Re: [R] [FORGED] Regression with factors ?
2016-07-13 20:09 GMT+02:00 Jeff Newmiller: > The formula interface as used in lm and nls searches for separate > coefficients for each variable.. it will take someone more clever than I to > figure out how to get the formula interface to think of two variables as > instances of one factor. > > However, R can do nonlinear optimization just fine: ... Hello Jeff, hello all, thank you. This appears to be what I was looking for. Is there a straightforward way to test for significance ? THX, stn __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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.
Re: [R] [FORGED] Regression with factors ?
The formula interface as used in lm and nls searches for separate coefficients for each variable.. it will take someone more clever than I to figure out how to get the formula interface to think of two variables as instances of one factor. However, R can do nonlinear optimization just fine: ## # as if read in using read.csv( fname, as.is=TRUE ) dta <- data.frame( y = observed_data$y , p1 = as.character( observed_data$p1 ) , p2 = as.character( observed_data$p2 ) , stringsAsFactors = FALSE ) lvls <- with( dta, unique( c( p1, p2 ) ) ) dta$p1f <- factor( dta$p1, levels = lvls ) dta$p2f <- factor( dta$p2, levels = lvls ) idxvmult <- length( lvls ) + 1L idxvoffs <- length( lvls ) + 2L # all values in a numeric vector # x = c( valice, vbob, ..., vmult, voffs ) calcY <- function( x ) { vmult <- x[ idxvmult ] voffs <- x[ idxvoffs ] vp1 <- x[ dta$p1f ] vp2 <- x[ dta$p2f ] vmult * ( voffs - ( vp1 - vp2 )^2 ) } optfcn <- function( x ) { sum( ( dta$y - calcY( x ) ) ^ 2 ) } oresult <- optim( par = rep( 1, idxvoffs ), optfcn) result <- list( multiplier = oresult$par[ idxvmult ] , offset = oresult$par[ idxvoffs ] , values = data.frame( lvls = lvls , values = oresult$par[ seq.int( length( lvls ) ) ] ) ) result #- I highly recommend reading the help page for optim and the CRAN Task View on optimization [1] [1] https://cran.r-project.org/web/views/Optimization.html On Wed, 13 Jul 2016, stn021 wrote: Is this what is intended? observed_data$p1ab <- persons$ability[ match(observed_data$p1, persons$name) ] observed_data$p2ab <- persons$ability[ match(observed_data$p2, persons$name) ] Hello David, thank you for your answer. The code in my previous post was intended as an answer to the question in an earlier post about example-data, quote: Would you like me to make a complete example dataset with more records and noise ? Yes. And preferably do it with R code. I should have re-stated this connection in the post. The code generates a matrix 'observed_data' which is the data the experimenter would get during the experiment. This matrix is output in the last line. All other output is only meant to document the generation-process. So the only thing visible to the experimenter before analysis is exactly that matrix 'observed_data' (usually in the form of some written documentation which is later entered into statistical software). Everything before that last line simulates those unknown parameters that the experiment is supposed to reveal. The unknown parameters are specifically - the matrix 'persons' - and the variable 'multiplyer' Both are supposed to be revealed by the analyis. p1ab and p2ab would therefore depend on the unknown parameters and could not be added to 'observed_data' before the analysis. Sorry again for omitting the back-reference. I would like to know: - how to get R to use p1 and p2 as levels of the same factor (=persons) instead of levels of two different factors. - how to get R to multiply the numerical levels of factors during the search for the solution. Factors cannot be multiplied before running lm() or some other package because before the analysis their numerical values are not known. THX, Stefan __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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. --- Jeff NewmillerThe . . Go Live... DCN:Basics: ##.#. ##.#. Live Go... Live: OO#.. Dead: OO#.. Playing Research Engineer (Solar/BatteriesO.O#. #.O#. with /Software/Embedded Controllers) .OO#. .OO#. rocks...1k __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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.
Re: [R] [FORGED] Regression with factors ?
