计算 R data.table 中前 3 行的总和（通过网格方块）

Question

我想为每个方格计算过去三天的降雨量，并将其作为新列添加到我的 data.table 中。明确地说，我想总结当前和前两 (2) 天的降雨量，对于每个气象网格方块

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

我的回答尝试：这条线曾经有效，但不再有效

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))

如果您有任何见解或建议，我将不胜感激。

非常感谢！

Answer 1

这是一个使用最新 data.table 版本 (v 1.9.6+)

的快速高效的解决方案

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

这里的基本思想是 shift rain 列两次，然后对行求和。

Answer 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

注意不要加载 dplyr，因为它屏蔽了 filter。如果需要dplyr，可以显式调用stats::filter。

Answer 3

你自己也快找到答案了。 rollsum （或 rollapply 在您的情况下）为您提供长度为 N-2 的向量，因此您只需用 NA 填充所需的单元格。可以这样简单地完成：roll<-c(NA,NA,rollsum(yourvector,k=3))

这是我的做法。我正在使用 {RcppRoll} 包中的 roll_sum，因为它更快并且更容易处理 NA。来自 data.table 的简单 by 参数可让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

Answer 4

一个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，可以在管道中使用 maggritr 中的 %<>% 符号：

library(magrittr)
weather %<>%
  group_by......

Answer 5

rollapply 的解决方案是这样的：

weather[, rain_3 := rollapplyr(rain, 3, sum, fill = NA_real_), by = square]

给予：

    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

更新

根据最初提出此问题后出现的动物园版本进行了简化。

Answer 6

聚会晚了，但是更新版本的 data.table 软件包（对我来说是 1.12.8）具有 frollsum 功能，可以比以前更干净地完成此操作（但非常多有效）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

计算 R data.table 中前 3 行的总和（通过网格方块）

calculating sum of previous 3 rows in R data.table (by grid-square)

r

filter

zoo

data.table

更新