遍历一列唯一值并创建多列

3
我试图分解我的以前的问题,并制定一个不同步骤的计划,以实现我最终想要的目标。目前,我正在尝试通过循环查找每个唯一源(如下表中所示的source列)是否开启机械系统。
例如,我已经获得了以下配置文件,告诉我在每个4个季节的典型工作日的哪些小时内系统处于开启状态。请注意,某些源在一天内的多个时段内都处于开启状态,因此您可以看到第二行重复了2个时段。

enter image description here

现在我要做的是创建一些样本日期,并通过每个唯一源进行循环,根据Profile表中提供的信息,仅表示特定小时的系统开启或关闭状态。到目前为止,我所做的是使用以下代码创建上面的表:

enter image description here

下面的代码将创建以上表格:

# create dates table
dates =data.frame(dates=seq(
  from=as.POSIXct("2010-1-1 0:00", tz="UTC"),
  to=as.POSIXct("2012-12-31 23:00", tz="UTC"),
  by="hour"))  

# add year month day hour weekday column

dates$year <- format(dates[,1], "%Y") # year
dates$month <- format(dates[,1], "%m") # month
dates$day <- format(dates[,1], "%d") # day
dates$hour <- format(dates[,1], "%H") # hour
dates$weekday <- format(dates[,1], "%a") # weekday

# set system locale for reproducibility

Sys.setlocale(category = "LC_TIME", locale = "en_US.UTF-8")

# calculate season column

d = function(month_day) which(lut$month_day == month_day)
lut <- data.frame(all_dates = as.POSIXct("2012-1-1") + ((0:365) * 3600 * 24),
                  season = NA)
lut <- within(lut, { month_day = strftime(all_dates, "%b-%d") })

lut[c(d("Jan-01"):d("Mar-15"), d("Nov-08"):d("Dec-31")), "season"] = "winter"
lut[c(d("Mar-16"):d("Apr-30")), "season"] = "spring"
lut[c(d("May-01"):d("Sep-27")), "season"] = "summer"
lut[c(d("Sep-28"):d("Nov-07")), "season"] = "autumn"
rownames(lut) = lut$month_day

dates = within(dates, {
  season = lut[strftime(dates, "%b-%d"), "season"]
})

我现在想做的是在profile表中为Source列中的每个唯一值添加右侧的列,并根据以下标准估计数据集中每个小时系统是否开启或关闭。
我正在苦恼如何用多个if条件进行类似于vlookup的操作并将值粘贴到新列中的编程概念。例如,对于我的样本数据,循环应该创建2个程序,因为Source列只有2个唯一来源Stack 1Stack 2。棘手的部分是将需要类似以下内容的if语句:
例如,表2的第一行应将季节列的值与profile表匹配,并查看该小时是否落在该特定季节的系统开启期间。如果它在规定的时间内,则说“开启”,如果在外面就说“关闭”。因此,结果应该像下图所示的2个红色字体列一样:
冬天的例子: enter image description here 春天的例子: enter image description here 我已经用以下代码获得了列的唯一值:
values <- unique(profile$Source)

但现在使用for循环无法继续工作。

我想知道有没有人能够给我任何建议,如何使用循环在表2中创建2列唯一来源?

以下是我正在使用的典型每周“概要”数据表:

