如何将每日时间序列转换为平均每周？

5 回答

• 6

  mrdwab的answer才恰好工作，因为他们与OP共享一个时区（或其特征） . 为了显示：

  Lines <- 
      "24/07/2012  250.5   499
      23/07/2012  264.75  518.25
      20/07/2012  269.25  525.25
      19/07/2012  267 522.5
      18/07/2012  261.25  517
      17/07/2012  265.75  522.25
      16/07/2012  264.25  523.25
      13/07/2012  258.25  517
      12/07/2012  253.75  513
      11/07/2012  246.25  512.75
      10/07/2012  248 515
      09/07/2012  247 519.25
      06/07/2012  243.25  508.25
      05/07/2012  245 508.5
      04/07/2012  236 500.5
      03/07/2012  234 497.75
      02/07/2012  234.25  489.75
      29/06/2012  229 490.25
      28/06/2012  229.75  487.25
      27/06/2012  229.75  493
      26/06/2012  226.5   486
      25/06/2012  220 482.25
      22/06/2012  214.25  472.5
      21/06/2012  212 469.5
      20/06/2012  210.25  473.75
      19/06/2012  208 472.75
      18/06/2012  206.75  462.5
      15/06/2012  203 456.5
      14/06/2012  205.25  460.5
      13/06/2012  205.25  465.25
      12/06/2012  205.25  469
      11/06/2012  208 471.5
      08/06/2012  208 468.5
      07/06/2012  208 471.25
      06/06/2012  208 467
      05/06/2012  208 458.75
      04/06/2012  208 457.5
      01/06/2012  208 463.5
      31/05/2012  208 466.75
      30/05/2012  208 468
      29/05/2012  212.75  469.75
      28/05/2012  212.75  469.75
      25/05/2012  212.75  465.5"
  
  # Get R's timezone information (from ?Sys.timezone)
  tzfile <- file.path(R.home("share"), "zoneinfo", "zone.tab")
  tzones <- read.delim(tzfile, row.names = NULL, header = FALSE,
    col.names = c("country", "coords", "name", "comments"),
    as.is = TRUE, fill = TRUE, comment.char = "#")
  
  # Run the analysis on each timezone
  out <- list()
  library(xts)
  for(i in seq_along(tzones$name)) {
    tzn <- tzones$name[i]
    Sys.setenv(TZ=tzn)
    con <- textConnection(Lines)
    Source <- read.zoo(con, format="%d/%m/%Y")
    out[[tzn]] <- apply.weekly(Source, colMeans)
  }
  

  现在你可以运行 head(out,5) 并看到一些输出根据使用的时区而有所不同：

  head(out,5)
  $`Europe/Andorra`
                 V2      V3
  2012-05-27 212.75 467.625
  2012-06-03 208.95 465.100
  2012-06-10 208.00 467.400
  2012-06-17 205.10 462.750
  2012-06-24 212.90 474.150
  2012-07-01 229.85 489.250
  2012-07-08 241.05 506.850
  2012-07-15 254.10 516.200
  2012-07-22 265.60 521.050
  2012-07-23 250.50 499.000
  
  $`Asia/Dubai`
                 V2      V3
  2012-05-27 212.75 467.625
  2012-06-03 208.95 465.100
  2012-06-10 208.00 467.400
  2012-06-17 205.10 462.750
  2012-06-24 212.90 474.150
  2012-07-01 229.85 489.250
  2012-07-08 241.05 506.850
  2012-07-15 254.10 516.200
  2012-07-22 265.60 521.050
  2012-07-23 250.50 499.000
  
  $`Asia/Kabul`
                 V2      V3
  2012-05-27 212.75 467.625
  2012-06-03 208.95 465.100
  2012-06-10 208.00 467.400
  2012-06-17 205.10 462.750
  2012-06-24 212.90 474.150
  2012-07-01 229.85 489.250
  2012-07-08 241.05 506.850
  2012-07-15 254.10 516.200
  2012-07-22 265.60 521.050
  2012-07-23 250.50 499.000
  
  $`America/Antigua`
                  V2      V3
  2012-05-25 212.750 465.500
  2012-06-01 209.900 467.550
  2012-06-08 208.000 464.600
  2012-06-15 205.350 464.550
  2012-06-22 210.250 470.200
  2012-06-29 227.000 487.750
  2012-07-06 238.500 500.950
  2012-07-13 250.650 515.400
  2012-07-20 265.500 522.050
  2012-07-24 257.625 508.625
  
  $`America/Anguilla`
                  V2      V3
  2012-05-25 212.750 465.500
  2012-06-01 209.900 467.550
  2012-06-08 208.000 464.600
  2012-06-15 205.350 464.550
  2012-06-22 210.250 470.200
  2012-06-29 227.000 487.750
  2012-07-06 238.500 500.950
  2012-07-13 250.650 515.400
  2012-07-20 265.500 522.050
  2012-07-24 257.625 508.625
  

  更强大的解决方案是确保正确表示您的时区，方法是使用 Sys.setenv(TZ="<yourTZ>") 全局设置或 indexTZ(Source) <- "<yourTZ>" 为每个单独的对象设置它 .

  回复于 2024-04-28T05:51:33+08:00
• 0

  我运行了你的例子，如果我正确理解了问题， apply.weekly 函数将第一个星期五与你数据的第一个星期一聚合在一起 . 我不使用 xts 包，所以其他人必须提供更多的见解 . 我会将日期转换为日期向量，每周的星期日代表该周的每个观察 . ?strptime 汇总了我用于转换的代码 .

  # Get the year of the first observation
  start_year <- format(time(source)[1],"%Y")
  # Convert this into a date for the 1st of Jan in that year.
  start_date <- as.Date(strptime(paste(start_year, "1 1"), "%Y %d %m"))
  
  # Using the difftime function determine the distance (days) since the first day of the first year.
  jul_day <- as.numeric(difftime(time(source),start_date),units="days")
  # Get the date of the Monday before each observation and add it to the start of the year. 
  mondays <- start_date + (jul_day - (jul_day-1)%%7)
  # the %% calculates the remainder.
  # to check that it has worked convert the mondays vector into day names.
  format(mondays, "%A")
  
  # And now you can aggregate the observations using the mondays vector.
  source.w <- aggregate(source[,1:2], mondays, "mean")
  

  回复于 2024-04-28T05:51:33+08:00
• 3

  再看看我手边的问题 .

  使用 xts 库是直截了当的 .

  # say you have xts object name 'dat'
  ep <- endpoints(dat, on = 'weeks')                          # 
  period.apply(x = dat, INDEX = ep, FUN = mean)
  

  回复于 2024-04-28T05:51:33+08:00
• 1

  跟进约书亚乌尔里希的答案 .

  在我的系统（kUbuntu 12）上，以下内容未检索zone.tab文件

  tzfile <- file.path(R.home("share"), "zoneinfo", "zone.tab")
  

  但是，我能够找到zone.tab

  locate zone.tab
  

  由于某种原因（可能是文件权限），我无法直接指向该zone.tab文件，即写入：

  tzfile <- "usr/share/zoneinfo/zone.tab"
  

  回：

  Error in file(file, "rt") : cannot open the connection
  In addition: Warning message:
  In file(file, "rt") :
    cannot open file 'usr/share/zoneinfo/zone.tab': No such file or directory
  

  制作zone.tab的本地副本并指向该副本后问题已解决：

  tzfile <- "~/R/zone.tab"
  

  现在，如果您使用Google for zone.tab，您将在线找到zone.tab的副本，以防您的系统没有或已损坏或其他任何内容 . 这是一个这样的地方：

  http://www.ietf.org/timezones/data/zone.tab
  

  附：我<15所以我不能发表评论，这是我原本应该做的 .

  回复于 2024-04-28T05:51:33+08:00
• 2

  我能够重现您的问题，您可以使用 period.apply() 和自定义"endpoints"来解决它 .

