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.
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