Question

我有197个关于位置的关卡，我想通过创建一个新的变量＆＃34; INSIDE＆＃34;来简化这个问题。当位置是建筑物/家庭/等时存储1，当位置在外面时存储0。我试过grepl（）但是它给出了一个错误

data$Inside<-ifelse(grepl(data$Premise.Description,pattern = c("BUILDING","ROOM","AUTO","BALCONY","BANK","BAR","STORE","CHURCH","COLLEGE","CONDOMINIUM","CENTER","DAY CARE","SCHOOL","HOSPITAL","LIBRARY","PARLOR","OFFICE","MOSQUE","CLUB","PORCH","MALL","WAREHOUSE")),1,0)

警告讯息：在grepl（crime_3yr $ Premise.Description，pattern = c（＆＃34; BUILDING＆＃34;，：论证＆＃39;模式＆＃39;长度> 1，只使用第一个元素

我尝试过使用lapply（），但它也没有用。我希望输出像这样：

BUILDING   1
SHOP       1
Street     0

Answer 1

试试这段代码：

您的data.frame：

data<-data.frame(Premise.Description= c("BUILDING 1","MY ROOM","AUTO","BALCONY","OTHER"))

解决方案：

toMatch<-c("BUILDING","ROOM","AUTO","BALCONY","BANK","BAR","STORE","CHURCH","COLLEGE","CONDOMINIUM","CENTER","DAY CARE","SCHOOL","HOSPITAL","LIBRARY","PARLOR","OFFICE","MOSQUE","CLUB","PORCH","MALL","WAREHOUSE")
data$Inside<-grepl(paste(toMatch,collapse="|"), data$Premise.Description)
data
  Premise.Description Inside
1          BUILDING 1   TRUE
2             MY ROOM   TRUE
3                AUTO   TRUE
4             BALCONY   TRUE
5               OTHER  FALSE

Answer 2

您可能最好使用data.table：

library(data.table)
setDT(data)
data[
grepl(c("BUILDING","ROOM","AUTO","BALCONY","BANK","BAR","STORE","CHURCH","COLLEGE","CONDOMINIUM","CENTER","DAY CARE","SCHOOL","HOSPITAL","LIBRARY","PARLOR","OFFICE","MOSQUE","CLUB","PORCH","MALL","WAREHOUSE"), Premise),
  Inside := TRUE
]

Answer 3

grepl使用正则表达式而不是选项列表，试试这个：

data$Inside<-ifelse(grepl(data$Premise.Description,pattern = "BUILDING|ROOM|AUTO|BALCONY|BANK|BAR|STORE|CHURCH|COLLEGE|CONDOMINIUM|CENTER|DAY CARE|SCHOOL|HOSPITAL|LIBRARY|PARLOR|OFFICE|MOSQUE|CLUB|PORCH|MALL|WAREHOUSE"),1,0)

Answer 4

如果你想保持代码与你列出的代码类似，你需要查看正则表达式，这是grepl的模式部分所需要的。

data$Inside<-ifelse(grepl(data$Premise.Description,pattern = "BUILDING|ROOM|AUTO|BALCONY|BANK|BAR|STORE|CHURCH|COLLEGE|CONDOMINIUM|CENTER|DAY CARE|SCHOOL|HOSPITAL|LIBRARY|PARLOR|OFFICE|MOSQUE|CLUB|PORCH|MALL|WAREHOUSE"),1,0)

在R

4 个答案: