用置信区间绘制相关数据？答案

【问题标题】：Plotting correlation data with confidence intervals?用置信区间绘制相关数据？
【发布时间】：2021-04-02 11:15:05
【问题描述】：

我想得到一个如下图所示的情节，我正在尝试@EDi 在此讨论中建议的基本解决方案How can I plot data with confidence intervals? 但我只得到一条线。由于某种原因，曲线和区间都没有出现在图表上。可能是什么原因？..

data1= structure(list(age = c(24L, 27L, 24L, 22L, 35L, 21L, 22L, 21L, 
28L, 38L, 24L, 25L, 19L, 26L, 29L, 24L, 19L, 29L, 28L, 20L, 24L, 
21L, 23L, 27L, 18L, 41L, 21L, 23L, 21L, 27L, 23L, 22L, 19L, 23L, 
33L, 26L, 17L, 18L, 25L, 18L), score = c(-6.0645124170511, 4.3940252995831, 
4.15580269131775, 1.1691679274712, -4.32827856995513, -0.2668521177591, 
8.91061981860061, 1.44416362490212, 3.39306437298507, 4.37743935782333, 
4.00814596065344, 3.38813584234337, -4.25848923986889, 5.20144422164507, 
-2.84703031978998, 1.38670515581247, 2.17503671423042, -0.8341918646001, 
6.63401697099899, -1.85160878674671, 3.87051319922875, 0.883889127851464, 
-1.1317506003907, 0.327451161805888, 7.16166723663285, 10.221595241833, 
-0.473906061363301, 4.96930361012877, -9.52463189435209, 0.319670180437333, 
5.61710360920224, 7.54367918513063, -3.61072956084597, 3.01758121182583, 
3.03415512263794, 1.34523469737787, -4.4845445737846, -3.22655995899929, 
0.735028502754514, 2.77863366523645)), row.names = c(NA, -40L
), class = "data.frame")

这是代码，我看不出与上述讨论中的 EDi 示例有任何区别

plot(score ~ 
       age, 
     data = data1)

# model
mod <- lm(score ~ age, 
          data = data1)


# predicts + interval
newx <- seq(min(data1$score), 
            max(data1$score), 
            length.out=40)


preds <- predict(mod, data1 = data.frame(x=newx),
                 interval = 'confidence')


# plot
plot(score ~ 
       age, 
     data = data1, 
     type = 'n')


# add fill
polygon(c(rev(newx), newx), 
        c(rev(preds[ ,3]), preds[ ,2]), 
        col = 'grey', 
        border = NA)


# model
abline(mod)


# intervals
lines(newx, preds[ ,3], lty = 'dashed', col = 'red')
lines(newx, preds[ ,2], lty = 'dashed', col = 'red')

当我使用 EDi 生成的数据尝试相同的代码时，它可以工作...

【问题讨论】：

标签： r plot correlation

【解决方案1】：

您从 age（x 轴，自变量）预测 score（y 轴，因变量）所以您需要记住这一点：

newx <- seq(min(data1$age), max(data1$age), length.out=40)
preds <- predict(mod, newdata = data.frame(age=newx), interval = 'confidence')

参数是newdata 而不是data1，您必须包含一个与自变量同名的变量，age 而不是x。现在你的其余代码应该可以工作了：

plot(score ~ age, data = data1, type = 'n')
polygon(c(rev(newx), newx), c(rev(preds[ ,3]), preds[ ,2]), col = 'grey', border = NA)
abline(mod)
lines(newx, preds[ ,3], lty = 'dashed', col = 'red')
lines(newx, preds[ ,2], lty = 'dashed', col = 'red')

【讨论】：