我想编写一个执行以下操作的函数:
filter_and_compute <- function(df,
condition,
col_to_modify,
calculus) {
df %>%
filter( condition ) %>%
mutate( col_to_modify = calculus ) %>%
rbind(df %>% filter(! condition))
}
但以字符串作为参数调用(df
除外)
filter_and_compute(mtcars, "cyl == 4", "mpg", "2 * hp")
我制作了这个版本,但是没有用,因为我操纵列 名称为字符串:
filter_and_compute <- function(df,
filter_condition,
column_to_modify,
calculus) {
enq_filter <- enquo(filter_condition)
enq_col_to_mod <- enquo(column_to_modify)
enq_calculus <- enquo(calculus)
(
df
%>% filter(!! enq_filter)
%>% mutate(!! enq_col_to_mod := !! enq_calculus)
%>% rbind( df %>% filter( ! (!! enq_filter)))
)
}
filter_and_compute(mtcars, cyl == 4, mpg, 2 * cyl + hp)
我知道rlang::sym
,但是我无法了解如何将它用于不可预测的表达式,例如2 * cyl + hp
...
答案 0 :(得分:2)
您可以将eval
与parse_expr
一起使用
library(dplyr)
library(rlang)
filter_and_compute <- function(df,
condition,
col_to_modify,
calculus) {
df %>%
filter(eval(parse_expr(condition))) %>%
mutate(!! col_to_modify := eval(parse_expr(calculus))) %>%
rbind(df %>% filter(! eval(parse_expr(condition))))
}
filter_and_compute(mtcars, "cyl == 4", "mpg", "2 * hp")
# mpg cyl disp hp drat wt qsec vs am gear carb
#1 186.0 4 108.0 93 3.85 2.320 18.61 1 1 4 1
#2 124.0 4 146.7 62 3.69 3.190 20.00 1 0 4 2
#3 190.0 4 140.8 95 3.92 3.150 22.90 1 0 4 2
#4 132.0 4 78.7 66 4.08 2.200 19.47 1 1 4 1
#5 104.0 4 75.7 52 4.93 1.615 18.52 1 1 4 2
#6 130.0 4 71.1 65 4.22 1.835 19.90 1 1 4 1
#7 194.0 4 120.1 97 3.70 2.465 20.01 1 0 3 1
#8 132.0 4 79.0 66 4.08 1.935 18.90 1 1 4 1
#9 182.0 4 120.3 91 4.43 2.140 16.70 0 1 5 2
#10 226.0 4 95.1 113 3.77 1.513 16.90 1 1 5 2
#11 218.0 4 121.0 109 4.11 2.780 18.60 1 1 4 2
#12 21.0 6 160.0 110 3.90 2.620 16.46 0 1 4 4
#....