  首先，您提供的数据采用其他人可以轻松阅读的格式 .

  temp = structure(list(V1 = structure(c(33L, 32L, 29L, 27L, 25L, 23L, 
  22L, 19L, 17L, 15L, 13L, 12L, 9L, 7L, 5L, 3L, 2L, 41L, 39L, 37L, 
  36L, 35L, 31L, 30L, 28L, 26L, 24L, 21L, 20L, 18L, 16L, 14L, 11L, 
  10L, 8L, 6L, 4L, 1L, 43L, 42L, 40L, 38L, 34L), .Label = c("01/06/2012", 
  "02/07/2012", "03/07/2012", "04/06/2012", "04/07/2012", "05/06/2012", 
  "05/07/2012", "06/06/2012", "06/07/2012", "07/06/2012", "08/06/2012", 
  "09/07/2012", "10/07/2012", "11/06/2012", "11/07/2012", "12/06/2012", 
  "12/07/2012", "13/06/2012", "13/07/2012", "14/06/2012", "15/06/2012", 
  "16/07/2012", "17/07/2012", "18/06/2012", "18/07/2012", "19/06/2012", 
  "19/07/2012", "20/06/2012", "20/07/2012", "21/06/2012", "22/06/2012", 
  "23/07/2012", "24/07/2012", "25/05/2012", "25/06/2012", "26/06/2012", 
  "27/06/2012", "28/05/2012", "28/06/2012", "29/05/2012", "29/06/2012", 
  "30/05/2012", "31/05/2012"), class = "factor"), V2 = c(250.5, 
  264.75, 269.25, 267, 261.25, 265.75, 264.25, 258.25, 253.75, 
  246.25, 248, 247, 243.25, 245, 236, 234, 234.25, 229, 229.75, 
  229.75, 226.5, 220, 214.25, 212, 210.25, 208, 206.75, 203, 205.25, 
  205.25, 205.25, 208, 208, 208, 208, 208, 208, 208, 208, 208, 
  212.75, 212.75, 212.75), V3 = c(499, 518.25, 525.25, 522.5, 517, 
  522.25, 523.25, 517, 513, 512.75, 515, 519.25, 508.25, 508.5, 
  500.5, 497.75, 489.75, 490.25, 487.25, 493, 486, 482.25, 472.5, 
  469.5, 473.75, 472.75, 462.5, 456.5, 460.5, 465.25, 469, 471.5, 
  468.5, 471.25, 467, 458.75, 457.5, 463.5, 466.75, 468, 469.75, 
  469.75, 465.5)), .Names = c("V1", "V2", "V3"), class = "data.frame", row.names = c(NA, 
  -43L))
  

  我们将清理并将对象转换为 xts 对象 .

  temp$V1 = as.Date(temp$V1, format="%d/%m/%Y")
  library(xts)
  temp.x = xts(temp[-1], order.by=temp$V1)
  

  现在 . 我们尝试 apply.weekly() 函数，但它没有给我们你想要的 .

  apply.weekly(temp.x, colMeans)
  #                V2      V3
  # 2012-05-28 212.75 467.625
  # 2012-06-04 208.95 465.100
  # 2012-06-11 208.00 467.400
  # 2012-06-18 205.10 462.750
  # 2012-06-25 212.90 474.150
  # 2012-07-02 229.85 489.250
  # 2012-07-09 241.05 506.850
  # 2012-07-16 254.10 516.200
  # 2012-07-23 265.60 521.050
  # 2012-07-24 250.50 499.000
  

  要使用 period.apply() ，您需要指定期间的终点（可以是不规则的） . 在这里，我们的第一个时期只是第一个日期，从那里开始，每五天一次 . 剩下几天，所以我们在最后一段时间结束时加上 nrow(temp.x) .

  ep = c(0, seq(1, nrow(temp.x), by = 5), nrow(temp.x))
  period.apply(temp.x, INDEX = ep, FUN = colMeans)
  #                 V2      V3
  # 2012-05-25 212.750 465.500
  # 2012-06-01 209.900 467.550
  # 2012-06-08 208.000 464.600
  # 2012-06-15 205.350 464.550
  # 2012-06-22 210.250 470.200
  # 2012-06-29 227.000 487.750
  # 2012-07-06 238.500 500.950
  # 2012-07-13 250.650 515.400
  # 2012-07-20 265.500 522.050
  # 2012-07-24 257.625 508.625
  

  回复于 2024-04-28T05:51:33+08:00