0

如果之前已经解决过这个问题,请提前道歉,但我已经尝试查看与 ddply、sapply 和 apply 相关的所有问题,但我终生无法解决这个问题......

我编写了一个函数 countMonths,它将计费周期中的日、月和总天数作为参数,并返回计费周期所属的日历月数:

countMonths <- function(day, month, cycle.days) {
  month.days <- c(31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31)
  if (month < 1 | month > 12 | floor(month) != month) {
    cat("Invalid month value, must be an integer from 1 to 12")
  } else if (day < 1 | day > month.days[month]) {
    cat("Invalid day value, must be between 1 and month.days[month]")
  } else if (cycle.days < 0) {
    cat("Invalid cycle.days value, must be >= 0")
  } else {
    nmonths <- 1
    day.ct <- cycle.days - day
    while (day.ct > 0) {
      nmonths <- nmonths + 1
      month <- ifelse(month == 1, 12, month - 1) # sets to previous month    
      day.ct  <- day.ct - month.days[month] # subtracts days of previous month
    }
    nmonths
  }
}

我想将此功能应用于包含客户计费记录的data.frame中的每一行,例如

> head(cons2[-1],10)
   kwh cycle.days  read.date row.index year month day kwh.per.day
1  381         29 2010-09-02         1 2010     9   2   13.137931
2  280         32 2010-10-04         2 2010    10   4    8.750000
3  282         29 2010-11-02         3 2010    11   2    9.724138
4  330         34 2010-12-06         4 2010    12   6    9.705882
5  371         30 2011-01-05         5 2011     1   5   12.366667
6  405         30 2011-02-04         6 2011     2   4   13.500000
7  441         32 2011-03-08         7 2011     3   8   13.781250
8  290         29 2011-04-06         8 2011     4   6   10.000000
9  296         29 2011-05-05         9 2011     5   5   10.206897
10 378         32 2011-06-06        10 2011     6   6   11.812500

> dput(head(cons2[-1],10))
structure(list(kwh = c(381L, 280L, 282L, 330L, 371L, 405L, 441L, 
290L, 296L, 378L), cycle.days = c(29L, 32L, 29L, 34L, 30L, 30L, 
32L, 29L, 29L, 32L), read.date = structure(c(1283385600, 1286150400, 
1288656000, 1291593600, 1294185600, 1296777600, 1299542400, 1302048000, 
1304553600, 1307318400), class = c("POSIXct", "POSIXt"), tzone = "UTC"), 
    row.index = 1:10, year = c(2010, 2010, 2010, 2010, 2011, 
    2011, 2011, 2011, 2011, 2011), month = c(9, 10, 11, 12, 1, 
    2, 3, 4, 5, 6), day = c(2L, 4L, 2L, 6L, 5L, 4L, 8L, 6L, 5L, 
    6L), kwh.per.day = c(13.1379310344828, 8.75, 9.72413793103448, 
    9.70588235294118, 12.3666666666667, 13.5, 13.78125, 10, 10.2068965517241, 
    11.8125)), .Names = c("kwh", "cycle.days", "read.date", "row.index", 
"year", "month", "day", "kwh.per.day"), row.names = c(NA, 10L
), class = "data.frame")

我尝试了几个选项,但没有一个效果很好。具体来说,我需要能够将给定 var 的值作为标量(或长度为 1 的向量)传递给数据帧中的每一行,但它们总是作为向量传递:

> cons2$tot.months <- countMonths(cons2$day, cons2$month, cons2$cycle.days)  
Warning messages:
1: In if (month < 1 | month > 12 | floor(month) != month) { :
  the condition has length > 1 and only the first element will be used
2: In if (day < 1 | day > month.days[month]) { :
  the condition has length > 1 and only the first element will be used
3: In if (cycle.days < 0) { :
  the condition has length > 1 and only the first element will be used
4: In while (day.ct > 0) { :
  the condition has length > 1 and only the first element will be used
5: In while (day.ct > 0) { :
  the condition has length > 1 and only the first element will be used

我终于能够使用 ddply 获得正确的结果,将每一行视为自己的组,但这需要很长时间:

cons2 <- ddply(cons2, .(account, year, month, day), transform,
               tot.months = countMonths(day, month, cycle.days)
)

有没有更好的方法将此函数应用于我的数据框的每一行?或者,作为一个相关问题,如何将数据框中的变量作为标量参数(给定行的值)而不是数据框中该 var 的所有值的向量传递?如果有人能指出我的想法在概念上哪里出了问题,我将特别感激。

4

1 回答 1

1

要使该函数正常工作,您可以使用mapply它将连续地将您的函数应用于您传递给它的所有向量的每个元素。所以你可以这样做:

mapply(countMonths,cons2$day,cons2$month,cons2$cycle.days)

正如我在评论中提到的,有更简单的方法可以做到这一点。例如,我认为这会起作用:

cons2$read.date=as.Date(cons2$read.date)
monnb <- function(d){ lt <- as.POSIXlt(as.Date(d, origin="1900-01-01"));  lt$year*12 + lt$mon }
mondf <- function(d1, d2)  monnb(d2) - monnb(d1) 
mondf(cons2$read.date-cons2$cycle.days,cons2$read.date) + 1

另外,我注意到您正在尝试捕获您的功能不起作用的所有条件,这很棒!有一个非常方便的函数stopifnot可以用于此目的:

countMonths <- function(day, month, cycle.days) {
  month.days <- c(31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31)
  stopifnot(month >=1 & month <= 12 & floor(month)==month & cycle.days >=0 & day >= 1 & day <= month.days[month]) 
  nmonths <- 1
  day.ct <- cycle.days - day
  while (day.ct > 0) {
    nmonths <- nmonths + 1
    month <- ifelse(month == 1, 12, month - 1) # sets to previous month    
    day.ct  <- day.ct - month.days[month] # subtracts days of previous month
  }
  nmonths
}

至于对您的函数的评论,我认为它有效,但它没有利用 R 中的向量运算。我从另一个答案中得到的函数非常巧妙,因为它允许您在一次,而不是依次循环遍历每一个。

于 2012-09-27T20:58:23.223 回答