> dput(profile)
structure(list(`Source no` = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L), Source = structure(c(1L, 
1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 
2L, 2L, 2L, 2L), .Label = c("Stack 1", "Stack 2"), class = "factor"), 
    Period = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
    1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L), Day = structure(c(2L, 
    6L, 7L, 5L, 1L, 3L, 4L, 2L, 6L, 7L, 5L, 1L, 3L, 4L, 2L, 6L, 
    7L, 5L, 1L, 3L, 4L), .Label = c("Fri", "Mon", "Sat", "Sun", 
    "Thu", "Tue", "Wed"), class = "factor"), `Spring On` = c(0L, 
    0L, 0L, 0L, 0L, 0L, 0L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 15L, 
    15L, 15L, 15L, 15L, 15L, 15L), `Spring Off` = c(23L, 23L, 
    23L, 23L, 23L, 23L, 23L, 7L, 7L, 7L, 7L, 7L, 7L, 7L, 18L, 
    18L, 18L, 18L, 18L, 18L, 18L), `Summer On` = structure(c(1L, 
    1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
    1L, 1L, 1L, 1L, 1L), .Label = "off", class = "factor"), `Summer Off` = structure(c(1L, 
    1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
    1L, 1L, 1L, 1L, 1L), .Label = "off", class = "factor"), `Autumn On` = structure(c(1L, 
    1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
    1L, 1L, 1L, 1L, 1L), .Label = "off", class = "factor"), `Autumn Off` = structure(c(1L, 
    1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
    1L, 1L, 1L, 1L, 1L), .Label = "off", class = "factor"), `Winter On` = structure(c(1L, 
    1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 
    2L, 2L, 2L, 2L, 2L), .Label = c("0", "off"), class = "factor"), 
    `Winter Off` = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
    1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L), .Label = c("23", 
    "off"), class = "factor")), .Names = c("Source no", "Source", 
"Period", "Day", "Spring On", "Spring Off", "Summer On", "Summer Off", 
"Autumn On", "Autumn Off", "Winter On", "Winter Off"), class = "data.frame", row.names = c(NA, 
-21L))

非常感谢


1
你的设置代码不起作用。请检查这一行代码:dates = data.frame(dates =seq(as.Date('2010-01-01'),as.Date('2012-12-31'),by = "hour")) - Pierre L
抱歉,我之前放错了代码,请现在查看已更正的代码,谢谢。dates =data.frame(dates=seq(from=as.POSIXct("2010-1-1 0:00", tz="UTC"), to=as.POSIXct("2012-12-31 23:00", tz="UTC"), by="hour")) - Achak
1
这行代码是什么意思?“如果当前小时与'season'列中的值相同,则查找周几、小时并返回系统是否开启。” - Pierre L
1
你应该在问题中发布你的示例所需的最终产品的外观。这样,就不会有任何混淆,以至于你希望最终得到什么。 - Pierre L
嗨,皮埃尔,举个例子,表2的第一行应该将季节列的值与“配置文件”表匹配,并查看该小时是否在该特定季节的时间段内系统将开启。如果它在规定的时间段内,则说“开”,如果在外面则说“关”。我只是发布另一个表格,以便我可以看到最终答案,谢谢。 - Achak
1个回答

6
为了实现从profiledates的数据传输,您需要转换profile数据,然后将其与dates连接。在以下步骤中,我使用了data.table包。 1) 加载data.table包并将数据集转换为数据表(增强的数据框架):
library(data.table)

setDT(profile)
setDT(dates)

2) 重新格式化 profile 数据集中的数值:

# set the 'off' values to NA
profile[profile=="off"] <- NA
# make sure that all the remaining values are numeric (which wasn't the case)
profile <- profile[, lapply(.SD, as.character), by=.(Source,Period,Day)][, lapply(.SD, as.numeric), by=.(Source,Period,Day)]

3) 为每个季节创建数据集,其中每小时一个(或两个)Source处于on状态。我只为春季和冬季创建了数据集,因为夏季和秋季只有off/NA值(我们稍后会处理这些):

pr.spring <- profile[, .(season = "spring",
                         hour = c(`Spring On`:(`Spring Off`-1))),
                     by=.(Source,Period,Day)]
pr.winter <- profile[!is.na(`Winter On`), .(season = "winter",
                                            hour = c(`Winter On`:(`Winter Off`-1))),
                     by=.(Source,Period,Day)]

请注意,我使用了Spring Off - 1。这是因为我假设堆栈在23:00关闭。通过使用-1,我包括了第22个小时,但不包括第23个小时。如果需要,您可以更改此设置。
4) 将步骤3中的数据集绑定在一起,并准备好进行dcast操作的结果数据集:
prof <- rbindlist(list(pr.spring,pr.winter))
prof <- prof[, .(weekday = Day, season, Source = gsub(" ",".",Source), hour = sprintf("%02d",hour))]

5) 将第4步的数据集转换为每个 Stack 都有列的数据集,并将 weekday 列更改为字符。在下一步的连接操作中,这是必需的,因为 dates 数据集中的 weekday 列也是字符列:

profw <- dcast(prof, weekday + season + hour ~ Source, value.var = "hour", fun.aggregate = length, fill = 0)
profw[, weekday := as.character(weekday)]

