我想计算每个网格方块过去三天的降雨量，并将其添加为我的data.table中的新列。为了清楚起见，我想总结一下目前和上一天的降雨量，每个气象网格广场
library ( zoo )
library (data.table)
# making the data.table
rain <- c(NA, NA, NA, 0, 0, 5, 1, 0, 3, 10) # rainfall values to work with
square <- c(1,1,1,1,1,1,1,1,1,2) # the geographic grid square for the rainfall measurement
desired_result <- c(NA, NA, NA, NA, NA, 5, 6, 6, 4, NA ) # this is the result I'm looking for (the last NA as we are now on to the first day of the second grid square)
weather <- data.table(rain, square, desired_result) # making the data.table
我尝试回答：这条线曾经有效，但不再有用
weather[, rain_3 := filter(rain, rep(1, 2), sides = 1), by = list(square)]
所以我在这里尝试另一种方法：
# this next line gets the numbers right, but sums the following values, not the preceeding ones.
weather$rain_3 <- rollapply(zoo(weather$rain), list(seq(-2,0)), sum)
# here I add in the by weather$ square, but still no success
weather$rain_3 <- rollapply(zoo(weather$rain), list(seq(-2,0)), sum, by= list(weather$square))
我非常感谢您提出的任何见解或建议。
非常感谢！
答案 0 :(得分：20)
这是使用最新data.table版本（v 1.9.6 +）的快速有效的解决方案
weather[, rain_3 := Reduce(`+`, shift(rain, 0:2)), by = square]
weather
# rain square desired_result rain_3
# 1: NA 1 NA NA
# 2: NA 1 NA NA
# 3: NA 1 NA NA
# 4: 0 1 NA NA
# 5: 0 1 NA NA
# 6: 5 1 5 5
# 7: 1 1 6 6
# 8: 0 1 6 6
# 9: 3 1 4 4
# 10: 10 2 NA NA
这里的基本想法是shift两次rain列，然后对行进行总结。
答案 1 :(得分：2)
weather[, rain_3 := filter(rain, rep(1, 3), sides = 1), by = list(square)]
#Error in filter(rain, rep(1, 3), sides = 1) :
# 'filter' is longer than time series
weather[, rain_3 := if(.N > 2) filter(rain, rep(1, 3), sides = 1) else NA_real_,
by = square]
# rain square desired_result rain_3
# 1: NA 1 NA NA
# 2: NA 1 NA NA
# 3: NA 1 NA NA
# 4: 0 1 NA NA
# 5: 0 1 NA NA
# 6: 5 1 5 5
# 7: 1 1 6 6
# 8: 0 1 6 6
# 9: 3 1 4 4
#10: 10 2 NA NA
请注意不加载dplyr，因为它会屏蔽filter。如果您需要dplyr，可以明确地致电stats::filter。
答案 2 :(得分：2)
你自己几乎得到了答案。 rollsum（或您的情况下为rollapply）为您提供长度为N-2的向量，因此您只需要用NA填充所需的单元格。它可以像这样简单地完成：roll<-c(NA,NA,rollsum(yourvector,k=3))
我是这样做的。我正在使用来自{RcppRoll}软件包的roll_sum，因为它更快，并且更容易处理NAs。 data.table中的简单by参数允许您按平方对结果进行分组。
library(RcppRoll)
weather[,rain_3:=if(.N>2){c(NA,NA,roll_sum(rain,n=3))}else{NA},by=square]
weather
rain square desired_result rain_3
1: NA 1 NA NA
2: NA 1 NA NA
3: NA 1 NA NA
4: 0 1 NA NA
5: 0 1 NA NA
6: 5 1 5 5
7: 1 1 6 6
8: 0 1 6 6
9: 3 1 4 4
10: 10 2 NA NA
答案 3 :(得分：2)
晚了聚会，但是data.table软件包的最新版本（对我来说是1.12.8）具有frollsum函数，它将比以前更干净地完成此操作（但非常有效）答案：
library (data.table)
# making the data.table
rain <- c(NA, NA, NA, 0, 0, 5, 1, 0, 3, 10) # rainfall values to work with
square <- c(1,1,1,1,1,1,1,1,1,2) # the geographic grid square for the rainfall measurement
desired_result <- c(NA, NA, NA, NA, NA, 5, 6, 6, 4, NA ) # this is the result I'm looking for (the last NA as we are now on to the first day of the second grid square)
weather <- data.table(rain, square, desired_result) # making the data.table
# using `frollsum`
weather[, rain3 := frollsum(rain, n = 3), by = square][]
#> rain square desired_result rain3
#> 1: NA 1 NA NA
#> 2: NA 1 NA NA
#> 3: NA 1 NA NA
#> 4: 0 1 NA NA
#> 5: 0 1 NA NA
#> 6: 5 1 5 5
#> 7: 1 1 6 6
#> 8: 0 1 6 6
#> 9: 3 1 4 4
#> 10: 10 2 NA NA
由reprex package（v0.3.0）于2020-07-09创建
答案 4 :(得分：0)
dplyr解决方案：
library(dplyr)
weather %>%
group_by(square) %>%
mutate(rain_3 = rain + lag(rain) + lag(rain, n = 2L))
结果：
Source: local data table [10 x 4]
rain square desired_result rain_3
(dbl) (dbl) (dbl) (dbl)
1 NA 1 NA NA
2 NA 1 NA NA
3 NA 1 NA NA
4 0 1 NA NA
5 0 1 NA NA
6 5 1 5 5
7 1 1 6 6
8 0 1 6 6
9 3 1 4 4
10 10 2 NA NA
如果要将rain3分配给数据集，可以使用管道中%<>%的{{1}}符号：
maggritr