将重复测量混合模型公式从SAS转换为R.

Question

关于更复杂的实验设计的混合模型有几个问题和帖子,所以我认为这个更简单的模型将有助于其他初学者以及我.

所以,我的问题是我想从sas proc混合程序中在R中制定重复测量ancova:

proc mixed data=df1;
FitStatistics=akaike
class GROUP person day;
model Y = GROUP X1 / solution alpha=.1 cl;
repeated / type=cs subject=person group=GROUP;
lsmeans GROUP;
run;

以下是使用R(下面)中创建的数据的SAS输出:

.           Effect       panel    Estimate       Error      DF    t Value    Pr > |t|     Alpha       Lower       Upper
            Intercept              -9.8693      251.04       7      -0.04      0.9697       0.1     -485.49      465.75
            panel        1         -247.17      112.86       7      -2.19      0.0647       0.1     -460.99    -33.3510
            panel        2               0           .       .        .         .             .           .           .
            X1                     20.4125     10.0228       7       2.04      0.0811       0.1      1.4235     39.4016

下面是我使用'nlme'包在R中制定模型的方法,但没有得到类似的系数估计:

## create reproducible example fake panel data set:
set.seed(94); subject.id = abs(round(rnorm(10)*10000,0))

set.seed(99); sds = rnorm(10,15,5);means = 1:10*runif(10,7,13);trends = runif(10,0.5,2.5)

this = NULL; set.seed(98)
for(i in 1:10) { this = c(this,rnorm(6, mean = means[i], sd = sds[i])*trends[i]*1:6)}
set.seed(97)
that = sort(rep(rnorm(10,mean = 20, sd = 3),6))

df1 = data.frame(day = rep(1:6,10), GROUP = c(rep('TEST',30),rep('CONTROL',30)),
                 Y = this,
                 X1 = that,
                 person = sort(rep(subject.id,6)))

## use package nlme
require(nlme)

## run repeated measures mixed model using compound symmetry covariance structure:
summary(lme(Y ~ GROUP + X1, random = ~ +1 | person,
            correlation=corCompSymm(form=~day|person), na.action = na.exclude,
            data = df1,method='REML'))

现在,R的输出,我现在意识到类似于以下输出lm():

                Value Std.Error DF    t-value p-value
(Intercept) -626.1622  527.9890 50 -1.1859379  0.2413
GROUPTEST   -101.3647  156.2940  7 -0.6485518  0.5373
X1            47.0919   22.6698  7  2.0772934  0.0764

我相信我对规范很接近,但不确定我缺少什么来使结果匹配(在合理范围内......).任何帮助,将不胜感激!

更新:使用下面答案中的代码,R输出变为:

> summary(model2)

滚动到底部查看参数估计 - 看!与SAS相同.

Linear mixed-effects model fit by REML
 Data: df1 
      AIC      BIC   logLik
  776.942 793.2864 -380.471

Random effects:
 Formula: ~GROUP - 1 | person
 Structure: Diagonal
        GROUPCONTROL GROUPTEST Residual
StdDev:      184.692  14.56864 93.28885

Correlation Structure: Compound symmetry
 Formula: ~day | person 
 Parameter estimate(s):
         Rho 
-0.009929987 
Variance function:
 Structure: Different standard deviations per stratum
 Formula: ~1 | GROUP 
 Parameter estimates:
    TEST  CONTROL 
1.000000 3.068837

Fixed effects: Y ~ GROUP + X1 

                Value Std.Error DF    t-value p-value
(Intercept)   -9.8706 251.04678 50 -0.0393178  0.9688
GROUPTEST   -247.1712 112.85945  7 -2.1900795  0.0647
X1            20.4126  10.02292  7  2.0365914  0.0811

Answer 1

请尝试以下:

model1 <- lme(
  Y ~ GROUP + X1,
  random = ~ GROUP | person,
  correlation = corCompSymm(form = ~ day | person),
  na.action = na.exclude, data = df1, method = "REML"
)
summary(model1)

在这种情况下,我认为random = ~ groupvar | subjvar选项提供R lme了类似的repeated / subject = subjvar group = groupvar选项结果SAS/MIXED.

编辑:

SAS /混合

R(修改后的模型2)

model2 <- lme(
  Y ~ GROUP + X1,
  random = list(person = pdDiag(form = ~ GROUP - 1)),
  correlation = corCompSymm(form = ~ day | person),
  weights = varIdent(form = ~ 1 | GROUP),
  na.action = na.exclude, data = df1, method = "REML"
)
summary(model2)

所以,我认为这些协方差结构非常相似(σ _G1 =τ _克² +σ ₁).

编辑2:

协变量估计(SAS/MIXED):

Variance            person          GROUP TEST        8789.23
CS                  person          GROUP TEST         125.79
Variance            person          GROUP CONTROL       82775
CS                  person          GROUP CONTROL       33297

所以

TEST group diagonal element
  = 125.79 + 8789.23
  = 8915.02
CONTROL group diagonal element
  = 33297 + 82775
  = 116072

其中对角元素=σ _K1 +σ _ķ ².

协变量估计(R lme):

Random effects:
 Formula: ~GROUP - 1 | person
 Structure: Diagonal
        GROUP1TEST GROUP2CONTROL Residual
StdDev:   14.56864       184.692 93.28885

Correlation Structure: Compound symmetry
 Formula: ~day | person 
 Parameter estimate(s):
         Rho 
-0.009929987 
Variance function:
 Structure: Different standard deviations per stratum
 Formula: ~1 | GROUP 
 Parameter estimates:
   1TEST 2CONTROL 
1.000000 3.068837 

所以

TEST group diagonal element
  = 14.56864^2 + (3.068837^0.5 * 93.28885 * -0.009929987) + 93.28885^2
  = 8913.432
CONTROL group diagonal element
  = 184.692^2  + (3.068837^0.5 * 93.28885 * -0.009929987) + (3.068837 * 93.28885)^2
  = 116070.5

其中对角元素=τ _克² +σ ₁ +σ _克².