Error In Solve.defaultr Lapack Routine Dgesv System Is Exactly Singular
here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company Business Learn more about hiring developers or posting ads with us Stack Overflow Questions Jobs Documentation Tags Users Badges Ask Question x Dismiss Join the Stack Overflow Community Stack Overflow is a community of 4.7 million programmers, just like you, helping each other. Join them; it only takes a minute: Sign up R inverting matrix with solve returning Error up vote 2 down vote favorite I am studying R programming. I am trying to inverting matrix. Below is what I have tried. x <- matrix(1:16, 4, 4) x # [,1] [,2] [,3] [,4] # [1,] 1 5 9 13 # [2,] 2 6 10 14 # [3,] 3 7 11 15 # [4,] 4 8 12 16 solve(x) # Error in solve.default(x) : # Lapack routine dgesv: system is exactly singular: U[3,3] = 0 solve(x) %*% x # Error in solve.default(x) : # Lapack routine dgesv: system is exactly singular: U[3,3] = 0 x %*% solve(x) # Error in solve.default(x) : # Lapack routine dgesv: system is exactly singular: U[3,3] = 0 I can not understand what 'singular' means. According to this link, it is said that if solve does not have second parameter, it inverts first parameter. I am fully confused, so need some explanation with example would be wonderful. r matrix matrix-inverse share|improve this question edited May 23 '15 at 9:31 user3710546 asked May 23 '15 at 9:23 Juneyoung Oh 1,35972152 Unfortunately for you, you are using a special case with your matrix "x". You would have more luck using a more general case (x <- matrix(rnorm(16),4,4)). –user3710546 May 23 '15 at 9:51 2 Possible duplicate of R solve:system is exactly singular –zx8754 Oct 8 '15 at 9:33 add a com
Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company Business Learn more about hiring developers or posting ads with us Cross Validated Questions Tags Users Badges Unanswered Ask Question _ Cross Validated is a question and answer site for people interested in statistics, machine learning, data analysis, data mining, and data visualization. Join them; it only takes a minute: Sign up Here's how it works: Anybody can http://stackoverflow.com/questions/30411024/r-inverting-matrix-with-solve-returning-error ask a question Anybody can answer The best answers are voted up and rise to the top Singularity issues in multinomial model using R up vote 3 down vote favorite I am trying to develop a mode choice model (4 modes: hov, transit, bike, walk) and below are two approaches I am using. I am having problems in both Approach 1 Mode choice as a function http://stats.stackexchange.com/questions/32585/singularity-issues-in-multinomial-model-using-r of price (cost of using the mode) price: generic variable trans1: choice variable (0,1) Dataset: work Command: > mode.choice <- mlogit(trans1 ~ price, work) Error: Error in solve.default(H, g[!fixed]) : Lapack routine dgesv: system is exactly singular P.S: Unlike previous posts, I don’t have NA’s in my dataset. Approach 2 Mode choice as a function of price and some alternative specific variables price: Generic Variable trans1: choice variable (0,1) hh1, hh2, hh3: alternative specific variables Dataset: work Command: > mode.choice <- mlogit(trans1 ~ price | hh1 + hh2 + hh3, work) Error: Error in solve.default(H, g[!fixed]) : Lapack routine dgesv: system is exactly singular I have tried different variables in both approaches but the singularity issue persists Help on any of these approaches would be greatly appreciated. r multinomial logit share|improve this question edited Jul 19 '12 at 9:00 chl♦ 37.5k6125243 asked Jul 19 '12 at 0:09 Transporter 1612 1 How is price coded in R (numeric or factor)? Did you check if any of your response categories didn't occur as a unique combination of your predictors? See also this related thread. –chl♦ Jul 19 '12 at 9:08 1 Highly probable that you have some perfectl
Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company Business http://stats.stackexchange.com/questions/56794/system-is-exactly-singular-in-r-function-boxcox-ar Learn more about hiring developers or posting ads with us Cross Validated Questions Tags Users Badges Unanswered Ask Question _ Cross Validated is a question and answer site for people interested in statistics, machine learning, data analysis, data https://stat.ethz.ch/pipermail/r-sig-ecology/2010-March/001141.html mining, and data visualization. Join them; it only takes a minute: Sign up Here's how it works: Anybody can ask a question Anybody can answer The best answers are voted up and rise to the top “system error in is exactly singular” in R function BoxCox.ar up vote 0 down vote favorite I'm trying to perform a Box-Cox transformation on some financial data (SPY). The BoxCox.ar function (in the package TSA) gives me the following error: Error in solve.default(res$hessian * length(x)) : Lapack routine dgesv: system is exactly singular: U[2,2] = 0 Here's the code I'm using: data = ts(read.csv("http://ichart.finance.yahoo.com/table.csv?s=SPY&d=3&e=21&f=2013&g=d&a=0&b=29&c=1993&ignore=.csv")[,7]) # (Normally I would reverse the data here!) plot(data) BoxCox.ar(data) How can I fix this? error in solve.defaultr I tried smoothing the data with a moving average but the error persisted. Is it possible to perturb the data to avoid the singular system? r time-series data-transformation finance share|improve this question edited Apr 21 '13 at 23:29 Glen_b♦ 149k19246512 asked Apr 21 '13 at 23:15 tba 1083 Possibly the clearly non-stationary data could lead to a problem. It's a bad idea to call your data data (not least because there's already a function by that name). You should at least call it something slightly descriptive like SPY93.13. If you have the package fortunes installed, fortune(77) is to the point. –Glen_b♦ Apr 21 '13 at 23:43 I can't advise on the specific problem here, but it is hard to see that you need any heavy machinery. Volume might be easier to analyse a log scale; the other variables are better left as they come. A glance at some basic descriptive statistics and graphs gives as much guidance as is needed here. –Nick Cox Apr 22 '13 at 0:34 @Glen_b Can you elaborate on why non-stationarity is an issue with Box-Cox? The power transformation is designed to eliminate heteroskedasticity. –tba Apr 22 '13 at 2:01 @NickCox The stock price isn't stationary, so I'm not sure how to analyze it without a transformation. Traditional financial models such as Black-Scholes assume
Messages sorted by: [ date ] [ thread ] [ subject ] [ author ] Thanks for the replies (including the last one from Scott as I was writing this reply). I'll summarize better what I did and what happens when I try weights and offsets (both implemented in the zeroinfl() function of package pscl). [Scott, no, I do not have to use the densities, but I would like to account for the surface area in the model. As well, yes, I would like to compare among body parts so I will look into the random effects - later.] I am using AIC for model selection (although most of the NB models are quite similar. In that case, I select the simplest model). I test each model with lrtest() as well to simplify the models. The result works great with using counts directly and I am happy with the results. The best model for this species is below and the estimates line up nicely with the data: zinb3<-zeroinfl(caligus_elongatus~location_on_body | 1, dist="negbin", link="logit", data=sturg) As suggested, weighting by surface area sounds like a great way to solve the issue. When I add weights = surface_area as: zinb3<-zeroinfl(caligus_elongatus~location_on_body | 1, weights=surface_area, dist="negbin", link="logit", data=sturg) I get the following errors: Error in solve.default(as.matrix(fit$hessian)) : Lapack routine dgesv: system is exactly singular In addition: Warning messages: 1: In eval(expr, envir, enclos) : non-integer #successes in a binomial glm! 2: In glm.fit(Z, as.integer(Y0), weights = weights, family = binomial(link = linkstr)) : fitted probabilities numerically 0 or 1 occurred There are two factor levels of location_on_body that have no counts so I removed those, but the same error occurs (the errors above are actually without those two factor levels - and I made sure to refactor the factor to remove empty levels). When I add in an offset, as suggested by Scott, the model runs. It does not run if I log(caligus_elongatus) counts - possibly I am interpreting the log( E( count)) = linear predictor + log( size) model setup as suggested by Scott? zinb3<-zeroinfl(caligus_elongatus~location_on_body | 1, offset=log(surface_area), dist="negbin", link="logit", data=sturg) Finally, I tried surface area as a covariate as it seems reasonable to assume that the effect of surface area is related to counts and this would be one way to mitigate the relationship. Please cor