6) 将两个数据集合并在一起,并用0填充缺失值(记住我在第3步中说过:“我们稍后会处理这些”):

dates.new <- profw[dates, on=c("weekday", "season", "hour")][is.na(Stack.1), `:=` (Stack.1 = 0, Stack.2 = 0)]

生成的数据集现在对于dates数据集中的每个日期都有一个堆叠列,其中1 ="on",而0 = "off"


从生成的数据集中获取的快照:

> dates.new[weekday=="Fri" & hour=="03" & month %in% c("03","04","09")]
    weekday season hour Stack.1 Stack.2               dates year month day
 1:     Fri winter   03       1       1 2010-03-05 03:00:00 2010    03  05
 2:     Fri winter   03       1       1 2010-03-12 03:00:00 2010    03  12
 3:     Fri spring   03       1       0 2010-03-19 03:00:00 2010    03  19
 4:     Fri spring   03       1       0 2010-03-26 03:00:00 2010    03  26
 5:     Fri spring   03       1       0 2010-04-02 03:00:00 2010    04  02
 6:     Fri spring   03       1       0 2010-04-09 03:00:00 2010    04  09
 7:     Fri spring   03       1       0 2010-04-16 03:00:00 2010    04  16
 8:     Fri spring   03       1       0 2010-04-23 03:00:00 2010    04  23
 9:     Fri spring   03       1       0 2010-04-30 03:00:00 2010    04  30
10:     Fri summer   03       0       0 2010-09-03 03:00:00 2010    09  03
11:     Fri summer   03       0       0 2010-09-10 03:00:00 2010    09  10
12:     Fri summer   03       0       0 2010-09-17 03:00:00 2010    09  17
13:     Fri summer   03       0       0 2010-09-24 03:00:00 2010    09  24
14:     Fri winter   03       1       1 2011-03-04 03:00:00 2011    03  04
15:     Fri winter   03       1       1 2011-03-11 03:00:00 2011    03  11
16:     Fri spring   03       1       0 2011-03-18 03:00:00 2011    03  18
17:     Fri spring   03       1       0 2011-03-25 03:00:00 2011    03  25
18:     Fri spring   03       1       0 2011-04-01 03:00:00 2011    04  01
19:     Fri spring   03       1       0 2011-04-08 03:00:00 2011    04  08
20:     Fri spring   03       1       0 2011-04-15 03:00:00 2011    04  15
21:     Fri spring   03       1       0 2011-04-22 03:00:00 2011    04  22
22:     Fri spring   03       1       0 2011-04-29 03:00:00 2011    04  29
23:     Fri summer   03       0       0 2011-09-02 03:00:00 2011    09  02
24:     Fri summer   03       0       0 2011-09-09 03:00:00 2011    09  09
25:     Fri summer   03       0       0 2011-09-16 03:00:00 2011    09  16
26:     Fri summer   03       0       0 2011-09-23 03:00:00 2011    09  23
27:     Fri autumn   03       0       0 2011-09-30 03:00:00 2011    09  30
28:     Fri winter   03       1       1 2012-03-02 03:00:00 2012    03  02
29:     Fri winter   03       1       1 2012-03-09 03:00:00 2012    03  09
30:     Fri spring   03       1       0 2012-03-16 03:00:00 2012    03  16
31:     Fri spring   03       1       0 2012-03-23 03:00:00 2012    03  23
32:     Fri spring   03       1       0 2012-03-30 03:00:00 2012    03  30
33:     Fri spring   03       1       0 2012-04-06 03:00:00 2012    04  06
34:     Fri spring   03       1       0 2012-04-13 03:00:00 2012    04  13
35:     Fri spring   03       1       0 2012-04-20 03:00:00 2012    04  20
36:     Fri spring   03       1       0 2012-04-27 03:00:00 2012    04  27
37:     Fri summer   03       0       0 2012-09-07 03:00:00 2012    09  07
38:     Fri summer   03       0       0 2012-09-14 03:00:00 2012    09  14
39:     Fri summer   03       0       0 2012-09-21 03:00:00 2012    09  21
40:     Fri autumn   03       0       0 2012-09-28 03:00:00 2012    09  28

1
嗨@Jaap,你的解释对我来说非常清晰,我有一个很好的想法来处理我的主要数据集,它有更多的来源和年份。这是理解如何入手的好开端。再次感谢。 - Achak

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