将日历周转换为日期 [英] convert calendar weeks into daily dates

问题描述

我得到了一个包含两列的列表，即2015年的日历周和一个值：

I got a list with two columns, calendar weeks for 2015 and a value:

calender week   Value
        KW 1    14000
        KW 2    24000

有 - 不幸的是 - 文件中没有更多信息我收到了。然而，我需要分析的是一个每日时间系列，因此我需要将KW 1等转换成一个具有相应日期的列（注意：日历周总是从星期一开始！）：

There is - unfortunately - no more information in the file I received. What I need for analysis, though, is a daily time series, hence I need to convert KW 1 etc into a column with the respective dates (Note: the calendar week always starts with Monday!):

calender week   date        Value
      KW 1      29-12-13    2000
      KW 1      30-12-13    2000
      KW 1      31-12-13    2000
      KW 1      01-01-14    2000
      KW 1      02-01-14    2000
      KW 1      03-01-14    2000
      KW 1      04-01-14    2000
      KW 2      05-01-14    3000
      KW 2      06-01-14    3000
      KW 2      07-01-14    3000
      KW 2      08-01-14    3000
      KW 2      09-01-14    3000
      KW 2      10-01-14    3000
      KW 2      11-01-14    3000

并且该值简单地除以7（=日历周中的天数）。

And the value is simply devided by 7 (= number of days in a calendar week).

推荐答案

lubridate 帮你。

似乎这个年份在你的例子中不是变量，所以我假设所有的日期都在2014年（或者最后一个日期） 2013年的几天，从第1周开始12月30日）。如果您不熟悉 lubridate ，以下内容将包含许多您不知道的功能。使用？获得关于它们的帮助（例如？ymd ）。

It seems that the year is not variable in your example, so I assume that all the dates are in 2014 (or in the last few days of 2013, since week 1 starts on December 30th). If you are not familiar with lubridate, the following will contain many functions that are unknown to you. Use ? to get help about them (e.g. ?ymd).

第一步是获取一年中第一个星期的星期一。当然，你可以查找它，但 lubridate 可用于计算：

The first step is to get the Monday of the first week of the year. Of course, you could look it up, but lubridate can be used to calculate it:

library(lubridate)
start_date <- ymd("20140201")
week(start_date) <- 1
wday(start_date) <- "Monday"
start_date
## [1] "2013-12-30 UTC"

这首先选择2014年的任意一天，然后将星期设置为1，将星期设置为星期一。现在我可以通过添加适当的周数获取任意日历周的第一天：

This first picks an arbitrary day in 2014 and then sets the week to 1 and the weekday to Monday. Now I can get the first day of any calendar week by adding the appropriate number of weeks:

start_date + weeks(2)
## [1] "2014-01-13 UTC"

数据集为三周：

data <- data.frame(week  = paste("KW", 1:3), value = c(14000, 21000,  28000))
data
##   week value
## 1 KW 1 14000
## 2 KW 2 21000
## 3 KW 3 28000

转换为所需格式的操作如下：

And the conversion to your desired format works as follows:

weeks <- rep(data$week, each = 7)
weeks_num = as.numeric(gsub("KW *", "", weeks))
intervals <- weeks(weeks_num - 1) + days(0:6)
dates <- as.Date(start_date + intervals)
values <- rep(data$value, each = 7)/7
new_data <- data.frame(week = weeks, date = dates, value = values)
new_data
##    week       date value
## 1  KW 1 2013-12-30  2000
## 2  KW 1 2013-12-31  2000
## 3  KW 1 2014-01-01  2000
## 4  KW 1 2014-01-02  2000
## 5  KW 1 2014-01-03  2000
## 6  KW 1 2014-01-04  2000
## 7  KW 1 2014-01-05  2000
## 8  KW 2 2014-01-06  3000
## 9  KW 2 2014-01-07  3000
## 10 KW 2 2014-01-08  3000
## 11 KW 2 2014-01-09  3000
## 12 KW 2 2014-01-10  3000
## 13 KW 2 2014-01-11  3000
## 14 KW 2 2014-01-12  3000
## 15 KW 3 2014-01-13  4000
## 16 KW 3 2014-01-14  4000
## 17 KW 3 2014-01-15  4000
## 18 KW 3 2014-01-16  4000
## 19 KW 3 2014-01-17  4000
## 20 KW 3 2014-01-18  4000
## 21 KW 3 2014-01-19  4000

这个工作原理如下：

• 首先，我准备日期。每个日历周重复七次（每天一次）。然后删除KW零件，将星期转换为数字。之后，我使用 lubridate 函数 weeks（）和 days（）来构造所有所需的时间间隔，因为 start_date 。

• First I prepare the dates. Each calendar week is repeated seven times (once for each day). Then the "KW " part is removed and the weeks are converted to numeric. Afterwards, I use the lubridate functions weeks() and days() to construct all the required time intervals since start_date.

最后，我将结果放到一个新的数据框架中。

Finally, I put the results into a new data frame.

最后一句话：这是一个复杂的解决方案。如果你确定，没有一周会丢失，更容易简单地产生一系列日期，甚至不考虑日历周的列如下：

A final remark: This is a complicated solution. If you are certain, that no week will be missing, it is easier to simply produce a series of dates without even considering the column for the calendar weeks as follows:

dates <- as.Date(start_date + days(0:(7*nrow(data) - 1)))
identical(new_data$date, dates)
## [1] TRUE

所以这里我只需要计算周数 nrows（），并产生一个时间间隔序列 days（），然后添加到 start_date 。
然而，如果日历周可能丢失，您应该使用上面更一般的解决方案。

So here I simply count the number of weeks with nrows(), and produce a sequence of time intervals with days() that I then add to start_date. However, if calendar weeks could be missing, you should use the more general solution above.

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