fastLm() 比 lm() 慢得多

本文介绍了fastLm() 比 lm() 慢得多的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

fastLm() 比 lm() 慢得多.基本上，我只是用相同的公式和数据调用 lm() 和 fastLm() ，但是 fastLm() 似乎比lm().这可能吗?我就是不知道怎么会这样?

fastLm() is much slower than lm(). Basically, I just call lm() and fastLm() with the same formula and data, but fastLm() seems to be much slower than lm(). Is this possible? I just don't know how could this happen?

dim(dat) #[1] 87462 90 ## library(Rcpp) library(RcppEigen) library(rbenchmark) benchmark(fastLm(formula(mez),data=dat),lm(formula(mez),data=dat)) test replications elapsed relative user.self sys.self user.child sys.child 1 fastLm(formula(mez), data = dat) 100 195.81 7.079 189.36 6.27 NA NA 2 lm(formula(mez), data = dat) 100 27.66 1.000 24.52 3.02 NA NA summary(mez) Call: lm(formula = totalActualVal ~ township + I(TotalFinishedSF^2) + mainfloorSF + nbrFullBaths + township + range + qualityCodeDscr + TotalFinishedSF:range + nbrBedRoom + PCT_HISP, data = dat) Residuals: Min 1Q Median 3Q Max -2607622 -53820 -2893 40704 3116043 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 2.418e+05 3.211e+03 75.307 < 2e-16 *** township1S 1.907e+04 1.239e+03 15.385 < 2e-16 *** township2N -7.540e+04 1.467e+03 -51.383 < 2e-16 *** township3N -9.482e+04 1.482e+03 -63.976 < 2e-16 *** I(TotalFinishedSF^2) 1.415e-02 3.923e-04 36.063 < 2e-16 *** mainfloorSF 6.754e+01 1.233e+00 54.793 < 2e-16 *** nbrFullBaths 5.261e+03 7.542e+02 6.977 3.05e-12 *** range71 -2.802e+04 5.172e+03 -5.418 6.03e-08 *** range72 -5.599e+04 7.615e+03 -7.353 1.96e-13 *** range73 -6.414e+04 1.067e+04 -6.010 1.86e-09 *** rangeothers -6.571e+04 2.662e+03 -24.687 < 2e-16 *** qualityCodeDscrEXCELLENT 5.800e+05 4.170e+03 139.090 < 2e-16 *** qualityCodeDscrEXCELLENT + 8.453e+05 9.713e+03 87.027 < 2e-16 *** qualityCodeDscrEXCELLENT++ 8.929e+05 1.013e+04 88.149 < 2e-16 *** qualityCodeDscrEXCEPTIONAL 1 1.134e+06 8.336e+03 136.005 < 2e-16 *** qualityCodeDscrEXCEPTIONAL 2 1.536e+06 1.411e+04 108.884 < 2e-16 *** qualityCodeDscrEXCEPTIONAL 3 2.061e+06 4.679e+04 44.040 < 2e-16 *** qualityCodeDscrFAIR -3.288e+04 3.760e+03 -8.744 < 2e-16 *** qualityCodeDscrGUT 5.931e+04 1.142e+03 51.941 < 2e-16 *** qualityCodeDscrLOW -1.394e+05 1.799e+04 -7.748 9.45e-15 *** qualityCodeDscrVERY GOOD 2.106e+05 2.242e+03 93.925 < 2e-16 *** qualityCodeDscrVERY GOOD + 3.126e+05 4.406e+03 70.942 < 2e-16 *** qualityCodeDscrVERY GOOD ++ 4.042e+05 3.839e+03 105.275 < 2e-16 *** nbrBedRoom 2.334e+04 5.874e+02 39.739 < 2e-16 *** PCT_HISP -1.571e+03 5.162e+01 -30.426 < 2e-16 *** range70 :TotalFinishedSF 3.997e+01 2.363e+00 16.919 < 2e-16 *** range71 :TotalFinishedSF 1.300e+02 2.990e+00 43.490 < 2e-16 *** range72 :TotalFinishedSF -2.289e+01 4.598e+00 -4.978 6.42e-07 *** range73 :TotalFinishedSF -4.111e+01 6.797e+00 -6.048 1.47e-09 *** rangeothers:TotalFinishedSF -6.331e+00 2.215e+00 -2.859 0.00426 ** Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 129100 on 87432 degrees of freedom Multiple R-squared: 0.8296, Adjusted R-squared: 0.8295 F-statistic: 1.468e+04 on 29 and 87432 DF, p-value: < 2.2e-16

推荐答案

RcppArmadillo 有更好的示例脚本，其中不同的版本定时:

The RcppArmadillo has a better example script in which different version are timed:

edd@max:~/git/rcpparmadillo/inst/examples(master)$ Rscript fastLm.r test replications relative elapsed 4 fLmSEXP(X, y) 5000 1.000 0.174 2 fLmTwoCasts(X, y) 5000 1.017 0.177 3 fLmConstRef(X, y) 5000 1.029 0.179 1 fLmOneCast(X, y) 5000 1.069 0.186 6 fastLmPureDotCall(X, y) 5000 1.218 0.212 5 fastLmPure(X, y) 5000 1.908 0.332 8 lm.fit(X, y) 5000 2.207 0.384 7 fastLm(frm, data = trees) 5000 29.609 5.152 9 lm(frm, data = trees) 5000 36.977 6.434 edd@max:~/git/rcpparmadillo/inst/examples(master)$

最后两个使用公式——这清楚地表明如果您追求速度，则不想使用公式，因为解析公式比实际运行要花费更长的时间回归.你可以为 RcppEigen 设置类似的东西，结果会相似.

The last two use a formula -- and this clearly shows that you do not want to use a formula if you are after speed as deparsing the formula takes a lot longer than actually running the regression. You could set something similar up for RcppEigen, the results will be similar.