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我想将列表对象转换为zoo然后应用于rollapply该zoo对象。下面复制的简短示例（我有 90,000 个此类文件要处理，使用 UNIX :)）。假设我的列表有两个数据框。

1）我想将每个数据框中的日期转换为这种格式：

dates <- as.Date(paste0(mylist$year, "-", mylist$month, "-", mylist$day), format="%Y-%m-%d")

z <- zoo(mylist, order.by=mylist[,1])

我知道lapply可以做到这一点，但我尝试过但没有成功。

一旦我得到我的zoo对象，我想使用rollapply：

library(hydroTSM)#for daily2annual function but aggregate can do 

    x.3max <- rollapply(data=zooobject, width=3, FUN=sum, fill=NA, partial= TRUE,
                         align="center")
    # Maximum value per year of 3-day total rainfall for each one of the simulations
    z.3max.annual <- daily2annual(z.3max,  FUN=max,na.rm=TRUE)#dates=1

上面的代码所做的是将一个 3 天的窗口集中在 zooobject 中数据框的每一列上，并对这些值求和。提取 3 天总数中每年的最大值。

      mylist<- list(a,a)
mylist<-lapply(mylist, function(x) x[x[["Month"]] %in% c(12,1,2),])# extract data for DJF for individual sites


    library(zoo)
       a= structure(list(Year = c(1975L, 1975L, 1975L, 1975L, 1975L, 1975L
), Month = c(1L, 1L, 1L, 1L, 1L, 1L), Site = structure(c(1L, 
1L, 1L, 1L, 1L, 1L), .Label = "G100", class = "factor"), Day = 1:6, 
    sim01 = c(28.49, 29.04, 27.62, 28.43, 28.69, 29.16), sim02 = c(29.49, 
    30.04, 28.62, 29.43, 29.69, 30.16), sim03 = c(30.49, 31.04, 
    29.62, 30.43, 30.69, 31.16), sim04 = c(31.49, 32.04, 30.62, 
    31.43, 31.69, 32.16), sim05 = c(32.49, 33.04, 31.62, 32.43, 
    32.69, 33.16), sim06 = c(33.49, 34.04, 32.62, 33.43, 33.69, 
    34.16), sim07 = c(34.49, 35.04, 33.62, 34.43, 34.69, 35.16
    ), sim08 = c(35.49, 36.04, 34.62, 35.43, 35.69, 36.16), sim09 = c(36.49, 
    37.04, 35.62, 36.43, 36.69, 37.16), sim10 = c(37.49, 38.04, 
    36.62, 37.43, 37.69, 38.16), sim11 = c(38.49, 39.04, 37.62, 
    38.43, 38.69, 39.16), sim12 = c(39.49, 40.04, 38.62, 39.43, 
    39.69, 40.16), sim13 = c(40.49, 41.04, 39.62, 40.43, 40.69, 
    41.16), sim14 = c(41.49, 42.04, 40.62, 41.43, 41.69, 42.16
    ), sim15 = c(42.49, 43.04, 41.62, 42.43, 42.69, 43.16), sim16 = c(43.49, 
    44.04, 42.62, 43.43, 43.69, 44.16), sim17 = c(44.49, 45.04, 
    43.62, 44.43, 44.69, 45.16), sim18 = c(45.49, 46.04, 44.62, 
    45.43, 45.69, 46.16), sim19 = c(46.49, 47.04, 45.62, 46.43, 
    46.69, 47.16), sim20 = c(47.49, 48.04, 46.62, 47.43, 47.69, 
    48.16)), .Names = c("Year", "Month", "Site", "Day", "sim01", 
"sim02", "sim03", "sim04", "sim05", "sim06", "sim07", "sim08", 
"sim09", "sim10", "sim11", "sim12", "sim13", "sim14", "sim15", 
"sim16", "sim17", "sim18", "sim19", "sim20"), row.names = c(NA, 
6L), class = "data.frame")

输出应类似于：

Year Site Sim01... 
1975 G100 ...
1976 G100 ...
1977 G100 ...

只需要 c(12,1,2) 月份中的值。

4

2 回答 2

1

这会生成一个动物园对象列表，Lz然后rollapply对列表中的每个组件执行L2。最后L3汇总一年中max的每一列。

library(zoo)

mylist <- list(a, a) # a is given at bottom of question

Lz <- lapply(mylist, read.zoo, index = 1:3, format = "%Y %m %d")
L2 <- lapply(Lz, rollapply, 3, sum, partial = TRUE)
L3 <- lapply(L2, function(z) aggregate(z, as.numeric(format(time(z), "%Y")), max))

给予：

> L3

[[1]]
     sim01 sim02 sim03 sim04 sim05  sim06  sim07  sim08  sim09  sim10  sim11
1975 86.28 89.28 92.28 95.28 98.28 101.28 104.28 107.28 110.28 113.28 116.28
      sim12  sim13  sim14  sim15  sim16  sim17  sim18  sim19  sim20
1975 119.28 122.28 125.28 128.28 131.28 134.28 137.28 140.28 143.28

[[2]]
     sim01 sim02 sim03 sim04 sim05  sim06  sim07  sim08  sim09  sim10  sim11
1975 86.28 89.28 92.28 95.28 98.28 101.28 104.28 107.28 110.28 113.28 116.28
      sim12  sim13  sim14  sim15  sim16  sim17  sim18  sim19  sim20
1975 119.28 122.28 125.28 128.28 131.28 134.28 137.28 140.28 143.28
于 2015-03-04T22:50:27.110 回答
0

解决了

lst1 <- lapply(list.files(pattern=".csv"),function(x) read.table(x,header=TRUE,sep="")) # read all files and data and replace -999.9 with NA

lst2<-lapply(lst1, function(x) x[x[["Month"]] %in% c(6,7,8),])#c(6,7,8) extract data for DJF for individual sites
names(lst2)<-list.files(pattern=".csv")
lapply(lst2,tail,4)
lst3<-lapply(lst2, function(x) x[!(names(x) %in% c("Site"))])
Lz <- lapply(lst3, read.zoo, index = 1:3, format = "%Y %m %d")

L2 <- lapply(Lz, rollapply, 3, sum, partial = TRUE)
L3 <- lapply(L2, function(z) aggregate(z, as.numeric(format(time(z), "%Y")), max))

mapply(
  write.table,
  x=L3, file=paste(names(L3), "csv", sep="."),
  MoreArgs=list(row.names=FALSE, sep=",")
) # write files to folder keeping the list names as file names
于 2015-03-05T03:19:57.213 回答