如何减少 R 中的残差标准差？
how to reduce the residual standard deviation in R?
subject <- factor(rep(c(1,2,3,4,5,6),each=12))
dep <- c(0.3763244126,0.2185001692,0.4191841742,0.9812978664,0.7429273683,0.6254715486,0.6200958213,0.4693300191,0.1779032899,0.0035873980,0.8821949826,0.4818012617,0.0008013437,0.6280700732,0.7126500814,0.4984349359,0.2457996449,0.3085733312,0.5903398243,0.3704800352,0.8215325437,0.0445236221,0.1849731791,0.3670945817,0.0022268933,0.1630332691,0.9734050406,0.2638539758,0.8550054496,0.9413964085,0.4548943471,0.0440815873,0.5222098769,0.6553600784,0.6853486744,0.0571945074,0.0923124240,0.6976544929,0.9257440316,0.5658967043,0.0636543627,0.1038574059,0.0662497468,0.9165439918,0.5200087291,0.9528015053,0.5347318368,0.1848373057,0.9948602219,0.9633110918,0.1482162909,0.9000614029,0.0898618386,0.7975253051,0.8334557347,0.8629821099,0.0001795699,0.2488384889,0.6382902598,0.1103540359,0.2199716354,0.2737281912,0.5694398067,0.7940423761,0.4906677457,0.5191186895,0.4770589883,0.2823238128,0.2458788699,0.6363522802,0.0306954833,0.6979198116)
f1 <- factor(rep(c("Female","Male","Female","Male","Male","Female"), each=12))
f2 <- c(0.098788608,0.934606288,0.145045152,0.841969882,0.498234471,0.562897249,0.359740488,0.082046687,0.183987342,0.082418820,0.173424633,0.799291329,0.041450568,0.686708743,0.352092230,0.823550310,0.650857094,0.331705763,0.659111451,0.745187314,0.066165065,0.870759966,0.154977488,0.031703774,0.065251788,0.707452073,0.564604314,0.224798417,0.656363138,0.047954841,0.500513114,0.923316812,0.706629266,0.561530974,0.670860932,0.414969178,0.709973062,0.452946384,0.187624344,0.278656351,0.562138433,0.655193272,0.014868182,0.518697012,0.414113229,0.273464316,0.844080831,0.962636550,0.952739605,0.728627219,0.761122951,0.309977150,0.755239042,0.208627128,0.481429897,0.376021223,0.753871400,0.164842337,0.921081061,0.859677311,0.600462073,0.119193708,0.276722102,0.854752641,0.962710853,0.956277061,0.228313179,0.920405764,0.001594131,0.104930433,0.241548888,0.549643015)
f3 <- factor(rep(c("day1","day2","day3","day4"),each=3, times=6))
data <- data.frame(sub=subject, dep=dep, f1=f1, f2=f2, f3=f3)
m <- lmer(dep ~ f1*f2*f3 + (1|sub), data=data)
在此 post 中，whuber 创建了一个新数据集，其中拟合系数与上述模型相同，但剩余标准偏差降低了十分之一，如第二张图所示。
实现该结果的代码是什么？
要从模型中恢复原始数据点但只有 1/10 的残差标准差，您可以使用：
data$dep2 <- predict(m) + 0.1 * residuals(m)
subject <- factor(rep(c(1,2,3,4,5,6),each=12))
dep <- c(0.3763244126,0.2185001692,0.4191841742,0.9812978664,0.7429273683,0.6254715486,0.6200958213,0.4693300191,0.1779032899,0.0035873980,0.8821949826,0.4818012617,0.0008013437,0.6280700732,0.7126500814,0.4984349359,0.2457996449,0.3085733312,0.5903398243,0.3704800352,0.8215325437,0.0445236221,0.1849731791,0.3670945817,0.0022268933,0.1630332691,0.9734050406,0.2638539758,0.8550054496,0.9413964085,0.4548943471,0.0440815873,0.5222098769,0.6553600784,0.6853486744,0.0571945074,0.0923124240,0.6976544929,0.9257440316,0.5658967043,0.0636543627,0.1038574059,0.0662497468,0.9165439918,0.5200087291,0.9528015053,0.5347318368,0.1848373057,0.9948602219,0.9633110918,0.1482162909,0.9000614029,0.0898618386,0.7975253051,0.8334557347,0.8629821099,0.0001795699,0.2488384889,0.6382902598,0.1103540359,0.2199716354,0.2737281912,0.5694398067,0.7940423761,0.4906677457,0.5191186895,0.4770589883,0.2823238128,0.2458788699,0.6363522802,0.0306954833,0.6979198116)
f1 <- factor(rep(c("Female","Male","Female","Male","Male","Female"), each=12))
f2 <- c(0.098788608,0.934606288,0.145045152,0.841969882,0.498234471,0.562897249,0.359740488,0.082046687,0.183987342,0.082418820,0.173424633,0.799291329,0.041450568,0.686708743,0.352092230,0.823550310,0.650857094,0.331705763,0.659111451,0.745187314,0.066165065,0.870759966,0.154977488,0.031703774,0.065251788,0.707452073,0.564604314,0.224798417,0.656363138,0.047954841,0.500513114,0.923316812,0.706629266,0.561530974,0.670860932,0.414969178,0.709973062,0.452946384,0.187624344,0.278656351,0.562138433,0.655193272,0.014868182,0.518697012,0.414113229,0.273464316,0.844080831,0.962636550,0.952739605,0.728627219,0.761122951,0.309977150,0.755239042,0.208627128,0.481429897,0.376021223,0.753871400,0.164842337,0.921081061,0.859677311,0.600462073,0.119193708,0.276722102,0.854752641,0.962710853,0.956277061,0.228313179,0.920405764,0.001594131,0.104930433,0.241548888,0.549643015)
f3 <- factor(rep(c("day1","day2","day3","day4"),each=3, times=6))
data <- data.frame(sub=subject, dep=dep, f1=f1, f2=f2, f3=f3)
m <- lmer(dep ~ f1*f2*f3 + (1|sub), data=data)
在此 post 中，whuber 创建了一个新数据集，其中拟合系数与上述模型相同，但剩余标准偏差降低了十分之一，如第二张图所示。 实现该结果的代码是什么？
要从模型中恢复原始数据点但只有 1/10 的残差标准差，您可以使用：
data$dep2 <- predict(m) + 0.1 * residuals(m)