Question

如果以下是我的数据，我如何才能获得截距和 beta 估计值的 95% 上下置信区间？

data_long <- structure(list(ID = c("00721-2-C", "00724-1-C", "00724-1-C", 
"00733-1-C", "00735-2-C", "00735-2-C", "00735-2-C", "00735-2-C", 
"00738-1-C", "00738-1-C", "00721-2-C", "00724-1-C", "00724-1-C", 
"00733-1-C", "00735-2-C", "00735-2-C", "00735-2-C", "00735-2-C", 
"00738-1-C", "00738-1-C", "00721-2-C", "00724-1-C", "00724-1-C", 
"00733-1-C", "00735-2-C", "00735-2-C", "00735-2-C", "00735-2-C", 
"00738-1-C", "00738-1-C"), study_group = structure(c(2L, 1L, 
1L, 1L, 2L, 2L, 2L, 2L, 1L, 1L, 2L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 
1L, 1L, 2L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 1L, 1L), .Label = c("Intervention group", 
"Control group"), class = "factor"), timepoint = structure(c(1L, 
1L, 2L, 1L, 3L, 4L, 1L, 2L, 2L, 1L, 1L, 1L, 2L, 1L, 3L, 4L, 1L, 
2L, 2L, 1L, 1L, 1L, 2L, 1L, 3L, 4L, 1L, 2L, 2L, 1L), .Label = c("Baseline", 
"6 months", "24 months", "48 months"), class = "factor"), variable = structure(c(1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 
2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L), .Label = c("Energy_kcal", 
"Protein_g", "MUFA_g"), class = "factor"), value = c(1268.406744, 
997.5694419165, 1402.338548, 2228.254514, 1969.496113, 1429.53627, 
1847.53429, 1390.102819, 1484.28248, 2005.2722320005, 58.485134107, 
35.8161015647956, 74.273355041, 59.7441761, 60.049667397, 57.924105381, 
73.246959056, 51.814332349, 56.547926964, 60.8630037867644, 9.683201376, 
11.0929851920735, 12.213572797, 24.875988218, 25.29578509, 20.701295571, 
26.720893066, 18.564045616, 18.833261276, 30.9237008583813)), row.names = c(1L, 
4L, 5L, 17L, 22L, 23L, 24L, 25L, 28L, 29L, 3608L, 3611L, 3612L, 
3624L, 3629L, 3630L, 3631L, 3632L, 3635L, 3636L, 25250L, 25253L, 
25254L, 25266L, 25271L, 25272L, 25273L, 25274L, 25277L, 25278L
), class = "data.frame")

我重复线性回归的代码是：

library(dplyr)
library(broom)
library(tidyr)

data_long %>% 
  group_by(variable) %>% 
  do({
    my_mdl <- lm(value ~ timepoint, .)  
    my_tidy <- tidy(my_mdl)
    my_tidy_wide <- pivot_wider(my_tidy, names_from = "term", values_from = everything())
    my_glance <- glance(my_mdl)
    bind_cols(my_tidy_wide, my_glance)
  }) %>% 
  select(variable, 
         starts_with("estimate"), 
         starts_with("std.error"), 
         starts_with("p.value"), 
         AIC, 
         adj.r.squared)


# A tibble: 5 x 16
# Groups:   variable [5]
  variable `estimate_(Inte~ estimate_timepo~ estimate_timepo~ estimate_timepo~ `std.error_(Int~ std.error_timep~ std.error_timep~ std.error_timep~ `p.value_(Inter~ p.value_timepoi~
  <fct>               <dbl>            <dbl>            <dbl>            <dbl>            <dbl>            <dbl>            <dbl>            <dbl>            <dbl>            <dbl>
1 Energy_~           1624.            45.9            -198.            346.              168.             215.             241.             343.          0.0000713            0.838
2 Protein~             59.1           -1.49              1.76            0.929             5.18             6.61             7.42            10.6         0.0000270            0.829
3 MUFA_g               20.8           -0.139            -4.26            4.50              3.25             4.15             4.65             6.62        0.000685             0.974
4 Sugar_g              82.3           -1.29            -18.5            46.2              10.6             13.5             15.2             21.6         0.000241             0.927
5 Sweets_g             34.7           -2.32            -23.0            52.8              11.7             14.9             16.7             23.8         0.0249               0.882
# ... with 5 more variables: p.value_timepoint2 <dbl>, p.value_timepoint3 <dbl>, p.value <dbl>, AIC <dbl>, adj.r.squared <dbl>

我是否也可以将解决方案用于混合效果模型或代码如何更改？

谢谢！

Answer 1

这有效：

data_long %>% 
  group_by(variable) %>% 
  do({
    my_mdl <- lm(value ~ timepoint, .)  
    my_tidy <- tidy(my_mdl, conf.int = TRUE)
    my_tidy_wide <- pivot_wider(my_tidy, names_from = "term", values_from = everything())
    my_glance <- glance(my_mdl)
    bind_cols(my_tidy_wide, my_glance)
  }) %>% 
  select(variable, 
         starts_with("estimate"), 
         starts_with("std.error"), 
         starts_with("p.value"), 
         starts_with("conf.low"), 
         starts_with("conf.high"), 
         AIC, 
         BIC,
         adj.r.squared,
         df.residual)

95% CI 的重复回归估计

1 个答案: