使用Tidyeval进行程序回归建模 [英] Programmatic regression modelling with tidyeval
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问题描述
我正在尝试使用tidyeval进行编程.
I am trying to get my head around programming using tidyeval.
我想编写一个函数来为选定的结果变量运行逻辑回归模型:
I want to write a function to run logistic regression models for selected outcome variables:
library(tidyverse)
set.seed(1234)
df <- tibble(id = 1:1000,
group = sample(c("Group 1", "Group 2", "Group 3"), 1000, replace = TRUE),
died = sample(c(0,1), 1000, replace = TRUE))
myfunc <- function(data, outcome){
enquo_var <- enquo(outcome)
fit <- tidy(glm(!!enquo_var ~ group, data=data,
family = binomial(link = "logit")),
exponentiate = TRUE, conf.int=TRUE)
fit
}
myfunc(df, died)
但是得到:
!enquo_outcome错误:参数类型无效
Error in !enquo_outcome : invalid argument type
(请注意实际情况涉及更复杂的功能).
(note real scenario involves more complex function).
这可能吗?
推荐答案
我们需要为 glm
创建一个公式以将其提取.一种选择是 paste
We need to create a formula for glm
to pick it up. One option is paste
myfunc <- function(data, outcome){
enquo_var <- enquo(outcome)
fit <- tidy(glm(paste(quo_name(enquo_var), "group", sep="~"), data=data,
family = binomial(link = "logit")),
exponentiate = TRUE, conf.int=TRUE)
fit
}
myfunc(df, died)
# term estimate std.error statistic p.value conf.low conf.high
#1 (Intercept) 0.8715084 0.1095300 -1.2556359 0.20924801 0.7026185 1.079852
#2 groupGroup 2 0.9253515 0.1550473 -0.5003736 0.61681204 0.6826512 1.253959
#3 groupGroup 3 1.3692735 0.1557241 2.0181864 0.04357185 1.0095739 1.859403
如果我们还需要使用tidyverse函数
If we also need to use the tidyverse functions
myfunc <- function(data, outcome){
quo_var <- quo_name(enquo(outcome))
fit <- tidy(glm(rlang::expr(!! rlang::sym(quo_var) ~ group), data=data,
family = binomial(link = "logit")),
exponentiate = TRUE, conf.int=TRUE)
fit
}
myfunc(df, died)
# term estimate std.error statistic p.value conf.low conf.high
#1 (Intercept) 0.8715084 0.1095300 -1.2556359 0.20924801 0.7026185 1.079852
#2 groupGroup 2 0.9253515 0.1550473 -0.5003736 0.61681204 0.6826512 1.253959
#3 groupGroup 3 1.3692735 0.1557241 2.0181864 0.04357185 1.0095739 1.859403
或者可以使用注释 get_expr
中的@lionel
myfunc <- function(data, outcome){
quo_var <- enquo(outcome)
fit <- tidy(glm(rlang::expr(!! rlang::get_expr(quo_var) ~ group), data=data,
family = binomial(link = "logit")),
exponentiate = TRUE, conf.int=TRUE)
fit
}
myfunc(df, died)
# term estimate std.error statistic p.value conf.low conf.high
#1 (Intercept) 0.8715084 0.1095300 -1.2556359 0.20924801 0.7026185 1.079852
#2 groupGroup 2 0.9253515 0.1550473 -0.5003736 0.61681204 0.6826512 1.253959
#3 groupGroup 3 1.3692735 0.1557241 2.0181864 0.04357185 1.0095739 1.859403
或者@lionel建议的一种更紧凑的方法,它避免了 enquo/quo_name/sym
转换,而是直接在 enexpr
myfunc <- function(data, outcome){
fit <- tidy(glm(rlang::expr(!! rlang::enexpr(outcome) ~ group), data=data,
family = binomial(link = "logit")),
exponentiate = TRUE, conf.int=TRUE)
fit
}
myfunc(df, died)
# term estimate std.error statistic p.value conf.low conf.high
#1 (Intercept) 0.8715084 0.1095300 -1.2556359 0.20924801 0.7026185 1.079852
#2 groupGroup 2 0.9253515 0.1550473 -0.5003736 0.61681204 0.6826512 1.253959
#3 groupGroup 3 1.3692735 0.1557241 2.0181864 0.04357185 1.0095739 1.859403
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