> On Jul 13, 2016, at 8:01 AM, David Winsemiuswrote: > > >> On Jul 13, 2016, at 6:48 AM, stn021 wrote: >> >> Hello, >> >> so here a numerical example in R-code. Code is appended below. >> >> The output should be >> 1) the numerical values of the abilities of the persons >> 2) the multiplyer >> >> >> Please note that >> >> 1) I have used non-linear optimization to solve this problem and got >> the expected result, though not with R but other software. >> >> 2) I have applied lm() to this problem, even before I posted the >> question. I am well aware of the syntax of formulas. I my last posting >> I wrote the formula "freehand" so I made the previously mentioned >> errors. Sorry about that. >> >> >> >> Unfortunately the formulas with I() as well as multiplying variables >> before running R does not work here. I() does not apply to factors (R >> tells me) and multiplying in advance also works only for continuous >> variables, not for factors, because there is no known numerical value >> to multiply. >> >> The latter is actually what my question is about, along with the >> question on how to get R to treat two columns as two instances of the >> same factor. >> >> >> Just to be sure I used R to check if the data really counts as a >> factor according to R-terminology. It really is a factor, see code >> below. >> >> >> >> This is the code for generating the example-data: >> >> # --- # >> pnames= c( "alice" , "bob" , "charlie" , "don" , "eve" , "freddy" >> , "grace" , "henry" ) >> pcount= length( pnames ) >> >> # abilities = runif( pcount ) >> abilities = (1:pcount) / 10 >> >> persons = data.frame( name = pnames , ability = abilities ) >> persons >> >> # random subset of possible combinations and extra df >> combinations = combn( nrow( persons ) , 2 ) ; >> combinations = cbind( combinations,combinations,combinations,combinations ) >> combinations = combinations[ , runif(ncol(combinations))<0.5 ] >> ccount = ncol( combinations ) >> >> observed_data = data.frame( >> idx1 = combinations[1,] >> , idx2 = combinations[2,] >> , p1 = ( persons$name[combinations[1,] ] ) >> , p2 = ( persons$name[combinations[2,] ] ) >> ) >> >> abilities_data = data.frame( >> a1 = persons$ability[ combinations[1,] ] >> , a2 = persons$ability[ combinations[2,] ] >> ) >> >> # y = result of cooperation of each pair >> multiplyer = runif(1) + 1 >> offset = 1 >> cat( "multiplyer = " , multiplyer , "\n" ) >> cat( "offset = " , offset , "\n" ) >> >> y0 = multiplyer * ( offset - ( abilities_data$a1 - abilities_data$a2 ) ^ 2 ) >> noise = .05 * rnorm( ccount ) >> >> # check variables are really factors : >> str( observed_data$p1 ) >> dput( observed_data$p1 ) >> >> observed_data = data.frame( y = round( y0+noise,3 ) , observed_data ) >> observed_data >> >> # --- # > > Is this what is intended? > >> observed_data$p1ab <- persons$ability[ match(observed_data$p1, persons$name) >> ] >> observed_data$p2ab <- persons$ability[ match(observed_data$p2, persons$name) >> ] >> head(observed_data) > y idx1 idx2p1 p2 p1ab p2ab > 1 1.14916 alice freddy 0.1 0.6 > 2 1.00617 alice grace 0.1 0.7 > 3 1.52923 bob charlie 0.2 0.3 > 4 1.40425 bob eve 0.2 0.5 > 5 1.20526 bob freddy 0.2 0.6 > 6 1.18727 bob grace 0.2 0.7 > > >> lm( y ~ I( (p1ab -p2ab)^2 ), data=observed_data) > > Call: > lm(formula = y ~ I((p1ab - p2ab)^2), data = observed_data) > > Coefficients: > (Intercept) I((p1ab - p2ab)^2) > 1.506 -1.435 > >> separate_term <- lm( y ~ I( (p1ab -p2ab)^2 ), data=observed_data) >> summary(separate_term) > > Call: > lm(formula = y ~ I((p1ab - p2ab)^2), data = observed_data) > > Residuals: > Min1QMedian3Q Max > -0.116249 -0.030996 0.002633 0.032765 0.136282 > > Coefficients: > Estimate Std. Error t value Pr(>|t|) > (Intercept) 1.505890.01067 141.08 <2e-16 *** > I((p1ab - p2ab)^2) -1.435270.05863 -24.48 <2e-16 *** > --- > Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 > > Residual standard error: 0.05304 on 44 degrees of freedom > Multiple R-squared: 0.9316, Adjusted R-squared: 0.93 > F-statistic: 599.2 on 1 and 44 DF, p-value: < 2.2e-16 > > You could also have compared 2 models differing only with rest to the > includion of an interaction term that was the squared difference in abilities: > >> full <- lm( y ~ p1ab + p2ab + I( (p1ab -p2ab)^2 ), data=observed_data) >> reduced <- lm( y ~ p1ab + p2ab , data=observed_data) >> anova(full,reduced) > Analysis of Variance Table > > Model 1: y ~ p1ab + p2ab + I((p1ab - p2ab)^2) > Model 2: y ~ p1ab + p2ab > Res.Df RSS Df Sum of Sq FPr(>F) > 1 42 0.11823
Re: [R] [FORGED] Regression with factors ?
> Is this what is intended? > >> observed_data$p1ab <- persons$ability[ match(observed_data$p1, persons$name) >> ] >> observed_data$p2ab <- persons$ability[ match(observed_data$p2, persons$name) >> ] Hello David, thank you for your answer. The code in my previous post was intended as an answer to the question in an earlier post about example-data, quote: > > Would you like me to make a complete example dataset with more records and > > noise ? > Yes. And preferably do it with R code. I should have re-stated this connection in the post. The code generates a matrix 'observed_data' which is the data the experimenter would get during the experiment. This matrix is output in the last line. All other output is only meant to document the generation-process. So the only thing visible to the experimenter before analysis is exactly that matrix 'observed_data' (usually in the form of some written documentation which is later entered into statistical software). Everything before that last line simulates those unknown parameters that the experiment is supposed to reveal. The unknown parameters are specifically - the matrix 'persons' - and the variable 'multiplyer' Both are supposed to be revealed by the analyis. p1ab and p2ab would therefore depend on the unknown parameters and could not be added to 'observed_data' before the analysis. Sorry again for omitting the back-reference. I would like to know: - how to get R to use p1 and p2 as levels of the same factor (=persons) instead of levels of two different factors. - how to get R to multiply the numerical levels of factors during the search for the solution. Factors cannot be multiplied before running lm() or some other package because before the analysis their numerical values are not known. THX, Stefan __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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.
Re: [R] [FORGED] Regression with factors ?
> On Jul 13, 2016, at 6:48 AM, stn021wrote: > > Hello, > > so here a numerical example in R-code. Code is appended below. > > The output should be > 1) the numerical values of the abilities of the persons > 2) the multiplyer > > > Please note that > > 1) I have used non-linear optimization to solve this problem and got > the expected result, though not with R but other software. > > 2) I have applied lm() to this problem, even before I posted the > question. I am well aware of the syntax of formulas. I my last posting > I wrote the formula "freehand" so I made the previously mentioned > errors. Sorry about that. > > > > Unfortunately the formulas with I() as well as multiplying variables > before running R does not work here. I() does not apply to factors (R > tells me) and multiplying in advance also works only for continuous > variables, not for factors, because there is no known numerical value > to multiply. > > The latter is actually what my question is about, along with the > question on how to get R to treat two columns as two instances of the > same factor. > > > Just to be sure I used R to check if the data really counts as a > factor according to R-terminology. It really is a factor, see code > below. > > > > This is the code for generating the example-data: > > # --- # > pnames= c( "alice" , "bob" , "charlie" , "don" , "eve" , "freddy" > , "grace" , "henry" ) > pcount= length( pnames ) > > # abilities = runif( pcount ) > abilities = (1:pcount) / 10 > > persons = data.frame( name = pnames , ability = abilities ) > persons > > # random subset of possible combinations and extra df > combinations = combn( nrow( persons ) , 2 ) ; > combinations = cbind( combinations,combinations,combinations,combinations ) > combinations = combinations[ , runif(ncol(combinations))<0.5 ] > ccount = ncol( combinations ) > > observed_data = data.frame( > idx1 = combinations[1,] > , idx2 = combinations[2,] > , p1 = ( persons$name[combinations[1,] ] ) > , p2 = ( persons$name[combinations[2,] ] ) > ) > > abilities_data = data.frame( > a1 = persons$ability[ combinations[1,] ] > , a2 = persons$ability[ combinations[2,] ] > ) > > # y = result of cooperation of each pair > multiplyer = runif(1) + 1 > offset = 1 > cat( "multiplyer = " , multiplyer , "\n" ) > cat( "offset = " , offset , "\n" ) > > y0 = multiplyer * ( offset - ( abilities_data$a1 - abilities_data$a2 ) ^ 2 ) > noise = .05 * rnorm( ccount ) > > # check variables are really factors : > str( observed_data$p1 ) > dput( observed_data$p1 ) > > observed_data = data.frame( y = round( y0+noise,3 ) , observed_data ) > observed_data > > # --- # Is this what is intended? > observed_data$p1ab <- persons$ability[ match(observed_data$p1, persons$name) ] > observed_data$p2ab <- persons$ability[ match(observed_data$p2, persons$name) ] > head(observed_data) y idx1 idx2p1 p2 p1ab p2ab 1 1.14916 alice freddy 0.1 0.6 2 1.00617 alice grace 0.1 0.7 3 1.52923 bob charlie 0.2 0.3 4 1.40425 bob eve 0.2 0.5 5 1.20526 bob freddy 0.2 0.6 6 1.18727 bob grace 0.2 0.7 > lm( y ~ I( (p1ab -p2ab)^2 ), data=observed_data) Call: lm(formula = y ~ I((p1ab - p2ab)^2), data = observed_data) Coefficients: (Intercept) I((p1ab - p2ab)^2) 1.506 -1.435 > separate_term <- lm( y ~ I( (p1ab -p2ab)^2 ), data=observed_data) > summary(separate_term) Call: lm(formula = y ~ I((p1ab - p2ab)^2), data = observed_data) Residuals: Min1QMedian3Q Max -0.116249 -0.030996 0.002633 0.032765 0.136282 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 1.505890.01067 141.08 <2e-16 *** I((p1ab - p2ab)^2) -1.435270.05863 -24.48 <2e-16 *** --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 0.05304 on 44 degrees of freedom Multiple R-squared: 0.9316,Adjusted R-squared: 0.93 F-statistic: 599.2 on 1 and 44 DF, p-value: < 2.2e-16 You could also have compared 2 models differing only with rest to the includion of an interaction term that was the squared difference in abilities: > full <- lm( y ~ p1ab + p2ab + I( (p1ab -p2ab)^2 ), data=observed_data) > reduced <- lm( y ~ p1ab + p2ab , data=observed_data) > anova(full,reduced) Analysis of Variance Table Model 1: y ~ p1ab + p2ab + I((p1ab - p2ab)^2) Model 2: y ~ p1ab + p2ab Res.Df RSS Df Sum of Sq FPr(>F) 1 42 0.11823 2 43 0.17315 -1 -0.05492 19.509 6.892e-05 *** -- David > > > 2016-07-11 19:16 GMT+02:00 Jeff Newmiller : >> Your clarification is promising. A reproducible example is always >> preferred, though never a
Re: [R] [FORGED] Regression with factors ?
Hello, so here a numerical example in R-code. Code is appended below. The output should be 1) the numerical values of the abilities of the persons 2) the multiplyer Please note that 1) I have used non-linear optimization to solve this problem and got the expected result, though not with R but other software. 2) I have applied lm() to this problem, even before I posted the question. I am well aware of the syntax of formulas. I my last posting I wrote the formula "freehand" so I made the previously mentioned errors. Sorry about that. Unfortunately the formulas with I() as well as multiplying variables before running R does not work here. I() does not apply to factors (R tells me) and multiplying in advance also works only for continuous variables, not for factors, because there is no known numerical value to multiply. The latter is actually what my question is about, along with the question on how to get R to treat two columns as two instances of the same factor. Just to be sure I used R to check if the data really counts as a factor according to R-terminology. It really is a factor, see code below. This is the code for generating the example-data: # --- # pnames= c( "alice" , "bob" , "charlie" , "don" , "eve" , "freddy" , "grace" , "henry" ) pcount= length( pnames ) # abilities = runif( pcount ) abilities = (1:pcount) / 10 persons = data.frame( name = pnames , ability = abilities ) persons # random subset of possible combinations and extra df combinations = combn( nrow( persons ) , 2 ) ; combinations = cbind( combinations,combinations,combinations,combinations ) combinations = combinations[ , runif(ncol(combinations))<0.5 ] ccount = ncol( combinations ) observed_data = data.frame( idx1 = combinations[1,] , idx2 = combinations[2,] , p1 = ( persons$name[combinations[1,] ] ) , p2 = ( persons$name[combinations[2,] ] ) ) abilities_data = data.frame( a1 = persons$ability[ combinations[1,] ] , a2 = persons$ability[ combinations[2,] ] ) # y = result of cooperation of each pair multiplyer = runif(1) + 1 offset = 1 cat( "multiplyer = " , multiplyer , "\n" ) cat( "offset = " , offset , "\n" ) y0 = multiplyer * ( offset - ( abilities_data$a1 - abilities_data$a2 ) ^ 2 ) noise = .05 * rnorm( ccount ) # check variables are really factors : str( observed_data$p1 ) dput( observed_data$p1 ) observed_data = data.frame( y = round( y0+noise,3 ) , observed_data ) observed_data # --- # 2016-07-11 19:16 GMT+02:00 Jeff Newmiller: > Your clarification is promising. A reproducible example is always preferred, > though never a guarantee. I expect to be somewhat preoccupied this week so > responses may be rather delayed, but the less setup we have to the more > likely that someone on the list will tackle it. > > Re an answer: If you can make the example simple enough that you can tell us > what the right numerical result will be, we will have a better chance of > understanding what you are after. E.g. if you start with a solution and use > it to create sample input data with then you don't need to actually solve it > to illustrate what you are after. [1] > > Note that I am not aware of any package dedicated to this type of problem, so > unless someone else responds otherwise then you will likely have to use > bootstrapping or your own statistical analysis (Bayesian?) of the result. > > [1] > http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example > -- > Sent from my phone. Please excuse my brevity. > > On July 11, 2016 7:28:41 AM PDT, stn021 wrote: >>Hello, >> >>thank you for the replies. Sorry about the html-email, I forgot. >>Should be OK with this email. >> >> >>Don't be fooled be the apparent simplicity of the problem. I have >>tried to reduce it to only a single relatively simple question. >> >>The idea here is to model cooperation of two persons. The model is >>about one specific aspect of that cooperation, namely that two persons >>with similar abilities may be able to produce better results that two >>very different persons. >> >>That is only one part of the model with other parts modeling for >>example the fact that of course two persons with a higher degree of >>ability will produce better results per se. >> >> >>It is not classic regression with factors. That can be easily done by >>something like lm( y ~ (p1-p2)^2 ). >> >>This expands to lm( y ~ p1^2 - 2*p1*p2 + p2^2 ). This contains a >>multiplicagtions and for lm() this implies interactions between the >>factor-levels and produces one parameter for each combination of >>factor-levels that occurs in the data. That is not what the question >>is about. >> >>Also p1 and p2 are different levels of the same factor, while for lm() >>it would be two different factors with different levels. >> >> >>As for the sensical part:
Re: [R] [FORGED] Regression with factors ?
Your clarification is promising. A reproducible example is always preferred, though never a guarantee. I expect to be somewhat preoccupied this week so responses may be rather delayed, but the less setup we have to the more likely that someone on the list will tackle it. Re an answer: If you can make the example simple enough that you can tell us what the right numerical result will be, we will have a better chance of understanding what you are after. E.g. if you start with a solution and use it to create sample input data with then you don't need to actually solve it to illustrate what you are after. [1] Note that I am not aware of any package dedicated to this type of problem, so unless someone else responds otherwise then you will likely have to use bootstrapping or your own statistical analysis (Bayesian?) of the result. [1] http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example -- Sent from my phone. Please excuse my brevity. On July 11, 2016 7:28:41 AM PDT, stn021wrote: >Hello, > >thank you for the replies. Sorry about the html-email, I forgot. >Should be OK with this email. > > >Don't be fooled be the apparent simplicity of the problem. I have >tried to reduce it to only a single relatively simple question. > >The idea here is to model cooperation of two persons. The model is >about one specific aspect of that cooperation, namely that two persons >with similar abilities may be able to produce better results that two >very different persons. > >That is only one part of the model with other parts modeling for >example the fact that of course two persons with a higher degree of >ability will produce better results per se. > > >It is not classic regression with factors. That can be easily done by >something like lm( y ~ (p1-p2)^2 ). > >This expands to lm( y ~ p1^2 - 2*p1*p2 + p2^2 ). This contains a >multiplicagtions and for lm() this implies interactions between the >factor-levels and produces one parameter for each combination of >factor-levels that occurs in the data. That is not what the question >is about. > >Also p1 and p2 are different levels of the same factor, while for lm() >it would be two different factors with different levels. > > >As for the sensical part: this has a real world application therefore >it makes sense. > >Also it is not so difficult to solve with non-linear optimization. I >was hoping to be able to use R for that purpose because then the >results could easily be checked with statistical tests. > >So my question is not "how to solve" but "how to solve with R". > > >As for the excess degrees of freedom, in real observations there would >of course be added noise due to either random variations or factors >not included in the model. So to generate a more reality-conforming >example I could add some random normal-distributed noise to the >dependent variable y. I previously left that part out because to me it >did not seem relevant. > > >Would you like me to make a complete example dataset with more records >and noise ? > > >The answer I look for would be the numerical values of the >factor-levels and numerical values for the multiplier (f) and the >offset (o), with p1 and p2 given as names (here: persons) and y given >as some level of achievement they reach by cooperating. > >y = f * ( o - ( p1 - p2 )^2 ) > >Is that what you meant by "answer" ? > > >THX >stefan > > > > >2016-07-10 2:27 GMT+02:00 Jeff Newmiller : >> >> I have seen less sensical questions. >> >> It would be nice if the example were a bit more complete (as in it >should have excess degrees of freedom and an answer) and less like a >homework problem (which are off topic here). It would of course also be >helpful if the OP were to conform to the Posting Guide, particularly in >respect to using plain text email. >> >> It looks like the kind of nonlinear optimization problem that >evolutionary algorithms are often applied to. It doesn't look (to me) >like a typical problem that factors get applied to in formulas though, >because multiple instances of the same factor variable are present. >> -- >> Sent from my phone. Please excuse my brevity. >> >> On July 9, 2016 4:59:30 PM PDT, Rolf Turner >wrote: >> >On 09/07/16 20:52, stn021 wrote: >> >> Hello, >> >> >> >> I would like to analyse a model like this: >> >> >> >> y = 1 * ( 1 - ( x1 - x2 ) ^ 2 ) >> >> >> >> x1 and x2 are not continuous variables but factors, so the >> >observation >> >> contain the level. >> >> Its numerical value is unknown and is to be estimated with the >model. >> >> >> >> >> >> The observations look like this: >> >> >> >> yx1 x2 >> >> 0.96 Alice Bob >> >> 0.84 Alice Charlie >> >> 0.96 Bob Charlie >> >> 0.64 Dave Alice >> >> etc. >> >> >> >> Each person has a numerical value. Here for example Alice = 0.2 >and >> >Bob = >> >> 0.4 >> >> >> >> Then y = 0.96 = 1* ( 1- ( 0.2-0.4 ) ^ 2 ) , see first observation.
Re: [R] [FORGED] Regression with factors ?
> On Jul 11, 2016, at 7:28 AM, stn021wrote: > > Hello, > > thank you for the replies. Sorry about the html-email, I forgot. > Should be OK with this email. > > > Don't be fooled be the apparent simplicity of the problem. I have > tried to reduce it to only a single relatively simple question. It would be useful to know whether this is a design effort and the data is not yet recorded or this is an analysis effort for data that is "in the can". > > The idea here is to model cooperation of two persons. The model is > about one specific aspect of that cooperation, namely that two persons > with similar abilities may be able to produce better results that two > very different persons. > > That is only one part of the model with other parts modeling for > example the fact that of course two persons with a higher degree of > ability will produce better results per se. > > > It is not classic regression with factors. That can be easily done by > something like lm( y ~ (p1-p2)^2 ). No. The caret "^" is an interaction operator in the formula context (not a power operator) and the minus sign causes variable removal. Read: ?formula If you want to create a calculated value that is the squared difference of two variables, then you need to do it either with `I` or in the dataframe before submission to the regression function. > > This expands to lm( y ~ p1^2 - 2*p1*p2 + p2^2 ). Used in a formula, p1^2 is exactly equal to p1. > This contains a > multiplicagtions and for lm() this implies interactions between the > factor-levels and produces one parameter for each combination of > factor-levels that occurs in the data. That is not what the question > is about. > > Also p1 and p2 are different levels of the same factor, while for lm() > it would be two different factors with different levels. Given your apparent lack of knowledge about R's formula syntax, we are also now unclear if you are using the word "factor" in the colloquial sense or as a technical term for discrete (factor) variables in R. What kind of values can p1 and p2 take? > As for the sensical part: this has a real world application therefore > it makes sense. > > Also it is not so difficult to solve with non-linear optimization. I > was hoping to be able to use R for that purpose