如何在R中选取与「状态非'Open'且类型为BMW/VW/Benz」相反的DataFrame行？

Got it, let's break this down step by step! First, let's recap your original filter logic: you're selecting rows where:

• status is not equal to 'Open', AND
• type is one of 'BMW', 'VW', or 'Benz'

To reverse this (i.e., select all rows that don't match the above combined condition), you have two straightforward approaches—either negate the entire original condition directly, or use De Morgan's laws to rewrite the reverse logic more explicitly.

Approach 1: Negate the Entire Original Condition #

You can wrap your original filter logic in a ! (not) operator to flip the selection.

Base R Example #

If your DataFrame is named df:

# Original filter logic stored for clarity
original_condition <- (df$status != 'Open') & (df$type %in% c('BMW', 'VW', 'Benz'))

# Reverse filter: select rows where original condition is FALSE
reverse_df <- df[!original_condition, ]

Or condensed into one line:

reverse_df <- df[!((df$status != 'Open') & (df$type %in% c('BMW', 'VW', 'Benz'))), ]

dplyr Example #

If you prefer tidyverse syntax:

library(dplyr)

reverse_df <- df %>%
  filter(!((status != 'Open') & (type %in% c('BMW', 'VW', 'Benz'))))

Approach 2: Rewrite with De Morgan's Laws #

De Morgan's laws tell us that !(A & B) is equivalent to !A | !B. Applying that to your original condition:

• !A = status == 'Open' (opposite of "status isn't Open")
• !B = !(type %in% c('BMW', 'VW', 'Benz')) (opposite of "type is in the brand list")

So the reverse condition becomes: status is 'Open' OR type is NOT in the brand list (or both). This version is often more readable at a glance.

Base R Example #

reverse_df <- df[(df$status == 'Open') | !(df$type %in% c('BMW', 'VW', 'Benz')), ]

dplyr Example #

library(dplyr)

reverse_df <- df %>%
  filter(status == 'Open' | !(type %in% c('BMW', 'VW', 'Benz')))

Both approaches produce identical results—pick whichever feels more intuitive for your code context!

内容的提问来源于stack exchange，提问作者user5879741