如何在rpy2中使用R'with'运算符

Question

I am doing an ordinal logistic regression, and following the guide here for the analysis: R Data Analysis Examples: Ordinal Logistic Regression

My dataframe (consult) looks like:

           n  raingarden  es_score  consult_case
garden_id                                       
27436      7           0         3             0
27437      1           0         0             1
27439      1           1         1             1
37253      1           0         3             0
37256      3           0         0             0

I am at the part where I need to to create graph to test the proportional odds assumption, with the command in R as follows:

(s <- with(dat, summary(es_score ~ n + raingarden + consult_case, fun=sf)))

(es_score is an ordinal ranked score with values between 0 - 4; n is an integer; raingarden and consult_case, binary values of 0 or 1)

I have the sf function:

sf <- function(y) {
     c('Y>=1' = qlogis(mean(y >= 1)),
       'Y>=2' = qlogis(mean(y >= 2)),
       'Y>=3' = qlogis(mean(y >= 3)))
}

in a utils.r file that I access as follows:

from rpy2.robjects.packages import STAP
with open('/file_path/utils.r', 'r') as f:
    string = f.read()
sf = STAP(string, "sf")

And want to do something along the lines of:

R = ro.r
R.with(work_case_control, R.summary(formula, fun=sf))

The major problem is that the R with operator is seen as a python keyword, so that even if I access it with ro.r.with it is still recognized as a python keyword. (As a side note: I tried using R's apply method instead, but got an error that TypeError: 'SignatureTranslatedAnonymousPackage' object is not callable ... I assume this is referring to my function sf ?)

I also tried using the R assignment methods in rpy2 as follows:

R('sf = function(y) { c(\'Y>=1\' = qlogis(mean(y >= 1)), \'Y>=2\' = qlogis(mean(y >= 2)), \'Y>=3\' = qlogis(mean(y >= 3)))}')

R('s <- with({0}, summary(es_score~raingarden + consult_case, fun=sf)'.format(consult))

but ran into issues where the dataframe column names were somehow causing the error: RRuntimeError: Error in (function (file = "", n = NULL, text = NULL, prompt = "?", keep.source = getOption("keep.source"), : <text>:1:19: unexpected symbol 1: s <- with( n raingarden

I could of course do this all in R, but I have a very involved ETL script in python, and would thus prefer to keep everything in python using rpy2 (I did try this using mord for scipy-learn to run my regreession, but it is pretty primitive).

Any suggestions would be most welcome right now.

EDIT

I tried various combinations @Parfait's suggestions, and qualifying the fun argument is syntactically incorrect, as per PyCharm interpreter (see image with red highlighting at end): ... it doesn't matter what the qualifier is, either, I always get an error that SyntaxError: keyword can't be an expression.

On the other hand, with no qualifier, there is no syntax error: , but I do get the error TypeError: 'SignatureTranslatedAnonymousPackage' object is not callable when using the function sf as obtained:

from rpy2.robjects.packages import STAP with open('/Users/gregsilverman/development/python/rest_api/rest_api/scripts/utils.r', 'r') as f: string = f.read() sf = STAP(string, "sf")

With that in mind, I created a package in R with the function sf , imported it, and tried various combos with the only one producing no error, being: print(base._with(consult_case_control, R.summary(formula, fun=gms.sf))) (gms is a reference to the package in RI made).

The output though makes no sense:

Length Class Mode 3 formula call

I am expecting a table ala the one on the UCLA site. Interesting. I am going to try recreating my analysis in R, just for the heck of it. I still would like to complete it in python though.

Answer 1

Consider bracketing the with call and be sure to qualify all arguments including fun :

ro.r['with'](work_case_control, ro.r.summary(formula, ro.r.summary.fun=sf))

Alternatively, import R's base package. And to avoid conflict with Python's named method with() translate the R name:

from rpy2.robjects.packages import importr

base = importr('base', robject_translations={'with': '_with'})

base._with(work_case_control, ro.r.summary(formula, ro.r.summary.fun=sf))

And be sure to properly create your formula. Consider using R's stats packages' as.formula to build from string. Notice too another translation is made due to naming conflict:

stats = importr('stats', robject_translations={'format_perc': '_format_perc'})

formula = stats.as_formula('es_score ~ n + raingarden + consult_case')