将自然年转换为水文年时间序列Rasterr

给定一个具有自然年（

Jan-Dec

）的

rasterbrick

，如何将其转换为水文年（从10月1日开始，到9月30日结束）？与问题相似（但不完全相同）

样本数据：

     nl <- 768
     s <- brick(nrows = 510,  ncols = 1068,
                   xmn = -180, xmx = 180, ymn = -90, ymx = 90,
                   crs = "+proj=longlat +datum=WGS84",
                   nl = nl,values=TRUE)
        dates <- seq(as.Date("1950-01-01"), as.Date("2013-12-31"), by = "month")
        s <- setZ(s, dates)
vals <- 1:ncell(s)
s[]=vals

nlI刚才对此表示感谢

to_wateryear <- function(thedate){
  thedate = as.Date(thedate)
  year = as.integer(substr(thedate, 1, 4))
  cutoff =  cutoff = as.Date(paste(year, '-10-01', sep=''))
  return(ifelse(thedate < cutoff, 
           year*1000 + thedate - as.Date(paste(year - 1, '-09-30', sep='')),
           (year + 1)*1000 + thedate - as.Date(paste(year, '-09-30', sep=''))))
}

编辑根据您的评论，听起来您真正想要的是按水年分割您的rasterbrick，而不是按水文年表示日期

library(raster)
nl <- 768
s <- brick(nrows = 510,  ncols = 1068, 
  xmn = -180, xmx = 180, ymn = -90, ymx = 90,
  crs = "+proj=longlat +datum=WGS84",
  nl = nl,values=TRUE)

dates <- seq(as.Date("1950-01-01"), as.Date("2013-12-31"), by = "month")
# use water year
s <- setZ(s, to_wateryear(dates))
vals <- 1:ncell(s)
s[]=vals

# split by water year
hyears = unique(getZ(s) %/% 1000)
res = vector("list", length(hyears))
for(i in 1:length(hyears))
  res[[i]] = s[[which(getZ(s) %/% 1000 == hyears[i])]]
names(res) = paste0("HY", hyears)

库（光栅）
nl这可能会很有帮助。你能在我的rasterbrick上测试它吗？我确实提供了上面的示例数据。我到电脑前会试试的。但是，请尝试让我知道它是否适合你。谢谢。我尝试了以下方法，但得到的是数字而不是日期：
s是的。。。您希望如何表示水年份的日期？也许您应该添加一个您期望的输出示例。类似这样的示例，但在rasterbrick
https://gist.github.com/cvitolo/e2ec4887f376e3b7d912
library(raster)
nl <- 768
s <- brick(nrows = 510,  ncols = 1068, 
  xmn = -180, xmx = 180, ymn = -90, ymx = 90,
  crs = "+proj=longlat +datum=WGS84",
  nl = nl,values=TRUE)

dates <- seq(as.Date("1950-01-01"), as.Date("2013-12-31"), by = "month")
# use water year
s <- setZ(s, to_wateryear(dates))
vals <- 1:ncell(s)
s[]=vals

# split by water year
hyears = unique(getZ(s) %/% 1000)
res = vector("list", length(hyears))
for(i in 1:length(hyears))
  res[[i]] = s[[which(getZ(s) %/% 1000 == hyears[i])]]
names(res) = paste0("HY", hyears)