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基于另一列更改列值,但仅限于第一列和第二列(R)中的某些条件

grepl dataframe r

Jaccar · 技术社区 · 7 年前

我有一个数据帧。

city <- as.character(c("London", "Unknown", "Birmingham", "Bristol", "Unknown", "Unknown", "Unknown", "Unknown"))
city_details <- as.character(c("London", "Camden", "Birmingham", "Outside London", "Camden Town", "Westminster", "London", "Birmingham"))
city_data <- data.frame(city, city_details)

因此,我可以替换其中一些:

city_data$city[grepl("Camden|Westminster", city_data$city_details)] <- 'London'

然而,在城市细节中写“伦敦”的更难,因为也有一个写“伦敦以外”,所以我不想随便挑一个有“伦敦”图案的。

出于这个目的,我并不是在寻找一种只包含精确匹配的方法(因为这对于我的真实数据不太合适)。

所以我只想对未知的城市值进行替换。

city_data <- within(city_data, city[city == "Unknown"] <- (city[grepl("London", city_details)] <- 'London'))

有人能帮忙吗?

3 回复 | 直到 7 年前

annhak 7 年前

我假设你只想在 city 是未知的 city_details 提到“伦敦”。在这种情况下,您可以使用以下方法:

city_data$city[(as.numeric(grepl("Unknown", city)) + as.numeric(grepl("London", city_details))) == 2] <- "London"

这能回答你的问题吗?

apitsch 7 年前

我建议如下:

one_hot <- grepl("Camden|Westminster|London", city_data$city_details) &
  city_data$city == "Unknown"
city_data$city[one_hot] <- "London"

city <- as.character(c("London", "Unknown", "Birmingham", "Bristol", "Unknown", "Unknown", "Unknown", "Unknown"))
city_details <- as.character(c("London", "Camden", "Birmingham", "Outside London", "Camden Town", "Westminster", "London", "Tottenham"))
city_data <- data.frame(city, city_details)

> city_data
        city   city_details
1     London         London
2    Unknown         Camden
3 Birmingham     Birmingham
4    Bristol Outside London
5    Unknown    Camden Town
6    Unknown    Westminster
7    Unknown         London
8    Unknown      Tottenham

> one_hot <- grepl("Camden|Westminster|London", city_data$city_details) &
+   city_data$city == "Unknown"
> city_data$city[one_hot] <- "London"
> city_data
        city   city_details
1     London         London
2     London         Camden
3 Birmingham     Birmingham
4    Bristol Outside London
5     London    Camden Town
6     London    Westminster
7     London         London
8    Unknown      Tottenham

Jaccar 7 年前

我还想出了下面的方法,这对我来说似乎更整洁、更直观。无需转换为数字。

city_data$city[grepl("Unknown", city_data$city) & 
               grepl("London|Camden|Westminster", city_data$city_details)] <- "London"