Question

我试图查看如何为我一直在运行的增长曲线模型绘制R中的二次增长。型号：

m1 <- lmer(score ~ Time + Group + Time_Sqaure + 
                      (1 + School | Subject), data=df, REML = FALSE)
tab_model(m1)

时间（B = 9.58，p <.01）和时间平方（B =-0.51，p <.01）以及组（B = 2.77，p <.01）的差异都很大。

如果我使用plot_model，它会为我提供适合每个组的最佳拟合线。

plot_model(m1, type = "pred", terms = c("Time", "Group"))

有没有一种方法可以绘制拟合曲线或二次增长曲线，以显示增长速度随着时间的推移而变慢？

谢谢！

Answer 1

要让sjPlot::plot_model理解发生了什么，您必须将Time_Square输入I(Time^2)而不是单独的预测变量。

鉴于df$Time_Square <- df$Time^2，以下两个模型应为您提供相同的结果：

m1 <- lmer(score ~ Time + Group + Time_Square + 
                      (1 + School | Subject), data=df, REML = FALSE)
m2 <- lmer(score ~ Time + Group + I(Time^2) + 
                      (1 + School | Subject), data=df, REML = FALSE)

但是，在第二个模型中，很明显两次输入了预测变量Time，因此在使用sjPlot::plot_model(...)进行绘制时可以考虑该变量。

为了确保这一点，我使用以下模拟数据进行了测试：

library(dplyr)

grps <- 2 #number of groups
subj <- 100 #number of subjects within group
obs <- 10 #number of observations/times per subjects

b_0 <- 0 #overall intercept
b_1 <- 9.58 #linear time effect
b_2 <- -0.51 #quadratic time effect

sd_b0 <- 0.4 #SD of random intercept per subject
sd_b1 <- 3 #SD of random slope per subject
sd_b3 <- 1 #SD of group effect (you can simulate more than 2 groups)
sd_resid <- 10 #SD of residuals

df <- list(Group = factor(rep(letters[1:grps], each=obs*subj)),
           Subject = factor(rep(1:subj, times=grps, each=obs)),
           Time = rep(1:obs, times=subj*grps)
           ) %>% as.data.frame()
df$TimeSq <- df$Time^2

subj_b0 <- rnorm(subj, b_0, sd_b0) %>% rep(times=grps, each=obs)
subj_b1 <- rnorm(subj, b_1, sd_b1) %>% rep(times=grps, each=obs)
grp_m <- rnorm(grps, 0, sd_b3) %>% rep(times=, each=subj*obs)

df$Score <- with(df, subj_b0 + Time*subj_b1 + (Time^2)*b_2 + grp_m + rnorm(grps*subj*obs, 0, sd_resid))

fit1 <- lme4::lmer(Score ~ Time + I(Time^2) + Group + (Time | Subject), data=df)

sjPlot::plot_model(fit1, type="pred", terms=c("Time"))

用二次增长绘制增长曲线

1 个答案: