时间戳

首页 > 解决方案 > 在 ggplot2 的意大利面条图中添加黄土回归线、中位数 (IQR) 或平均值 (SD)

r - 在 ggplot2 的意大利面条图中添加黄土回归线、中位数 (IQR) 或平均值 (SD)

问题描述

我对连续结果重复测量数据。

我想要做的是绘制这个变量随年龄变化的进程。首先，我制作了一个意大利面条情节如下：

mydata <- structure(list(ID = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 
3L, 3L, 3L, 3L, 3L, 4L, 4L, 5L, 5L, 5L, 6L, 6L, 6L, 6L, 6L, 7L, 
7L, 7L, 7L, 7L, 8L, 8L, 9L, 9L, 10L, 10L, 11L, 11L, 11L, 11L, 
12L, 12L, 13L, 13L, 14L, 14L, 14L, 14L, 14L, 15L, 15L, 16L, 16L, 
17L, 17L, 17L, 17L, 17L, 18L, 18L, 19L, 19L, 20L, 20L, 21L, 21L, 
22L, 22L, 22L, 22L, 22L, 23L, 23L, 24L, 24L, 24L, 24L), .Label = c("2", 
"3", "4", "7", "8", "13", "14", "20", "21", "22", "24", "25", 
"27", "29", "30", "31", "34", "36", "37", "38", "39", "40", "48", 
"49", "50", "51", "52", "54", "58", "60", "61", "65", "74", "75", 
"76", "77", "80", "81", "82", "83", "84", "86", "87", "88", "92", 
"94", "95", "96", "103", "104", "105", "114", "115", "116", "117", 
"119", "125", "126", "127", "132", "134", "135", "137", "138", 
"141", "142", "145", "152", "153", "154", "157", "159", "160", 
"162", "164", "165", "171", "172", "179", "180", "184", "185", 
"189", "194", "195", "197", "198", "202", "203", "205", "209", 
"213", "221", "253", "255", "258", "262", "271", "273", "277", 
"279", "310", "315", "320"), class = "factor"), date_ct = structure(c(15923, 
16122, 16715, 16902, 17086, 18003, 16150, 16841, 16421, 16764, 
16951, 17135, 18011, 16622, 18247, 16582, 16752, 18045, 16729, 
16862, 17042, 17226, 18102, 16568, 16736, 16916, 17100, 18040, 
16743, 16841, 16589, 16729, 16526, 16729, 16619, 16862, 17042, 
17226, 16407, 18437, 16512, 16953, 16457, 16946, 17112, 17310, 
17989, 16573, 16841, 15923, 16752, 16505, 16729, 16909, 17107, 
18038, 16540, 16743, 15951, 16122, 16624, 18202, 16623, 18221, 
16694, 16715, 16902, 17086, 18037, 16451, 16743, 16421, 16736, 
16909, 17100), class = "Date"), age = c(56.6, 57.1, 58.8, 59.3, 
59.8, 62.3, 43.2, 45.1, 52, 52.9, 53.4, 53.9, 56.3, 58.5, 63, 
57.4, 57.9, 61.4, 57.8, 58.2, 58.7, 59.2, 61.6, 52.4, 52.8, 53.3, 
53.8, 56.4, 70.8, 71.1, 61.4, 61.8, 59.2, 59.8, 61.5, 62.2, 62.7, 
63.2, 48.9, 54.5, 54.2, 55.4, 50.1, 51.4, 51.8, 52.4, 54.3, 55.4, 
56.1, 48.6, 50.9, 64.2, 64.8, 65.3, 65.8, 68.4, 68.3, 68.8, 66.7, 
67.1, 60.5, 64.8, 56.5, 60.9, 62.7, 62.8, 63.3, 63.8, 66.4, 49, 
49.8, 61, 61.8, 62.3, 62.8), continuous_outcome = c(1636.4, 544.1, 
1408, 1594.7, 1719.4, 2345.9, 115.3, 226, 2678.2, 3451.6, 3702.7, 
3632.7, 5805, 155.2, 1095, 992.2, 296.6, 2020.4, 3708.6, 2710.7, 
2934.2, 3080.4, 4489.7, 3459.4, 4965.3, 5553.1, 5037.8, 7315.7, 
29980.8, 35407.5, 2263.2, 2060.6, 3220.7, 4467.1, 5902.3, 6407.2, 
5947.1, 6271.6, 306, 689.3, 1430.6, 1672.1, 9.9, 58.7, 69.9, 
125.3, 39.5, 3842.5, 5136.3, 216.6, 332.4, 5719.3, 5386, 5490.7, 
5268.2, 6166.7, 12520.6, 12981.8, 2896.1, 2976.8, 5495.6, 6470.6, 
4235.5, 7603.5, 3887, 3344.5, 2885.7, 3324.1, 6401, 1942.2, 2000.9, 
2401.7, 2231.5, 2749.7, 2741.7)), row.names = c(NA, -75L), class = c("tbl_df", 
"tbl", "data.frame"))

# A normal spaghetti plot:
sp <- 
  ggplot(soquestiondata, aes(x=age, y=continuous_outcome, group=ID, color=ID)) + 
  geom_point() + 
  geom_line() + 
  xlab("Age (years)") + 
  ylab("Continuous outcome") +
  theme(legend.position = "none") 
sp

然后，我想添加一条回归线来查看随年龄增长的情况。为此，我想使用黄土线，因为进展似乎不是线性的。我试过了：

# Adding geom_smooth:
sp_loess <- 
  ggplot(soquestiondata, aes(x=age, y=continuous_outcome, group=ID, color=ID)) + 
  geom_point() + 
  geom_line() + 
  xlab("Age (years)") + 
  ylab("Continuous outcome") +
  theme(legend.position = "none") + 
  geom_smooth(method="loess", formula=y~x)
sp_loess

但是，这会产生很多我不太理解的警告。

然后我想也许只添加一个中位数，但这也没有用：

# Adding the median:
median_df <- 
  soquestiondata %>% 
  summarise(median=median(continuous_outcome, na.rm=TRUE))

sp_median <- 
  ggplot(soquestiondata, aes (x = age, y = continuous_outcome, group = ID, color = ID)) + 
  geom_line() + 
  geom_point() + 
  stat_summary(fun.df="median_df", color="red") + 
  theme(legend.position = "none")
sp_median

这会产生以下警告消息（图中没有中位数）：警告消息：#Computation failed in stat_summary(): #object 'median_df' of mode 'function' was not found

任何有助于使我的图表更上一层楼的帮助将不胜感激！谢谢！

标签： rggplot2graphsummarylongitudinal

解决方案

由于您的数据是ID通过color和/或groupaes 分组的，geom_smooth因此将尝试为每个ID.

  1. 如果您想要一条回归线显示 和 之间的关系continuous_outcome，age您必须使用 eg覆盖分组ID。geom_smoothgroup=1

  2. stat_summary需要一个汇总函数，而不是汇总数据框（顺便说一句：stat_summary 没有参数 fun.df）。使用您的汇总数据框，您可以使用您想要geom的添加中位数。下面我使用 stat_summary 计算中位数并通过hline.

library(ggplot2)

sp <-
  ggplot(soquestiondata, aes(x = age, y = continuous_outcome, group = ID, color = ID)) +
  geom_point() +
  geom_line() +
  xlab("Age (years)") +
  ylab("Continuous outcome") +
  theme(legend.position = "none")

sp  + 
  geom_smooth(aes(group = 1), method="loess", formula=y~x) +
  stat_summary(aes(x = 60, yintercept = ..y.., group = 1), fun = "median", color = "black", geom = "hline")

推荐阅读