R语言：筛选minimum_nights>90后提取非UK区域的host_location值

完全可行，你可以在已处理的数据集基础上，通过筛选条件轻松提取host_location列中不属于"UK"的记录或值。以下是两种常用实现方式：

方法1：使用dplyr（tidyverse工具链） #

假设你之前用tidyverse完成了数据预处理，后续只需添加筛选步骤即可：

library(dplyr)

# 先执行你已有的预处理操作
airbnb_processed <- airbnb1 %>%
  select(host_location, minimum_nights, revenue) %>%
  filter(minimum_nights > 90) %>%
  mutate(host_location = ifelse(is.na(host_location), "unknown", host_location))

# 提取host_location不为"UK"的完整记录
airbnb_non_uk <- airbnb_processed %>%
  filter(host_location != "UK")

# 如果只需要提取host_location列的非UK值
non_uk_locations <- airbnb_processed %>%
  filter(host_location != "UK") %>%
  pull(host_location)

方法2：使用基础R语法 #

如果你习惯用基础R操作，代码如下：

# 基础R版的预处理操作
airbnb_processed_base <- airbnb1[, c("host_location", "minimum_nights", "revenue")]
airbnb_processed_base <- airbnb_processed_base[airbnb_processed_base$minimum_nights > 90, ]
airbnb_processed_base$host_location[is.na(airbnb_processed_base$host_location)] <- "unknown"

# 提取非UK的完整记录
airbnb_non_uk_base <- airbnb_processed_base[airbnb_processed_base$host_location != "UK", ]

# 单独提取host_location列的非UK值
non_uk_locations_base <- airbnb_processed_base$host_location[airbnb_processed_base$host_location != "UK"]

额外提示：灵活匹配场景 #

如果host_location里存在"United Kingdom"、"GB"等其他指代英国的表述，可通过正则匹配来排除这类记录：

# 忽略大小写，排除包含UK或United Kingdom的位置
airbnb_non_uk_flexible <- airbnb_processed %>%
  filter(!grepl("UK|United Kingdom", host_location, ignore.case = TRUE))

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