Question

我有一个数据框df，如下所述。

a <- c(1:6)
b <- c("Audi,BMW,Skoda, Rackets,Toy,Football",
       "Suzuki,Kawasaki,Ducati,Aprilia,Baseball, Rugby",
       "Mazda, Ford, chevrolet,Mercedes,Gloves,Helmet",
       "Lemon,Yamaha,Table,Kawasaki,Chair,Fruits", 
       "Ford, chevrolet,Bread,Ducati,Tesla,Hyundai",
       "Honey,Apple,Alcohol,cake,Sweets, Mango")
       df <- data.frame(a,b)

*

我还有两个列表，分别包含汽车和自行车的品牌名称。

cars <- c("Audi","BMW","Ford","Skoda","Mazda","chevrolet","Mercedes","Volkswagen","Tesla","Hyundai","Lamborghini","Mini-Cooper","Lexus")
motorbike <- c("Yamaha","Suzuki","Kawasaki","Harley-Davidson","Ducati","Aprilia","KTM", "Triumph","Piaggio","Hyosung","Vespa","MV-Agusta")

我使用grepl和ifelse来匹配df $ b中两个列表中的单词，并为每个行分配一个匹配的值。

df$c<-ifelse(grepl(paste(cars, collapse="|"), df$b), "cars",
      ifelse(grepl(paste(motorbike, collapse="|"),df$b), "bikes","others"))

现在，我想提出一个条件，如果每行匹配4个或4个以上的单词，则仅在df $ c中分配一个值（car，bike）。我希望我的df像这样：

structure(list(a = 1:6, b = structure(c(1L, 6L, 5L, 4L, 2L, 3L
), .Label = c("Audi,BMW,Skoda, Rackets,Toy,Football", "Ford, chevrolet,Bread,Ducati,Tesla,Hyundai", 
"Honey,Apple,Alcohol,cake,Sweets, Mango", "Lemon,Yamaha,Table,Kawasaki,Chair,Fruits", 
"Mazda, Ford, chevrolet,Mercedes,Gloves,Helmet", "Suzuki,Kawasaki,Ducati,Aprilia,Baseball, Rugby"
), class = "factor"), c = c("others", "bikes", "cars", "others", 
"cars", "others")), row.names = c(NA, 6L), class = "data.frame")

Answer 1

这有帮助吗？当然，您可以删除amountcars和amountmotors列。而且，如果同时拥有> 3辆汽车和> 3辆电动机，您是否希望它永远不会发生？根据评论，我现在更新了答案。

library(stringr)
df$amountcars <- str_count(df$b, paste(cars, collapse="|"))
df$amountmotors <- str_count(df$b, paste(motorbike, collapse="|"))



df$c <- ifelse(df$amountcars > 3 & df$amountcars > df$amountmotors, "cars", ifelse(df$amountmotors > 3 & df$amountmotors > df$amountcars, "bikes", "others"))
df

  a                                              b amountcars amountmotors      c
1 1           Audi,BMW,Skoda, Rackets,Toy,Football          3            0 others
2 2 Suzuki,Kawasaki,Ducati,Aprilia,Baseball, Rugby          0            4  bikes
3 3  Mazda, Ford, chevrolet,Mercedes,Gloves,Helmet          4            0   cars
4 4       Lemon,Yamaha,Table,Kawasaki,Chair,Fruits          0            2 others
5 5     Ford, chevrolet,Bread,Ducati,Tesla,Hyundai          4            1   cars
6 6         Honey,Apple,Alcohol,cake,Sweets, Mango          0            0 others

根据评论，如果您喜欢9个字符串：首先用字符串创建所有向量：

cars <- c("Audi","BMW","Ford","Skoda","Mazda","chevrolet","Mercedes","Volkswagen","Tesla","Hyundai","Lamborghini","Mini-Cooper","Lexus")
motorbike <- c("Yamaha","Suzuki","Kawasaki","Harley-Davidson","Ducati","Aprilia","KTM", "Triumph","Piaggio","Hyosung","Vespa","MV-Agusta")

然后将它们放在列表中，并添加名称

list1 <- list(cars, motorbike)
names(list1) <- c("cars", "motorbike")

最后，运行以下代码：

df$d <- 
ifelse(apply(sapply(list1, function(x) str_count(df$b, paste0(x, collapse = "|"))), 1, max) > 3,
apply(sapply(list1, function(x) str_count(df$b, paste0(x, collapse = "|"))), 1, function(x) names(list1)[which.max(x)]),
"others")

基本上，它从向量之一计算最大字符串数，如果大于3，则分配适当的名称，否则分配“其他”。

使用grepl和ifelse进行基于条件的字符串匹配

1 个答案: