables:
$ yield : num 27 48.9 27.4 39.9 33 ...
$ variety: Factor w/ 10 levels "Svansota","No. 462",..: 3 3 3 3 3 3 7 7 7 7 ...
$ year : Factor w/ 2 levels "1932","1931": 2 2 2 2 2 2 2 2 2 2 ...
$ site : Factor w/ 6 levels "Grand Rapids",..: 3 6 4 5 1 2 3 6 4 5 ...
Notice how the levels for year are in the opposite order you would expect. The documentation for order discusses this very briefly:
For factors, this sorts on the internal codes, which is particularly appropriate for ordered factors.
I personally think this terribly confusing, but it is what it is. factor are very useful in most contexts, but incredibly dangerous in others if you're not careful. Having numbers represented as factors (as year was here) is particularly bad.
See ?factor for more details.
By default, sort doesn't know how to do anything with a data frame. You can sort the individual columns within a data frame, with something like df$x <- sort(df$x) but you almost certainly don't want to do that; it will just mess up your data.
You order the rows in the data frame by using order as in the example code you have there. This orders the rows by values in the column site, breaking ties with year, and then with variety.