r - 基于多个条件创建输出

Question

我有 2 个数据帧用于 2 个堆栈，提供有关潜在排放的信息。一个数据框给出了系统打开和关闭 4 个季节的时间框架。每个季节都在特定日期开始。第二个文件给了我堆栈的详细信息。

我正在尝试使用一些示例文件来测试如何执行此操作，到目前为止，我已经设法在堆栈溢出示例之后创建了一个函数，该函数允许我创建一个包含我想要的日期的数据框和一个包含每个日期的季节的列.我现在真的很纠结于编程概念，以了解如何组合 3 个数据框来创建我尝试设置的输出模板。

为了向您展示一个示例，我的示例输入是:

堆栈信息文件:



显示系统何时开启或关闭的季节性配置文件示例:



并且我所追求的输出应该以以下格式为每年创建数据框(只有黑色字体和红色文本仅用于解释值是什么):



我发现最困难的是，我每年的输出文件将有一个唯一的第一行，而第二行将针对每种污染物重复。从第 3 行开始，所有 8760 小时的每小时数据。这需要对下一个污染物重复。

到目前为止，我已经成功地创建了一个函数来帮助我为一年中的每一天分配季节。例如:

#function to create seasons
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

## create date data frame and assign seasons
dates = data.frame(dates =seq(as.Date('2010-01-01'),as.Date('2012-12-31'),by = 1))

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

这给了我一个日期数据框，我的其他 2 个示例数据框是(如图所示):

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)) -> profile

structure(list(SNAME = structure(1:2, .Label = c("Stack 1", "Stack 2"
), class = "factor"), ISVARY = c(1L, 4L), VELVOL = c(1L, 4L), 
    TEMPDENS = c(0L, 2L), `DUM 1` = c(999L, 999L), `DUM 2` = c(999L, 
    999L), NPOL = c(2L, 2L), `EXIT VEL` = c(26.2, 22.4), TEMP = c(341L, 
    328L), `STACK DIAM` = c(1.5, 2.5), W = c(0L, 15L), Nox = c(39, 
    33.3), Sox = c(15.5, 17.9)), .Names = c("SNAME", "ISVARY", 
"VELVOL", "TEMPDENS", "DUM 1", "DUM 2", "NPOL", "EXIT VEL", "TEMP", 
"STACK DIAM", "W", "Nox", "Sox"), class = "data.frame", row.names = c(NA, 
-2L)) -> stack_info

如果有人能给我关于如何继续编程部分的任何指导，那将非常有用，因为我不确定如何处理它以创建单独的输出文件作为 2010、2011 和 2012 年的数据框。

Answer 1

数据的组织方式并不适合处理。也许你看看 Hadley Wickhams papar 关于 tidy data .

根据您所需的输出，您需要一个数据帧，其行数等于特定机器(堆栈 n)打开的小时数。因此，我建议您创建一个包含给定年份每小时的数据框:

d.out = data.frame(dates = seq(from=as.POSIXct("2010-01-01"), by=3600, to= as.POSIXct("2010-12-31")))
d.out$year = as.numeric(format(d.out$dates, "%Y"))
d.out$month = as.numeric(format(d.out$dates, "%m"))
d.out$day = as.numeric(format(d.out$dates, "%d"))
d.out$hour = as.numeric(format(d.out$dates, "%H"))
d.out$weekday = as.character(format(d.out$dates, "%a"))
d.out$doj = as.numeric(format(d.out$dates, "%j"))
d.out$season = "Winter"
d.out$season[d.out$doj >= 75 & d.out$doj < 121] = "Spring"
d.out$season[d.out$doj >= 121 & d.out$doj < 271] = "Summer"
d.out$season[d.out$doj >= 271 & d.out$doj < 312] = "Autumn"

目标是将此数据框与您的个人资料数据框连接起来。在加入之前，profile-df 必须重新排列:

library(dplyr)
library(tidyr)

profile_new =
profile %>%
    gather(season, hour, -c(`Source no`, Source, Period, Day)) %>%
    extract(season, c("season", "status"), "(\\w+?)\\s(\\w+)") %>%
    filter(hour != "off") %>%
    mutate(Day = as.character(Day), hour=as.numeric(hour)) %>%
    spread(status, hour)

现在很容易将三个数据框连接起来，将创建输出所需的所有信息放在一起:

d.out %>%
    inner_join(profile_new, by=c("weekday"="Day", "season"="season")) %>%
    group_by(Source, dates, year, day, weekday, season, hour) %>%
    summarise(status = any(hour >= On & hour <= Off)) %>%
    inner_join(stack_info, by=c("Source"="SNAME")) %>%
    mutate(Nox = ifelse(status, Nox, 0),
             Sox = ifelse(status, Sox, 0)) %>%
    arrange(Source, year, dates, hour) %>%
    select(Source, year, day, weekday, season, hour, `EXIT VEL`, TEMP, `STACK DIAM`, W, Nox, Sox)

显然，这不是您发布的格式。从这里您可以将数据帧写入 csv(使用 append = TRUE 逐个堆栈)。