Question

我正在寻找一种更有效的方法，将每年与时间序列（2002-2016）逐月分开。我已经手工完成了，但需要花费很多时间。

mypath<-"D:/SNOWL"
myras<-list.files(path=mypath,pattern = glob2rx("*.tif$"), 
                    full.names = TRUE, recursive = TRUE)

> myras
   [1] "D:/SNOWL/MOYDSL10A1.A2002001.tif" "D:/SNOWL/MOYDSL10A1.A2002002.tif"
   [3] "D:/SNOWL/MOYDSL10A1.A2002003.tif" "D:/SNOWL/MOYDSL10A1.A2002004.tif"
   [5] "D:/SNOWL/MOYDSL10A1.A2002005.tif" "D:/SNOWL/MOYDSL10A1.A2002006.tif"
   [7] "D:/SNOWL/MOYDSL10A1.A2002007.tif" "D:/SNOWL/MOYDSL10A1.A2002008.tif"
   [9] "D:/SNOWL/MOYDSL10A1.A2002009.tif" "D:/SNOWL/MOYDSL10A1.A2002010.tif"
  [11] "D:/SNOWL/MOYDSL10A1.A2002011.tif" "D:/SNOWL/MOYDSL10A1.A2002012.tif"

serie<-orgTime(myras, nDays = "asIn", begin ="2002-01-01",end = "2016-12-31", pillow = 75, pos1 = 13, pos2 = 19)
filter<-serie$inputLayerDates 

> filter
   [1] "2002-01-01" "2002-01-02" "2002-01-03" "2002-01-04" "2002-01-05"
   [6] "2002-01-06" "2002-01-07" "2002-01-08" "2002-01-09" "2002-01-10"
  [11] "2002-01-11" "2002-01-12" "2002-01-13" "2002-01-14" "2002-01-15"
  [16] "2002-01-16" "2002-01-17" "2002-01-18" "2002-01-19" "2002-01-20"
  [21] "2002-01-21" "2002-01-22" "2002-01-23" "2002-01-24" "2002-01-25"
  [26] "2002-01-26" "2002-01-27" "2002-01-28" "2002-01-29" "2002-01-30"
  [31] "2002-01-31" "2002-02-01" "2002-02-02" "2002-02-03" "2002-02-04"
  [36] "2002-02-05" "2002-02-07" "2002-02-08" "2002-02-09" "2002-02-10"
  [41] "2002-02-11" "2002-02-12" "2002-02-13" "2002-02-14" "2002-02-15"

Answer 1

修改：

好的，让我们试试一个完整的示例，看看它是否适合您：

# Here we generate filenames as returned from `list.files`: rm(list = ls()) myras <- sapply(1:5465, function(i) paste0('D:/SNOWL/MOYDSL10A1.A',sample(2000:2016,1),sample(c(paste0('00',1:9),paste0('0',10:99),100:365),1),'.tif')) head(myras) # Let's extract the timestamps tstmps <- regmatches(myras,regexpr('[[:digit:]]{7}',myras)) head(tstmps,50) # And now convert the timestamps to dates dates <- as.Date(as.numeric(substr(tstmps,5,7)) - 1, origin = paste0(substr(tstmps,1,4),"-01-01")) head(dates,10) # Last step is to sort the files by month #check months print(month.name) myras_byM = sapply(month.name,function(x) myras[months(dates) == x]) head(myras_byM$January) head(myras_byM$February) head(myras_byM$March) head(myras_byM$April) head(myras_byM$May) head(myras_byM$June) head(myras_byM$July) head(myras_byM$August) head(myras_byM$September) head(myras_byM$October) head(myras_byM$November) head(myras_byM$December)

如果您有一致的命名约定，则可以轻松地从文件名中获取日期。

在您的情况下，我看到文件按年份和年份排序。因此，只需从文件名中删除日期，然后您可以根据需要过滤它。要做到这一点，我正在使用正则表达式。在这种情况下，我对日期和DOY字符串感兴趣，它应该始终是7个数字。因此，相应的RE是[[:digit:]]{7}，这意味着7个连续的数字。 regexpr找到匹配项，regmatches返回匹配项。

dts <- regmatches(myras,regexpr('[[:digit:]]{7}',myras))

然后你只需使用substring来提取你需要的数字（这种方法假定它总是4位数，然后是3位DOY）并将其转换为日期：

dts <-as.Date(as.numeric(substr(dts,5,7)) - 1, origin = paste0(substr(dts,1,4),"-01-01"))

这会为您提供示例中filter的变量。

如果您希望按月对整个时间序列进行排序，则可以使用内置名称month.name进行sapply或lapply。基函数months将返回给定日期的月份名称：

myras_byMonth <- sapply(month.name,function(x) myras[months(dts) == x])

希望我能正确理解你的问题，这就是你要找的东西。

最佳，

缬氨酸

按月汇总modis列表文件

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