在没有 for 循环的情况下计算随后几天之间的差异

Question

我有以下数据框：

time <- c("2004-01-01 01:30:00","2004-01-01 04:30:00","2004-01-01 07:30:00",
          "2004-01-01 10:30:00","2004-01-01 13:30:00","2004-01-01 16:30:00",
          "2004-01-01 19:30:00","2004-01-01 22:30:00","2004-01-02 01:30:00",
          "2004-01-02 04:30:00","2004-01-02 07:30:00","2004-01-02 10:30:00",
          "2004-01-02 13:30:00","2004-01-02 16:30:00","2004-01-02 19:30:00",
          "2004-01-02 22:30:00","2004-01-03 01:30:00","2004-01-03 04:30:00",
          "2004-01-03 07:30:00","2004-01-03 10:30:00")
d <- c(0.00, 0.00,152808.30, 739872.84, 82641.22, 83031.04, 83031.04, 82641.22, 0.00, 
       0.00, 267024.71,1247414.7, 151638.85, 151249.03, 151249.03, 152028.67, 0.00, 0.00, 
       296650.81,1355783.85)
dat <- data.frame(time = time, dat = d)

显示来自预测模型 3 天的太阳辐射累积（每天）。

要将太阳辐射的单位从 J/m2 转换为 W/m2，我需要计算每天不同预测时间之间的差异，然后除以 10800（预测时间）。这是我的尝试：

itime <- as.numeric(as.Date(dat$time))
utime <- unique(itime)
l <- list()
for(i in 1:length(utime)){
  idx <- itime == utime[i]
  dat2 <- dat[idx,]
  dat3 <- dat2[1,2]/10800
  for(ii in 2:nrow(dat2)){
    dat3[ii] <- (abs(dat2[ii,2] - dat2[ii-1,2]))/10800
  }
  df <- data.frame(dateTime = dat2$time,
                   dd = dat3)
  l[[i]] <- df
}
df1 <- do.call(rbind.data.frame, l)
df1[,1] <- as.POSIXct(df1[,1])

按预期执行。但是，我打算使用此代码的实际数据长度超过 100 天。因此，运行循环并不是最优的。

还有其他方法可以代替循环吗？

我试过了：

dat2 <- c(dat[1,2]/10800,rev(abs(diff(rev(dat[,2])))/10800))
df2 <- data.frame(time = as.POSIXct(dat[,1]), dd = dat2)

它给出了几乎相同的答案（与循环一样），但它还计算不同日期的时间步长之间的差异，而不是将计算隔离到各个日期。

plot(df1, type = 'l')
lines(df2, col = 'red')

如您所见，凌晨存在不匹配。

谁能推荐另一种方法？

Answer 1

对于您的列表l，您可以通过

获得相同的结果

dat <- data.frame(
time = c("2004-01-01 01:30:00","2004-01-01 04:30:00","2004-01-01 07:30:00",
          "2004-01-01 10:30:00","2004-01-01 13:30:00","2004-01-01 16:30:00",
          "2004-01-01 19:30:00","2004-01-01 22:30:00","2004-01-02 01:30:00",
          "2004-01-02 04:30:00","2004-01-02 07:30:00","2004-01-02 10:30:00",
          "2004-01-02 13:30:00","2004-01-02 16:30:00","2004-01-02 19:30:00",
          "2004-01-02 22:30:00","2004-01-03 01:30:00","2004-01-03 04:30:00",
          "2004-01-03 07:30:00","2004-01-03 10:30:00"),
dat = c(0.00, 0.00,152808.30, 739872.84, 82641.22, 83031.04, 83031.04, 82641.22, 0.00, 
       0.00, 267024.71,1247414.7, 151638.85, 151249.03, 151249.03, 152028.67, 0.00, 0.00, 
       296650.81,1355783.85)
)

dat$itime <- as.numeric(as.Date(dat$time))
utime <- unique(dat$itime)

daydat <- function(u) { 
  dat2 <- dat[dat$itime==u,]
  data.frame(dateTime = dat2$time, dd = c(dat2$dat[1], abs(diff(dat2$dat)))/10800)
}
l <- lapply(utime, daydat)

这是一个带有split()的版本：

dat$itime <- as.numeric(as.Date(dat$time))

daydat <- function(d) data.frame(dateTime = d$time, dd = c(d$dat[1], abs(diff(d$dat)))/10800)

L <- split(dat, dat$itime)
l <- lapply(L, daydat)

或者不创建dat$itime:

daydat <- function(d) data.frame(dateTime = d$time, dd = c(d$dat[1], abs(diff(d$dat)))/10800)
l <- lapply(split(dat, as.Date(dat$time)), FUN=daydat)

或使用by()

l2 <- unclass(by(dat, as.Date(dat$time), FUN=daydat))

如果你想在原始数据框中得到结果，你可以使用ave()

dat$dd <- ave(dat$dat, as.Date(dat$time), FUN=function(x) c(x[1], abs(diff(x)))/10800)

Answer 2

使用可以使用lag() from dplyr 和group_by()

library(dplyr)
df <- dat %>%
    mutate(date = as.Date(time)) %>%
    group_by(date) %>%
    mutate(before.dat = lag(dat, order_by=date)) %>%
    mutate(diff = abs(dat - before.dat)/10800) %>%
    select(time, date, dat, before.dat, diff)
df
#Source: local data frame [20 x 5]
#Groups: date [3]
#                time        date        dat before.dat          diff
#                <fctr>     <date>      <dbl>      <dbl>         <dbl>
#1  2004-01-01 01:30:00 2004-01-01       0.00         NA            NA
#2  2004-01-01 04:30:00 2004-01-01       0.00       0.00    0.00000000
#3  2004-01-01 07:30:00 2004-01-01  152808.30       0.00   14.14891667
#4  2004-01-01 10:30:00 2004-01-01  739872.84  152808.30   54.35782778
#5  2004-01-01 13:30:00 2004-01-01   82641.22  739872.84   60.85477963
#6  2004-01-01 16:30:00 2004-01-01   83031.04   82641.22    0.03609444
#7  2004-01-01 19:30:00 2004-01-01   83031.04   83031.04    0.00000000
#8  2004-01-01 22:30:00 2004-01-01   82641.22   83031.04    0.03609444
#9  2004-01-02 01:30:00 2004-01-02       0.00         NA            NA
#10 2004-01-02 04:30:00 2004-01-02       0.00       0.00    0.00000000
#11 2004-01-02 07:30:00 2004-01-02  267024.71       0.00   24.72451019
#12 2004-01-02 10:30:00 2004-01-02 1247414.70  267024.71   90.77685093
#13 2004-01-02 13:30:00 2004-01-02  151638.85 1247414.70  101.46072685
#14 2004-01-02 16:30:00 2004-01-02  151249.03  151638.85    0.03609444
#15 2004-01-02 19:30:00 2004-01-02  151249.03  151249.03    0.00000000
#16 2004-01-02 22:30:00 2004-01-02  152028.67  151249.03    0.07218889
#17 2004-01-03 01:30:00 2004-01-03       0.00         NA            NA
#18 2004-01-03 04:30:00 2004-01-03       0.00       0.00    0.00000000
#19 2004-01-03 07:30:00 2004-01-03  296650.81       0.00   27.46766759
#20 2004-01-03 10:30:00 2004-01-03 1355783.85  296650.81   98.06787407

基于 GGamba 评论的简化代码

dat %>%
    mutate(time = as.Date(time)) %>%
    group_by(time) %>%
    mutate(diff = (dat-lag(dat)) / 10800)