Hi It seems that some linearisation can be done by with(dat,plot(exp(Requests), log(1/Time)))
it can be fitted by fit<-lm(log(1/Time)~exp(Requests), data=dat) or library(MASS) fit.r<-lqs(log(1/Time)~exp(Requests), data=dat) coef(fit.r) (Intercept) exp(Requests) -7.137549922 -0.002378384 fff<- function(x) 1/(exp(-7.137549922 -0.002378384*exp(x))) points(dat$Requests, fff(dat$Requests), col=2) It is not perfect and especially points > dat[c(91,94),] Requests Time 91 5.103478 3618 94 5.262253 2908 are quite far from predicted model. Also it is probably better to use nonlinear regression ?nls Regards Petr > -----Original Message----- > From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- > project.org] On Behalf Of Manoj Srivastava > Sent: Monday, April 08, 2013 6:05 PM > To: r-help@r-project.org > Subject: Re: [R] fitting a hyperbola to data points > > On Mon, Apr 08 2013, PIKAL Petr wrote: > > Thanks for responding. > > > without data we can provide just basic help. > > fit<-lm(Time~I(1/Requests)) > > shall give you hyperbolic fit. > > > You can test if your data follow this assumption by plot(1/Requests, > > Time) which shall for straight line. > > > > anyway, when you want to provide data use > > dput(your.data) and copy console output directly to your mail. > > Pardon. I had put my data on pastebin, but here is the dput: > > dput(dat) > structure(list(Requests = c(0.045364248295124, 0.11341062073781, > 0.16633557708212, 0.20413911732806, 0.26462478172156, 0.31754973806587, > 0.37047469441018, 0.42339965075449, 0.47632460709881, 0.52168885539393, > 0.58217451978743, 0.62753876808255, 0.68802443247605, 0.73338868077118, > 0.79387434516468, 0.8392385934598, 0.8997242578533, 0.94508850614843, > 0.99801346249274, 1.050938418837, 1.1038633751814, 1.1567883315257, > 1.20971328787, 1.2626382442143, 1.3155632005586, 1.3684881569029, > 1.4214131132472, 1.4743380695915, 1.5272630259359, 1.5801879822802, > 1.6255522305753, 1.6860378949688, 1.7389628513131, 1.7843270996082, > 1.8448127640017, 1.897737720346, 1.9431019686412, 1.9960269249855, > 2.056512589379, 2.1018768376741, 2.1548017940184, 2.2152874584119, > 2.260651706707, 2.3135766630513, 2.3665016193957, 2.41942657574, > 2.4723515320843, 2.5252764884286, 2.5782014447729, 2.6311264011172, > 2.6764906494123, 2.7369763138058, 2.7899012701502, 2.8428262264945, > 2.8881904747896, 2.9486761391831, 2.9940403874782, 3.1603759645603, > 3.2057402128555, 3.266225877249, 3.3115901255441, 3.3645150818884, > 3.4174400382327, 3.4779257026262, 3.5232899509213, 3.5762149072656, > 3.62913986361, 3.6820648199543, 3.7349897762986, 3.7879147326429, > 3.8408396889872, 3.8937646453315, 3.9391288936266, 3.9996145580201, > 4.0525395143644, 4.0979037626596, 4.1583894270531, 4.2113143833974, > 4.2566786316925, 4.317164296086, 4.3625285443811, 4.4230142087746, > 4.4683784570698, 4.5213034134141, 4.5817890778076, 4.7330032387913, > 4.7934889031848, 4.8388531514799, 4.9976280205129, 5.0505529768572, > 5.1034779332015, 5.1564028895458, 5.2017671378409, 5.2622528022344 ), > Time = c(1289L, 1262L, 1272L, 1222L, 1243L, 1259L, 1266L, 1242L, 1249L, > 1232L, 1303L, 1236L, 1251L, 1263L, 1234L, 1226L, 1232L, 1246L, 1249L, > 1272L, 1263L, 1247L, 1253L, 1257L, 1267L, 1262L, 1284L, 1266L, 1269L, > 1268L, 1273L, 1261L, 1261L, 1264L, 1276L, 1283L, 1275L, 1277L, 1293L, > 1285L, 1284L, 1289L, 1290L, 1289L, 1300L, 1292L, 1300L, 1291L, 1297L, > 1310L, 1303L, 1311L, 1317L, 1345L, 1315L, 1327L, 1322L, 1597L, 1623L, > 1510L, 1372L, 1429L, 1371L, 1365L, 1357L, 1366L, 1402L, 1370L, 1407L, > 1373L, 1399L, 1420L, 1405L, 1393L, 1428L, 1394L, 1422L, 1425L, 1457L, > 1501L, 1426L, 1539L, 1492L, 1476L, 1493L, 1580L, 1670L, 1556L, 1661L, > 1593L, 3618L, 1903L, 1941L, 2908L)), .Names = c("Requests", "Time"), > class = "data.frame", row.names = c(NA, -94L)) > > > When plotting with plot(1/Requests, Time), I still get a > parabolic line, not a linear one. Perhaps there is a data transform I > can do to get this into a linear mode? > > I have also plotted the data (Time vs Requests and 1/Requests) > on a log and log-log scale, but the curve remains stubbornly curved. > This seems to argue against an exponential relationship, does it not? > > > thanks again. > > manoj > > > >> I am new to R, and I suspect I am missing something simple. > > >> I have a data set that performance data that correlates > >> request rate to response times > >> http://pastebin.com/Xhg0RaUp > >> There is some jitter in the data, but mostly it looks like a hockey > >> puck curve. It does not get converted into a straight line when I > >> tried log conversions, so it does not seem to be a power series > relationship. > > >> My expectation is that the data will fit a curve that is a > >> hyperbola, but I don't know how to formulate that regression. How > >> does one fit data to a general function > >> AX^2 + Bxy + Cy^2 +D = 0 > > >> I have tried polynomial functions and inverse functions > >> lm2 = lm(Time ~ Requests + I(Requests^2) + I(Requests^3)) but > that > >> does not appear to be close. > > >> Any pointers appreciated. > > -- > Many of us spend half our life wishing for things we could have if we > didn't spend half our time wishing. -- Alexander Woollcott Manoj > Srivastava <sriva...@acm.org> <http://www.golden-gryphon.com/> > 4096R/C5779A1C E37E 5EC5 2A01 DA25 AD20 05B6 CF48 9438 C577 9A1C ______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
