Question

我无法从mapply函数的输出中获取值而不是因子。下面的代码试图找到特定位置和时间的温度。

now <- Sys.time()

datetime<-as.POSIXct(seq(now, length.out = 5, by = "hours"), tz = "GMT")
location<-c("Chicago","Detroit","New York","Cleveland","Atlanta")
temperature<-c(2,9,4,8,59)

df1 <- data.frame(datetime, location, temperature)

# add 3 hours to System time
now <- Sys.time() + 3*60*60

datetime<-as.POSIXct(seq(now, length.out = 5, by = "hours"), tz = "GMT")
location<-c("Chicago","Cleveland","Chicago","Detroit","Atlanta")
temperature<-c(6,8,3,12,69)

df2 <- data.frame(datetime, location, temperature)

# Data Frame with Location and Time and Temperature data
loc_temp_df <- rbind(df1,df2)

# Add 2 hours to system time
now <- Sys.time() + 2*60*60

datetime<-as.POSIXct(seq(now, length.out = 7, by = "hours"), tz = "GMT")
location<-c("Chicago","Chicago","New York","Atlanta","New York", "Detroit", "Cleveland")

df <- data.frame(datetime, location)

# Function to find temperature at specific location closest to specific time
myfunction <- function(x,y) {
  same_location <- loc_temp_df[which(loc_temp_df$location == y),]
  time_at_location <- as.POSIXct(same_location$datetime, format = '%Y-%m-%d %H:%M:%S', tz = "GMT")
  index <- which.min(abs(difftime(x, time_at_location, "mins")))
  same_location[index,]
}

mapply(myfunction, df$datetime, df$location)

当我运行代码时，我得到的是返回的因子而不是值。知道我怎么能回报价值而不是因素？

  [,1]       [,2]       [,3]       [,4]       [,5]       [,6]       [,7]      
datetime    1436564283 1436564283 1436560683 1436567883 1436560683 1436575083 1436567883
location    factor,1   factor,1   factor,1   factor,1   factor,1   factor,1   factor,1  
temperature 6          6          4          59         4          12         8

我尝试使用lapply但无法以相同的方式使用它。

Answer 1

您可以使用Map中的SIMPLIFY=FALSE或mapply将其保留为list，然后rbind。

 do.call(rbind,Map(myfunction, df$datetime, df$location))

或者

 do.call(rbind, mapply(myfunction, df$datetime, df$location, SIMPLIFY=FALSE))

如何使用因子函数参数和输出从mapply返回值

1 个答案: