将黄土回归线、中位数 (IQR) 或均值 (SD) 添加到 ggplot2 中的意大利面条图
Add loess regression line, median (IQR) or mean (SD) to spaghetti plot in ggplot2
我有关于连续结果的重复测量数据。
我想做的是绘制此变量随年龄变化的进展图。首先我做了一个意大利面条图如下:
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着年龄的增长情况。为此,我想使用黄土线,因为进展看起来不是线性的。我试过:
# 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,但这也行不通:
# 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(图中没有中值):
警告信息:
#stat_summary()
计算失败:
未找到模式 'function' 的 #object 'median_df'
如果能帮助我的图表更上一层楼,我们将不胜感激!谢谢!
由于您的数据通过 color
and/or group
aes 按 ID
分组,geom_smooth
将尝试为每个 ID
.
如果你想要一条回归线显示 continuous_outcome
和 age
之间的关系,你必须使用 geom_smooth
中的 ID
覆盖分组例如group=1
.
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)。使用汇总数据框,您可以使用所需的 geom
添加中位数。下面我使用 stat_summary 计算中位数并通过 hline
.
显示它
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")
我有关于连续结果的重复测量数据。
我想做的是绘制此变量随年龄变化的进展图。首先我做了一个意大利面条图如下:
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着年龄的增长情况。为此,我想使用黄土线,因为进展看起来不是线性的。我试过:
# 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,但这也行不通:
# 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(图中没有中值):
警告信息:
#stat_summary()
计算失败:
未找到模式 'function' 的 #object 'median_df'
如果能帮助我的图表更上一层楼,我们将不胜感激!谢谢!
由于您的数据通过 color
and/or group
aes 按 ID
分组,geom_smooth
将尝试为每个 ID
.
如果你想要一条回归线显示
continuous_outcome
和age
之间的关系,你必须使用geom_smooth
中的ID
覆盖分组例如group=1
.
显示它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)。使用汇总数据框,您可以使用所需的geom
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")