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我希望（算术）平均每日数据，从而将我的每日时间序列转换为每周时间序列。

遵循此线程：如何使用 R 按列计算每周数据的平均值？，我正在使用xts库。

# Averages daily time series into weekly time series
# where my source is a zoo object
source.w <- apply.weekly(source, colMeans)

我遇到的问题是它平均了周二到下周一的系列数据。

我正在寻找从周一到周五平均我的每日数据的选项。

有什么提示吗？

这里还有一点：

    # here is part of my data, from a "blé colza.txt" file


    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



# Loads external libraries
library("zoo") # or require("zoo")
library("xts") # or require("xts")

# Loads data as a zoo object
source <- read.zoo("blé colza.txt", sep=",", dec=".", header=T, na.strings="NA",     format="%d/%m/%Y")

# Averages daily time series into weekly time series
# https://stackoverflow.com/questions/11129562/how-does-one-compute-the-mean-of-weekly-    data-by-column-using-r
source.w <- apply.weekly(source, colMeans)
4

5 回答 5

6

mrdwab的答案恰好起作用，因为它们与 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)
}

现在您可以运行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

Sys.setenv(TZ="<yourTZ>")更强大的解决方案是通过使用全局设置或indexTZ(Source) <- "<yourTZ>"为每个单独的对象设置它来确保正确表示您的时区。

于 2012-08-10T15:43:26.087 回答
3

我能够重现您的问题，您可以使用period.apply()自定义“端点”来解决它。

首先，您提供的数据以其他人可以轻松阅读的格式提供。

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))

我们会进行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。我们尝试了这个apply.weekly()功能，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()，您需要指定期间的端点（可以是不规则的）。在这里，我们的第一次月经只是第一次约会，从那里开始，每五天一次。这还剩下几天，所以我们添加nrow(temp.x)作为最后一个时期的结束。

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
于 2012-08-10T07:01:45.833 回答
2

我运行了您的示例，如果我正确理解了问题，则该apply.weekly函数将第一个星期五与您的数据的第一个星期一聚合。我不使用该xts软件包，因此其他人必须对此提供更多见解。我会将日期转换为日期向量，其中每周星期一的日期表示该周的每个观察值。?strptime总结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")
于 2012-08-10T03:04:08.753 回答
1

跟进约书亚乌尔里希的回答。

在我的系统（kUbuntu 12）上，以下内容没有检索到 zone.tab 文件

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

但是，我能够通过以下方式找到 zone.tab

locate zone.tab

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

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

回来：

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解决：

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

现在，如果你用谷歌搜索 zone.tab，你会在网上找到一份 zone.tab 的副本，以防你的系统没有它或者它已损坏或其他什么。这里是这样一个地方：

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

PS我<15，所以我不能发表评论，否则我会这样做。

于 2013-02-16T10:49:39.473 回答
1

再看看我手头的问题。

使用xts库非常简单。

# say you have xts object name 'dat'
ep <- endpoints(dat, on = 'weeks')                          # 
period.apply(x = dat, INDEX = ep, FUN = mean)
于 2013-04-20T21:20:34.957 回答