Question

我有一个像这样的数据。数据分别包含3年（2005-2006-2007）的降雨值。

数据：

mydata<-structure(list(X2005.hourly = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0.6, 0.2, 0, 0, 0.6, 0.2, 0.6, 0.6, 0, 0, 0, 0.6, 1.2, 1.8, 1.8, 
1.2, 1, 1.2, 1.6, 1.2, 1.4, 1, 1.2, 0.8, 0.8, 0.2, 0, 0, 1.6, 
0.4, 0, 0.4, 1.2, 0.8, 0.2, 0.4, 0.2, 0.4, 0.4, 0.2, 0.2, 0.2, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0.2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0.2, 0.2, 0.4, 0.4, 
0.2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0.2, 0.4, 1.2, 1.6, 2.2, 1.8, 0.4, 0.6, 0, 0.2, 0.2, 
0, 0, 0, 0, 0, 0, 0, 0, 0.2, 0.2, 0.2, 0, 0.6, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0.4, 0.2, 
0, 0, 0.8, 0, 0, 0.2, 0, 0, 0, 0, 0.2, 0.2, 0.2, 0, 0, 1.2, 4, 
0.2, 0, 0, 0, 0, 0, 0, 0, 1.6, 0.6, 0, 0, 0, 1.4, 0.6, 0.4, 0.8, 
0, 0, 0, 0, 0.2, 0, 0.2, 0, 0, 0, 0, 0.6, 0, 0, 0, 0, 0.2, 1.2, 
1, 1.8, 2, 1.8, 0.4, 0.6, 1.2, 1.8, 1.4, 1.2, 0.6, 0.4, 0, 0.2, 
0.2, 1.6, 1.6, 1.4, 0.8, 0.4, 0.2, 0.2, 0, 0.6, 0, 0.4, 0.4, 
0.2, 0.2, 0.2, 0.4, 0.4, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0.6, 0, 0, 0, 0.2, 0.2, 0.4, 0.4, 1.6, 1.8, 
1.2, 1.4, 1.4, 1.4, 0.6, 0.6, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0), X2006.hourly = c(0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 4.8, 
1, 0, 0, 0, 0, 0, 0, 0, 0.8, 0, 0, 0, 0, 0, 0, 6.6, 0.2, 0, 0, 
0.2, 0, 0, 0.2, 0, 1.4, 0, 1.2, 0, 0.2, 0, 0.2, 1, 0, 0, 0, 0, 
0, 0.2, 1.8, 2, 2.2, 0.6, 0.8, 0, 0, 0, 0.2, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0.2, 0, 0, 0.4, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0.2, 1.8, 5.8, 0.2, 0.4, 0, 3.8, 0.2, 1.2, 0.4, 0, 
0, 1, 0.4, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0.2, 0.6, 0.4, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0.8, 0.8, 1, 0.2, 0, 0.2, 0, 0, 0.6, 2.4, 
0.2, 0, 0.2, 0, 0, 1.6, 0.2, 2, 0, 1.2, 4.6, 0.6, 0, 0, 0, 0, 
0, 0, 0, 0, 0.4, 0, 0, 0, 0, 0, 0.2, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0.8, 0.4, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0.2, 0.2, 0, 0, 0, 0, 0.2, 0, 0, 0, 0, 0, 0, 0, 0, 
0.2, 1.8, 0.6, 0, 0, 0, 0.4, 2, 1, 0.8, 0.4, 0.2, 0, 0.2, 0, 
0.8, 0.8, 0.2, 0.2, 0, 0.4, 0.4, 0.6, 0.8, 4, 2, 0.8, 2.6, 1.4, 
0.6, 1, 1, 1, 1.2, 1.2, 0.8, 1.6, 1.8, 0.4, 0.2, 0.2, 0.2, 0.4, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0.6, 1.2, 0.6, 0.8, 
0.8, 1, 0.8, 1.2, 1.2, 0.8, 1, 2.2, 3.4, 2.6, 1.8, 1.6, 1.8, 
3.6, 3, 1.8, 3.2, 1.8, 1, 0.6, 0.4, 0, 1, 0.6, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 3.6, 3.8, 0.2, 0.4, 0, 0.2, 0.2, 0.4, 0.2, 
0.2, 0.2, 0.4, 0, 0, 0, 0.2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0.2, 0.2, 0, 0.2, 0.6, 0, 0.6, 0.4, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0.4, 1, 0, 0, 0, 0.2, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0.6, 0.4, 0.6, 0.4, 0.6, 1, 0.8, 0.2, 0, 0, 
0.2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0.2, 0.6, 1.4, 2.2, 0, 0.4, 1, 0.6, 0, 1.2, 1.6, 0.6, 0, 
0.2, 0, 0, 0, 0, 0, 0.2, 0, 0, 0.2, 0, 0, 0.2, 0, 0.2, 0, 0, 
0, 0, 0.2, 0, 0, 0, 0, 0, 0, 0.2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0.4, 0.2, 0, 1.2, 0.4, 1, 0.6, 0.8, 1, 0, 
0, 0, 0, 0), X2007.hourly = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0.6, 0.4, 1, 2.2, 2.6, 1.8, 0.8, 0.6, 0, 0, 0.2, 0, 0, 0, 
0, 0.4, 0, 0.2, 0.2, 0, 0, 0.2, 0, 0, 0, 0, 0.6, 0.4, 0.4, 0.6, 
0.4, 0.4, 0.2, 0, 0, 0.2, 0.2, 0, 0, 0, 0.2, 0, 0, 0.2, 0.4, 
0.4, 0, 1.4, 0.4, 0.8, 0.4, 0.2, 0.4, 0.6, 0.8, 1.6, 0.6, 0.4, 
0.6, 0.8, 1, 0.8, 1.4, 1.4, 2.2, 0.8, 0.2, 0.2, 0.2, 0, 0, 0.6, 
0.6, 0.4, 0.2, 0.4, 0.6, 0, 0, 0, 0, 0, 0, 0, 0, 0.4, 0.6, 1, 
1, 0.6, 1.4, 1.4, 0.6, 0.4, 0.6, 0.6, 0.8, 1, 1.8, 1.6, 0.6, 
0.4, 0.2, 0.2, 0, 0.2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0.4, 0.8, 1.6, 1.8, 1.6, 
1, 1, 0.4, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0.2, 0, 0, 0, 0, 0, 0.2, 0.4, 0, 0, 0.2, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0.4, 0, 0.2, 0, 0, 0, 0.2, 0.2, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0.2, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0.2, 0, 0.6, 0.6, 0.6, 0.8, 0.2, 0, 1, 0.2, 0.6, 1.6, 1, 0.6, 
0.6, 0.4, 0.6, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0.2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 3.4, 0, 0, 0, 0, 1.6, 1.4, 0.2, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0.2, 2.4, 1.2, 0, 0.2, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0.6, 3, 1.4, 0, 0, 0.2, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0.6, 0.2, 2.6, 3, 0.4, 0, 0, 0, 0, 0, 0, 0, 
0.2, 0, 0, 0, 0.2, 0, 0, 0, 0, 0.8, 0.2, 0.2, 0, 0, 0, 0, 0, 
0.2, 0, 0.2, 0, 0.4, 0.4, 0.8, 0.8, 1.2, 1, 0.6, 0.6, 0.4, 0.2, 
0.2, 0.4, 0.2, 0.2, 0.4, 0.2, 0.8, 0, 0, 0.2, 0, 0, 0.2, 0, 0.2, 
0, 0.2, 0, 0.2, 0.2, 0, 0, 0, 0)), .Names = c("X2005.hourly", "X2006.hourly", 
"X2007.hourly"), class = "data.frame", row.names = c(NA, -744L))

