正如你从注解中看到的，这可以作为一个标准的选择简洁地完成，但是有时逻辑向量会使事情更清晰。
假设您的 Dataframe 名为df

redcars <- df$color == "red"
fords <- df$company == "ford"
ford_or_nissan = fords | df$company == "nissan" # or alternatively
ford_or_nissan = df$company %in% c("ford","nissan")

这将为您提供三个可用于选择所需行的向量

df$car_total[ford_or_nissan] <- 0
df$car_total[fords | redcars] <- 0

使用逻辑运算符，您可以根据需要构建复杂的选择。

我喜欢使用data.table库

# Replace car total with 0 when company = ford OR company = nissan
dt[company %in% c("ford","nissan"), car_total := 0]

# Replace any car_total with 0 when company = ford OR color = red
dt[company == "ford" | color == "red", car_total := 0]

第一个问题：

year<-c(2000,2013,2012,2002,2000,2012,2009,2010,2000,2011,2014)
company<-c('toyota','honda','nissan','toyota','nissan','honda','toyota','ford','ford','nissan','ford')
color<-c('red','red','green','blue','green','red','red','yellow','green','blue','blue')
car_total<-as.integer(c(873,737,809,429,861,742,320,319,587,777,32))
df<-data.frame(year,company,color,car_total)
for (i in 1:nrow(df))
{
ifelse (df$company[i]=='ford', df$car_total[i]<-0, NA)
ifelse (df$company[i]=='nissan',df$car_total[i]<-0, NA)
}