Fabulous! Thank you so much! On Fri, Nov 13, 2009 at 2:32 PM, Laurent Gautier <lgaut...@gmail.com> wrote: > Guilherme Castelao wrote: >> >> Hello Laurent, >> >> I'm sorry about that. >> >> I used to do like this: >> >> import rpy >> start = rpy.r.list(...) >> lower = rpy.r.list(...) >> upper = rpy.r.list(...) >> control = rpy.r.nls_control(maxiter=100,tol = 1e-05,minFactor=1./4096) >> data = rpy.r.data_frame(x=x, x2=x2,y=y) # x,x2 and y are >> numpy.array >> rpy.set_default_mode(rpy.NO_CONVERSION) >> model = rpy.r.nls(formula, data = data, start = start, lower=lower, >> upper=upper, control=control, algorithm="port",trace=True) >> rpy.set_default_mode(rpy.BASIC_CONVERSION) >> coef=rpy.r.coef(model) >> >> >> Now I'm trying >> >> import rpy2.robjects as robjects >> start = robjects.r.list(...) >> lower = robjects.r.list(...) >> upper = robjects.r.list(...) >> control = ?!?!?! >> ... >> model = robjects.r.nls(formula, data = data ...) >> >> But I don't know how to define the nls.control ! > > Just like you would do for other function calls. > > control = robjects.r['nls.control'](maxtiter=100, <...>) > (you can read why at: > http://rpy.sourceforge.net/rpy2/doc/html/robjects.html#r-the-instance-of-r > ) > > > With 2.1.x, that's tentatively a little nicer: > > from rpy2.robjects.packages import importr > stats = importr("stats") > > control = stats.nls_control(maxiter=100, <...>) > > (more on importing packages at > http://rpy.sourceforge.net/rpy2/doc-dev/html/robjects.html#module-rpy2.robjects.packages) > > > > > L. > > > > >> Thanks! >> >> >> On Fri, Nov 13, 2009 at 3:45 AM, Laurent Gautier <lgaut...@gmail.com> >> wrote: >>> >>> Guilherme Castelao wrote: >>>> >>>> Hello all, >>>> >>>> I'm moving to rpy2 but I still didn't get the new syntax. I have two >>>> simple questions: >>>> >>>> >>>> - I can create the nls with no trouble as below, but I can't figure >>>> out how to create the nls.control with rpy2? >>>> >>>> >>>> >>>> model=robjects.r.nls(formula,data=dataf,start=start,control=control,trace=True) >>>> >>> It hard to tell without knowing how you created the objects used in the >>> function call. >>> >>>> - How would be a more elegant way to create a data frame then >>>> >>>> >>>> >>>> d={"x":robjects.FloatVector([4.17,5.58,5.18,6.11,4.50,4.61,5.17,4.53,5.33,5.14]),"y":robjects.FloatVector([4.81,4.17,4.41,3.59,5.87,3.83,6.03,4.89,4.32,4.69])} >>>> dataf = robjects.r['data.frame'](**d) >>>> >>> That way of creating a data.frame is correct, although not the only way. >>> >>> # with rpy2-2.1.x-dev >>> >>> import rpy2.robjects as robjects >>> >>> dataf = robjects.DataFrame(d) >>> >>> >>> If having to know the type of the R vectors to create is seen as >>> inelegant, >>> on can put rpy_classic to use: >>> >>> import rpy2.rpy_classic as rpy >>> >>> d={"x":rpy.seq2veq([4.17,5.58,5.18,6.11,4.50,4.61,5.17]), >>> "y":rpy.seq2veq([4.81,4.17,4.41,3.59,5.87,3.83,6.03])} >>> >>> >>> I have to remind myself to write more about creating data.frames in the >>> documentation. >>> >>> L. >>> >>> >>>> Thanks! >>>> >>>> >>>> >>>> ------------------------------------------------------------------------------ >>>> Let Crystal Reports handle the reporting - Free Crystal Reports 2008 >>>> 30-Day trial. Simplify your report design, integration and deployment - >>>> and >>>> focus on what you do best, core application coding. Discover what's new >>>> with >>>> Crystal Reports now. http://p.sf.net/sfu/bobj-july >>>> _______________________________________________ >>>> rpy-list mailing list >>>> rpy-list@lists.sourceforge.net >>>> https://lists.sourceforge.net/lists/listinfo/rpy-list >>> > >
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