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我想绘制一个二进制混合效果模型的结果在视觉表示中的一篇论文。如何为视觉表示绘制二元混合效果模型
我使用LME以适应混合模型:
M2 <- lme(Pass ~ zone.time + length + Fat,
random =~ 1 | Year)
通行证=二进制1/0 zone.time,长度&脂肪=连续
得到:
Linear mixed-effects model fit by maximum likelihood
Data: DF1
AIC BIC logLik
39.05604 47.25981 -13.52802
Random effects:
Formula: ~1 | Year
(Intercept) Residual
StdDev: 5.03879e-06 0.3857927
Fixed effects: Pass ~ zone.time + length + Fat
Value Std.Error DF t-value p-value
(Intercept) 4.549716 1.2384118 24 3.673832 0.0012
zone.time 0.299438 0.1239111 24 2.416559 0.0236
length -0.006718 0.0019492 24 -3.446603 0.0021
Fat -0.051460 0.0213211 24 -2.413563 0.0238
Correlation:
(Intr) zon.tm length
zone.time 0.045
length -0.979 -0.168
Fat -0.447 -0.191 0.330
Standardized Within-Group Residuals:
Min Q1 Med Q3 Max
-1.9097237 -0.7802111 -0.1410353 0.5683329 2.0908188
Number of Observations: 29
Number of Groups: 2
我然后去计算预测值和标准误差:
MyData <- expand.grid(zone.time = seq(1,3.6, length = 10),
length = seq(525, 740, length = 10),
Fat = seq(3.7, 17, length = 10))
X <- model.matrix(~zone.time + length + Fat, data = MyData)
提取参数和参数协方差矩阵
betas <- fixef(M2)
用于样本数据使用
betas<- structure(c(4.54971638246632, 0.299438350935228, -0.00671801197327911,-0.0514597408192487), .Names = c("(Intercept)", "zone.time", "length","Fat"))
。
Covbetas <- vcov(M2)
样本数据使用:
Covbetas <- structure(c(1.32212400759181, 0.0059001955657893, -0.00203725210229123,
-0.0101822039057957, 0.0059001955657893, 0.0132361635192455,
-3.50672281561515e-05, -0.000434188193496185, -0.00203725210229123,
-3.50672281561515e-05, 3.27522409259271e-06, 1.18250356798504e-05,
-0.0101822039057957, -0.000434188193496185, 1.18250356798504e-05,
0.000391886154502855), .Dim = c(4L, 4L), .Dimnames = list(c("(Intercept)",
"zone.time", "length", "Fat"), c("(Intercept)", "zone.time",
"length", "Fat")))
计算在预测器规模
MyData$eta <- X %*% betas
MyData$Pi <- exp(MyData$eta)/(1 + exp(MyData$eta))
计算上的预测器功能的规模的SE拟合值
MyData$se <- sqrt(diag(X %*% Covbetas %*% t(X)))
MyData$SeUp <- exp(MyData$eta + 1.96 *MyData$se)/(1 + exp(MyData$eta + 1.96 *MyData$se))
MyData$SeLo <- exp(MyData$eta - 1.96 *MyData$se)/(1 + exp(MyData$eta - 1.96 *MyData$se))
head(MyData)
这是计算预测值的正确方法?
我该如何绘制这个视觉呈现?
我应该使用像
library(effects)
plot(allEffects(M2, default.levels=50))
或GGPLOT2
我希望得到更多的东西,沿着传统的二元逻辑回归:http://ww2.coastal.edu/kingw/statistics/R-tutorials/images/menarche2.png – 2014-09-02 16:13:27
在你的例子中,你有年龄在x轴和y我猜测是逻辑回归的拟合值。我们有拟合的值,而不是逻辑回归,但我们现在可以忽略这个变量,你想在x轴上使用什么变量? – bstockton 2014-09-02 17:28:36
无论如何,我想我可以在 – 2014-09-02 18:49:19