也许不是最好的解决方案,但我们可以动态创建表达式并使用eval(parse()) 进行评估。我们将NA 的值分别替换为Inf/-Inf 中的Max/Min。
dfLimits$Max[is.na(dfLimits$Max)] <- Inf
dfLimits$Min[is.na(dfLimits$Min)] <- -Inf
subset(df, eval(parse(text = paste(dfLimits$Names, ">", dfLimits$Min, "&",
dfLimits$Names, "<", dfLimits$Max, collapse = " & "))))
# mpg cyl disp hp drat wt qsec vs am gear carb
#Hornet Sportabout 18.7 8 360.0 175 3.15 3.440 17.02 0 0 3 2
#Merc 450SE 16.4 8 275.8 180 3.07 4.070 17.40 0 0 3 3
#Merc 450SL 17.3 8 275.8 180 3.07 3.730 17.60 0 0 3 3
#Merc 450SLC 15.2 8 275.8 180 3.07 3.780 18.00 0 0 3 3
#Cadillac Fleetwood 10.4 8 472.0 205 2.93 5.250 17.98 0 0 3 4
#Lincoln Continental 10.4 8 460.0 215 3.00 5.424 17.82 0 0 3 4
#Pontiac Firebird 19.2 8 400.0 175 3.08 3.845 17.05 0 0 3 2
#Ferrari Dino 19.7 6 145.0 175 3.62 2.770 15.50 0 1 5 6
paste 的输出是:
paste(dfLimits$Names, ">", dfLimits$Min, "&",
dfLimits$Names, "<", dfLimits$Max, collapse = " & ")
#[1] "hp > 160 & hp < 220 & wt > 1.6 & wt < Inf"
如果对dplyr 解决方案感兴趣,我们可以在filter 和parse_expr 中使用相同的paste 表达式。
library(dplyr)
library(rlang)
df %>%
filter(eval(parse_expr(paste(dfLimits$Names, ">", dfLimits$Min, "&",
dfLimits$Names, "<", dfLimits$Max, collapse = " & "))))