because then the > results could easily be checked with statistical tests. > > So my question is not "how to solve" but "how to solve with R". > > > As for the excess degrees of freedom, in real observations there would > of course be added noise due to either random variations or factors > not included in the model. So to generate a more reality-conforming > example I could add some random normal-distributed noise to the > dependent variable y. I previously left that part out because to me it > did not seem relevant. Knowing the nature of the outcome variable is generally important in statistical design. > > > Would you like me to make a complete example dataset with more records > and noise ? Yes. And preferably do it with R code. > > The answer I look for would be the numerical values of the > factor-levels and numerical values for the multiplier (f) and the > offset (o), with p1 and p2 given as names (here: persons) and y given > as some level of achievement they reach by cooperating. > > y = f * ( o - ( p1 - p2 )^2 ) > > Is that what you meant by "answer" ? Not really. We would expect to see some data, at least dummy data, in a form that could be used for testing and demonstration. The nature of "f" is particularly unclear (in large part because the science or "reality" is not described.) Is it a function? The "o" is probably going to be returned as an "(Intercept)". You started out with `lm` which would have little to do with non-linear optimization. You then said it "would not be so difficult" to do non-linear optimization of "something" which was not really specified with any substance. Without data and code it still reads as a salad of fragments of terminology lacking reference to a well-described scientific substrate. An "answer" would be: Describe an experiment or a well designed set of observations with a specific outcome. Describe the hypotheses. Present code or data with a desired analysis plan. Ask for problems in R coding. An off-topic question would be: Help me design my psychology class project. -- David. > > > THX > stefan > > > > > 2016-07-10 2:27 GMT+02:00 Jeff Newmiller : >> >> I have seen less sensical questions. >> >> It would be nice if the example were a bit more complete (as in it should >> have excess degrees of freedom and an answer) and less like a homework >> problem (which are off topic here). It would of course also be helpful if >> the OP were to conform to the Posting Guide, particularly in respect to >> using plain text email. >> >> It looks like the kind of nonlinear optimization problem that evolutionary >> algorithms
Re: [R] [FORGED] Regression with factors ?
Hello, thank you for the replies. Sorry about the html-email, I forgot. Should be OK with this email. Don't be fooled be the apparent simplicity of the problem. I have tried to reduce it to only a single relatively simple question. The idea here is to model cooperation of two persons. The model is about one specific aspect of that cooperation, namely that two persons with similar abilities may be able to produce better results that two very different persons. That is only one part of the model with other parts modeling for example the fact that of course two persons with a higher degree of ability will produce better results per se. It is not classic regression with factors. That can be easily done by something like lm( y ~ (p1-p2)^2 ). This expands to lm( y ~ p1^2 - 2*p1*p2 + p2^2 ). This contains a multiplicagtions and for lm() this implies interactions between the factor-levels and produces one parameter for each combination of factor-levels that occurs in the data. That is not what the question is about. Also p1 and p2 are different levels of the same factor, while for lm() it would be two different factors with different levels. As for the sensical part: this has a real world application therefore it makes sense. Also it is not so difficult to solve with non-linear optimization. I was hoping to be able to use R for that purpose because then the results could easily be checked with statistical tests. So my question is not "how to solve" but "how to solve with R". As for the excess degrees of freedom, in real observations there would of course be added noise due to either random variations or factors not included in the model. So to generate a more reality-conforming example I could add some random normal-distributed noise to the dependent variable y. I previously left that part out because to me it did not seem relevant. Would you like me to make a complete example dataset with more records and noise ? The answer I look for would be the numerical values of the factor-levels and numerical values for the multiplier (f) and the offset (o), with p1 and p2 given as names (here: persons) and y given as some level of achievement they reach by cooperating. y = f * ( o - ( p1 - p2 )^2 ) Is that what you meant by "answer" ? THX stefan 2016-07-10 2:27 GMT+02:00 Jeff Newmiller: > > I have seen less sensical questions. > > It would be nice if the example were a bit more complete (as in it should > have excess degrees of freedom and an answer) and less like a homework > problem (which are off topic here). It would of course also be helpful if the > OP were to conform to the Posting Guide, particularly in respect to using > plain text email. > > It looks like the kind of nonlinear optimization problem that evolutionary > algorithms are often applied to. It doesn't look (to me) like a typical > problem that factors get applied to in formulas though, because multiple > instances of the same factor variable are present. > -- > Sent from my phone. Please excuse my brevity. > > On July 9, 2016 4:59:30 PM PDT, Rolf Turner wrote: > >On 09/07/16 20:52, stn021 wrote: > >> Hello, > >> > >> I would like to analyse a model like this: > >> > >> y = 1 * ( 1 - ( x1 - x2 ) ^ 2 ) > >> > >> x1 and x2 are not continuous variables but factors, so the > >observation > >> contain the level. > >> Its numerical value is unknown and is to be estimated with the model. > >> > >> > >> The observations look like this: > >> > >> yx1 x2 > >> 0.96 Alice Bob > >> 0.84 Alice Charlie > >> 0.96 Bob Charlie > >> 0.64 Dave Alice > >> etc. > >> > >> Each person has a numerical value. Here for example Alice = 0.2 and > >Bob = > >> 0.4 > >> > >> Then y = 0.96 = 1* ( 1- ( 0.2-0.4 ) ^ 2 ) , see first observation. > >> > >> How can this be done in R ? > > > > > >This question makes about as little sense as it is possible to imagine. > > > >cheers, > > > >Rolf Turner > __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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.
Re: [R] [FORGED] Regression with factors ?
I have seen less sensical questions. It would be nice if the example were a bit more complete (as in it should have excess degrees of freedom and an answer) and less like a homework problem (which are off topic here). It would of course also be helpful if the OP were to conform to the Posting Guide, particularly in respect to using plain text email. It looks like the kind of nonlinear optimization problem that evolutionary algorithms are often applied to. It doesn't look (to me) like a typical problem that factors get applied to in formulas though, because multiple instances of the same factor variable are present. -- Sent from my phone. Please excuse my brevity. On July 9, 2016 4:59:30 PM PDT, Rolf Turnerwrote: >On 09/07/16 20:52, stn021 wrote: >> Hello, >> >> I would like to analyse a model like this: >> >> y = 1 * ( 1 - ( x1 - x2 ) ^ 2 ) >> >> x1 and x2 are not continuous variables but factors, so the >observation >> contain the level. >> Its numerical value is unknown and is to be estimated with the model. >> >> >> The observations look like this: >> >> yx1 x2 >> 0.96 Alice Bob >> 0.84 Alice Charlie >> 0.96 Bob Charlie >> 0.64 Dave Alice >> etc. >> >> Each person has a numerical value. Here for example Alice = 0.2 and >Bob = >> 0.4 >> >> Then y = 0.96 = 1* ( 1- ( 0.2-0.4 ) ^ 2 ) , see first observation. >> >> How can this be done in R ? > > >This question makes about as little sense as it is possible to imagine. > >cheers, > >Rolf Turner __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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.
Re: [R] [FORGED] Regression with factors ?
On 09/07/16 20:52, stn021 wrote: Hello, I would like to analyse a model like this: y = 1 * ( 1 - ( x1 - x2 ) ^ 2 ) x1 and x2 are not continuous variables but factors, so the observation contain the level. Its numerical value is unknown and is to be estimated with the model. The observations look like this: yx1 x2 0.96 Alice Bob 0.84 Alice Charlie 0.96 Bob Charlie 0.64 Dave Alice etc. Each person has a numerical value. Here for example Alice = 0.2 and Bob = 0.4 Then y = 0.96 = 1* ( 1- ( 0.2-0.4 ) ^ 2 ) , see first observation. How can this be done in R ? This question makes about as little sense as it is possible to imagine. cheers, Rolf Turner -- Technical Editor ANZJS Department of Statistics University of Auckland Phone: +64-9-373-7599 ext. 88276 __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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.