r-访问查询的点参数的名称

Question

我有两个函数：

exclude <- function(df, ...) {
dots <- rlang::enquos(...)
for(i in 1:length(dots)) {
df <- exclude_cycle(df, dots[[i]])
}
return(df)
}
exclude_cycle <- function(df, condition) {
df <- dplyr::filter(df, !!condition)
print(paste(nrow(df), "records left after excluding", eval(condition), sep = " "))
return(df)
}

我希望print(paste(nrow(df), "records left after excluding", eval(condition), sep = " "))打印一个看起来像100000 records left after excluded age > 18的简单字符串，其中age > 18是我直接传递给exclude函数的过滤器：

exclude(df, age > 18)

我最终得到的输出看起来像：

[1] "100000 records left after excluding ~"         "100000 patients left after excluding age >= 18"

这与我想要的非常接近，但每个字符向量打印两个字符串，而不是1个。如何实现所需的输出？

Answer 1

您可以使用rlang::enexprs允许在单个函数中将多个条件传递给点。此外，如果您希望排除条件，请记住反转过滤器：

exclude <- function(df, ...) {
dots <- rlang::enexprs(...)
for(i in seq_along(dots)){
df <-  dplyr::filter(df, !(!!(dots[[i]])))
cat(nrow(df), "records left after excluding", as.character(dots[i]), "n")
}
}

例如：

df <- data.frame(letters = rep(LETTERS[1:3], 10),
numbers = 1:30)
exclude(df, letters == "A", numbers < 15)
#> 20 records left after excluding letters == "A" 
#> 11 records left after excluding numbers < 15 

Answer 2

这里有一个选项

exclude_cycle <- function(df, condition) {
df <- dplyr::filter(df, !!condition)
print(paste(nrow(df), "records left after excluding", 
as.list(eval(condition))[-1], sep = " "))
return(df)
}

-测试

exclude(df, age > 18)
#[1] "2 records left after excluding age > 18"
#  age
#1  42
#2  19

数据

df <- data.frame(age = c(42, 19, 3))