以及我之前使用的功能：

    myfun<- function(x) {
      y<-as.matrix(sapply(x, as.numeric)) 
#Aggregating hourly data to 6,12 and 24 hourly data      
h<-sapply(c(1, 6, 12, 24), function(hrs) colSums(matrix(y, ncol=length(x)/hrs)))
      hours<-sapply(list(h[[1]], h[[2]], h[[3]], h[[4]]), function(qq) list(qq))
      names(hours)<-c("hourly", "h.6", "h.12", "h.24")
#Calculating means of 1, 6, 12 and 24 hourly data      
means<-sapply(list(hours[[1]], hours[[2]], hours[[3]], hours[[4]]), function(ss) mean(ss, na.rm = T))
    #means of 1, 6, 12 and 24 hourly data 
      mean1<-means[[1]]
      mean6<-means[[2]]
      mean12<-means[[3]]
      mean24<-means[[4]]
      meanlist<-list(mean1, mean6, mean12, mean24)
      names(meanlist) <- c("mean1", "mean6", "mean12", "mean24")
      return(meanlist)
    }

我做什么： 我正在使用具有myfun功能的lapply，并且每年分别获得结果。而且这里没有麻烦。

results<-lapply(mydata, myfun)

我想要的东西：

如您所见，h.24函数中names(hours)内有myfun项。为了计算24 hourly data的平均值，函数使用从h.24数据中聚合的hourly数据。但是我想使用外部的24小时daily数据而不是h.24。在下面的数据中，daily分别有2005,2006,2007个值

adddata<-structure(list(X2005.daily = c(0, 0, 3.4, 21.4, 1.8, 1.4, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 8.6, 1.4, 0, 0, 0, 0, 0.6, 7, 6.4, 21.8, 
3.2, 0, 0, 0.6, 11.2, 0), X2006.daily = c(0, 8.6, 11.2, 8.4, 
0, 0, 0, 0, 0, 7.8, 7.6, 1.2, 10.2, 7, 0, 0, 1.6, 7.6, 23.6, 
3.2, 0, 9, 31.4, 10, 2.2, 1.6, 0, 4.8, 10, 1.2, 4.6), X2007.daily = c(0.6, 
9.4, 4.2, 7.2, 15, 15, 0.4, 0, 8.6, 0, 0.2, 0.8, 0, 0.6, 0.6, 
1.4, 8.2, 0.2, 0, 0, 0, 0, 5, 1.6, 4, 5.2, 0, 0, 7, 4.2, 7.2)), .Names = c("X2005.daily", 
"X2006.daily", "X2007.daily"), class = "data.frame", row.names = c(NA, 
-31L))

简而言之：：我想每年添加外部daily数据，然后我想计算1、6、12小时值的均值以及增加的{{1 }}每年的数据。因此，daily的结果将是相同的，但mean1, mean6, mean12将是增加的每日值的平均值。

Answer 1

如果我正确理解了您的问题，则应该可以：

results <- do.call(cbind,results) #convert your list to a data.frame for better overview
colnames(results) <- c("X2005","X2006","X2007")
results <- rbind(results,meanDaily=colMeans(adddata)) #rbind the colMeans of you adddata

results

          X2005        X2006        X2007
mean1     0.1193548    0.2336022    0.1430108   
mean6     0.716129     1.401613     0.8580645   
mean12    1.432258     2.803226     1.716129    
mean24    2.864516     5.606452     3.432258 
meanDaily 2.864516     5.574194     3.43871

评论中请求的其他答案：

library(zoo)

myfun <- function(df_hourly,df_daily)
{

  DF <- sapply(df_hourly,function(col){c(mean(rollapply(col,width=1,sum,by=1,align="left")),
                                mean(rollapply(col,width=6,sum,by=6,align="left")),
                                mean(rollapply(col,width=12,sum,by=12,align="left")),
                                mean(rollapply(col,width=24,sum,by=24,align="left")))})
  DF <- rbind(DF,colMeans(adddata))

  colnames(DF) <- c("yr2005","yr2006","yr2007")
  rownames(DF) <- c("mean.1hr","mean.6hr","mean.12hr","mean.24hr","mean.daily")
  return(DF)
}


myfun(mydata,adddata)

              yr2005    yr2006    yr2007
mean.1hr   0.1193548 0.2336022 0.1430108
mean.6hr   0.7161290 1.4016129 0.8580645
mean.12hr  1.4322581 2.8032258 1.7161290
mean.24hr  2.8645161 5.6064516 3.4322581
mean.daily 2.8645161 5.5741935 3.4387097

使用lapply函数来分隔由两列组成的列

1 个答案: