在观测值上绘制线性混合模型的预测值？
Plot predicted values of linear mixed model over the observed values?
我对如下所示的连续结果进行了重复测量：
library(magrittr)
library(ggplot2)
library(nlme)
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"))
通过探索性分析，我发现具有 2 节的样条 运行 以下线性混合模型可以确定年龄是否是我结果的预测因子：
model1 <-
lme(data=mydata,
fixed=continuous_outcome ~ age,
random=~1|ID)
我想找到一种方法来绘制此模型的预测值与我的结果的观察值。有帮助吗？
predict() 是基本答案。有更复杂的方法使用 ggeffects 包（但它似乎不适用于 lme 模型？）或 sjPlot（但叠加数据不是很容易吗？）
model1 <-
lme(data=mydata,
fixed=continuous_outcome ~ splines::ns(age, 2),
random=~1|ID)
构建预测数据框：此步骤并非总是必要的，但如果您的数据点不唯一或无序，或者如果您想要绘制非线性响应或置信区间[哪个通常是弯曲的]，或者如果您有一个非常大的数据集（在这种情况下可能不需要对整个数据集进行预测）；如果您有一个带有非焦点预测变量的更复杂的模型（effects/ggeffects 包旨在自动执行此步骤，通常也有必要）。
pframe <- data.frame(age=seq(min(mydata$age), max(mydata$age), length.out=101))
使用 newdata 指定要对其进行预测的值，并使用 level=0 在人口级别进行预测（这也意味着我们不需要在中给出 ID 值newlevels)
pframe$continuous_outcome <- predict(model1, newdata = pframe, level = 0)
ggplot(mydata, aes(age, continuous_outcome)) +
geom_point() +
geom_line(data=pframe)
我对如下所示的连续结果进行了重复测量：
library(magrittr)
library(ggplot2)
library(nlme)
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"))
通过探索性分析，我发现具有 2 节的样条 运行 以下线性混合模型可以确定年龄是否是我结果的预测因子：
model1 <-
lme(data=mydata,
fixed=continuous_outcome ~ age,
random=~1|ID)
我想找到一种方法来绘制此模型的预测值与我的结果的观察值。有帮助吗？
predict() 是基本答案。有更复杂的方法使用 ggeffects 包（但它似乎不适用于 lme 模型？）或 sjPlot（但叠加数据不是很容易吗？）
model1 <-
lme(data=mydata,
fixed=continuous_outcome ~ splines::ns(age, 2),
random=~1|ID)
构建预测数据框：此步骤并非总是必要的，但如果您的数据点不唯一或无序，或者如果您想要绘制非线性响应或置信区间[哪个通常是弯曲的]，或者如果您有一个非常大的数据集（在这种情况下可能不需要对整个数据集进行预测）；如果您有一个带有非焦点预测变量的更复杂的模型（effects/ggeffects 包旨在自动执行此步骤，通常也有必要）。
pframe <- data.frame(age=seq(min(mydata$age), max(mydata$age), length.out=101))
使用 newdata 指定要对其进行预测的值，并使用 level=0 在人口级别进行预测（这也意味着我们不需要在中给出 ID 值newlevels)
pframe$continuous_outcome <- predict(model1, newdata = pframe, level = 0)
ggplot(mydata, aes(age, continuous_outcome)) +
geom_point() +
geom_line(data=pframe)