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r - 遍历列的唯一值并创建多个列

转载 作者:行者123 更新时间:2023-12-04 11:34:57 26 4
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我试图分解我之前的问题,并制定了一个计划,以通过不同的步骤实现我最终寻找的目标。目前我正在尝试做一个循环来查明机械系统是否为每个唯一源打开或不打开,如下面第一个表中 source 列所示。

例如,我得到了以下配置文件,该配置文件告诉我系统在 4 个季节中每个季节的典型工作日的工作时间。请注意,某些来源在一天中出现不止一个时段,因此您可以看到堆栈 2 重复了 2 个时段。

enter image description here

我现在想要实现的是,我已经创建了一些样本日期,并且想循环遍历每个唯一的来源,并根据 配置文件表。到目前为止,我所做的是使用以下代码创建下表:

enter image description here

下面的代码将创建上表:

# 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"]
})

我现在要做的是在 profile 表中为 Source 列中的每个唯一值添加列,并根据以下标准进行估算系统在数据集中每小时开启或关闭。

我正在为如何使用多个 if 条件并在新列中粘贴值来执行类似于 vlookup 的编程概念而苦苦挣扎。例如,对于我的示例数据,循环应该创建 2 个程序,因为 Source 列只有 2 个唯一源 Stack 1Stack 2。棘手的一点是它的 if 语句需要像这样的东西:

例如,表 2 的第一行应将季节列的值与 profile 表相匹配,并查看该小时是否在系统将运行的特定季节内上。如果它落在规定的时间段内,则说“开”,如果在规定的时间段内,则说 off。所以结果应该看起来像下图中的这 2 个红色字体列:

冬天的一个例子: enter image description here

Spring 的一个例子: enter image description here我已设法使用以下代码获取列的唯一值:

values <- unique(profile$Source)

但现在它不再使用 for 循环了。

我只是想知道是否有人可以给我任何建议,告诉我如何进行循环以使用表 2 中的唯一来源创建另外 2 列?

下面是我使用的典型的每周“个人资料”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

最佳答案

为了实现从 profiledates 的数据传输,您必须转换 profile 数据,然后加入它与 日期。对于以下步骤,我使用了 data.table 包。

1) 加载 data.table 包并将数据集转换为 data.tables(增强型数据帧):

library(data.table)

setDT(profile)
setDT(dates)

2) 重新格式化profile 数据集中的值:

# 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,其中一个(或两个)Sourceon 每小时的值。我只在 Spring 和冬季这样做,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。那是因为我假设 Stack 在 23:00 关闭。通过使用 -1 我包括了第 22 个小时而不是第 23 个小时。如果需要,您可以更改此设置。

4) 将第 3 步中的数据集绑定(bind)在一起,并为 dcast 操作准备生成的数据集:

prof <- rbindlist(list(pr.spring,pr.winter))
prof <- prof[, .(weekday = Day, season, Source = gsub(" ",".",Source), hour = sprintf("%02d",hour))]

5) 将第 4 步中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。后面步骤中的连接操作需要后者,因为 dates 数据集中的 weekday 列也是一个字符列:

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

6) 将两个数据集连接在一起并用 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)]

生成的数据集现在具有 dates 数据集中每个日期的堆栈列,其中 1 ="on"0 = "off".


结果数据集的快照:

> 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

关于r - 遍历列的唯一值并创建多个列,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/32584744/

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