Question

我是R.的新手。我已经找到了我的问题的答案，但没有找到任何对我有用的东西。

我有以下data.table：

data2 <- fread("file.txt", header = T, sep= "|", quote = "^" stringsAsFactors=FALSE);

Accounts_filed_date Incorporation_Date 
         2015-08-25         1863-05-01   
         2016-06-28         1863-07-16    
         2016-06-29         1863-12-29  
         2015-12-10         1880-08-17

Y <- function(incorporation_date, accounts_filled_date)
{timee <-Sys.Date()-incorporation_date;timee <- as.numeric(timee);
  if(timee/365 <2){y=0; return(y)}
  else{x <- as.Date(format(accounts_filled_date, "%2016-%m-%d"));
   + x <- Sys.Date()- x;
   + x <- as.numeric(x);
   + if(x<274) {y <-timee/365; return(pmax(pmin(y,5),0))}
   + else{y<-timee/365-1; return(pmax(pmin(y,5),0))}}}

data2[, c("Y") := Y(Incorporation_Date, Accounts_filed_date), with = FALSE]

但是，这不起作用，因为incorporation_date和accounts_filled_date是向量。

我尝试了ifelse，但使用以下代码：

Y <- function(incorporation_date, accounts_filled_date) 
{timee <- Sys.Date()-incorporation_date;timee <- as.numeric(timee); 
timee<- timee/365;
ifelse(timee<2,y=0,
       x<-as.Date(format(accounts_filled_date, "%2016-%m-%d"));
       x <-Sys.Date()- x;
       x <- as.numeric(x);
       ifelse(x<274,y <-timee, y<- timee-1));
return(pmax(pmin(y,5),0))}

返回以下内容：

Error: unexpected ';' in "Y <- function(incorporation_date, accounts_filled_date) {timee <- Sys.Date()-incorporation_date;timee <- as.numeric(timee); timee <- timee/365;ifelse(timee<2,y=0,x

基本上，我的问题是：如何在ifelse语句的“false”部分中包含多个操作？

更新

Y <- function(incorporation_date, accounts_filled_date) 
 {timee <- Sys.Date()-incorporation_date;timee <- as.numeric(timee); 
 timee2<- timee/365;
 timee2 <- as.integer(timee2);
 x<-as.Date(format(accounts_filled_date, "%2016-%m-%d"));
   x <- Sys.Date()- x;
   x <- as.numeric(x);
ifelse**(timee2 <= 2,** y<-0, ifelse(x<274,y <-timee2, y<- timee2-1));
 return(pmax(pmin(y,100),0))}

除了时间＆lt; = 2，bit之外，这是有效的。我已经双重检查 - 它返回的值小于2，只是逻辑返回FALSE。有什么建议吗？

Answer 1

重新安排语句捕获逻辑：

Y <- function(incorporation_date, accounts_filed_date) 
 {timee <- Sys.Date()-incorporation_date;timee <- as.numeric(timee); 
  timee2<- timee/365;
   x<-as.Date(format(accounts_filed_date, "%2016-%m-%d"));
   x <- Sys.Date()- x;
   x <- as.numeric(x);
   y <- ifelse(timee2 <= 2, 0,ifelse(x < 274, timee2, timee2-1));
   return(pmax(pmin(y,5),0))}

R ifelse多于true section / ifelse中的命令当它实际为TRUE时返回False

1 个答案: