假设在 RI 中有以下向量：
[1 2 3 10 20 30]
如何执行一个操作，在每个索引处将 3 个连续元素相加，得到以下向量：
[6 15 33 60]
其中第一个元素 = 1+2+3，第二个元素 = 2+3+10 等等...？谢谢
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29831 次
6 回答
32
你所拥有的是一个向量，而不是一个数组。您可以使用rollapplyzoo 包中的函数来获得所需的内容。
> x <- c(1, 2, 3, 10, 20, 30)
> #library(zoo)
> rollapply(x, 3, sum)
[1] 6 15 33 60
查看?rollapply有关rollapply功能和使用方法的更多详细信息。
于 2013-10-05T17:50:57.307 回答
22
我整理了一个包来处理这些类型的“滚动”函数，它提供类似于zoo's 的功能rollapply，但在后端使用 Rcpp。查看CRAN 上的RcppRoll。
library(microbenchmark)
library(zoo)
library(RcppRoll)
x <- rnorm(1E5)
all.equal( m1 <- rollapply(x, 3, sum), m2 <- roll_sum(x, 3) )
## from flodel
rsum.cumsum <- function(x, n = 3L) {
tail(cumsum(x) - cumsum(c(rep(0, n), head(x, -n))), -n + 1)
}
microbenchmark(
unit="ms",
times=10,
rollapply(x, 3, sum),
roll_sum(x, 3),
rsum.cumsum(x, 3)
)
给我
Unit: milliseconds
expr min lq median uq max neval
rollapply(x, 3, sum) 1056.646058 1068.867550 1076.550463 1113.71012 1131.230825 10
roll_sum(x, 3) 0.405992 0.442928 0.457642 0.51770 0.574455 10
rsum.cumsum(x, 3) 2.610119 2.821823 6.469593 11.33624 53.798711 10
如果速度是一个问题，您可能会发现它很有用。
于 2013-10-05T18:47:54.543 回答
17
如果速度是一个问题，您可以使用卷积过滤器并切断末端：
rsum.filter <- function(x, n = 3L) filter(x, rep(1, n))[-c(1, length(x))]
甚至更快，将其写为两个累积和之间的差：
rsum.cumsum <- function(x, n = 3L) tail(cumsum(x) - cumsum(c(rep(0, n), head(x, -n))), -n + 1)
两者都只使用基本功能。一些基准：
x <- sample(1:1000)
rsum.rollapply <- function(x, n = 3L) rollapply(x, n, sum)
rsum.sapply <- function(x, n = 3L) sapply(1:(length(x)-n+1),function(i){
sum(x[i:(i+n-1)])})
library(microbenchmark)
microbenchmark(
rsum.rollapply(x),
rsum.sapply(x),
rsum.filter(x),
rsum.cumsum(x)
)
# Unit: microseconds
# expr min lq median uq max neval
# rsum.rollapply(x) 12891.315 13267.103 14635.002 17081.5860 28059.998 100
# rsum.sapply(x) 4287.533 4433.180 4547.126 5148.0205 12967.866 100
# rsum.filter(x) 170.165 208.661 269.648 290.2465 427.250 100
# rsum.cumsum(x) 97.539 130.289 142.889 159.3055 449.237 100
x此外，我想如果所有应用的权重都是整数而不是数字，所有方法都会更快。
于 2013-10-05T18:30:00.553 回答
13
仅使用基础 R 您可以执行以下操作：
v <- c(1, 2, 3, 10, 20, 30)
grp <- 3
res <- sapply(1:(length(v)-grp+1),function(x){sum(v[x:(x+grp-1)])})
> res
[1] 6 15 33 60
另一种比 sapply 更快的方法（与@flodel's 相当rsum.cumsum）如下：
res <- rowSums(outer(1:(length(v)-grp+1),1:grp,FUN=function(i,j){v[(j - 1) + i]}))
这是更新的 florel 基准：
x <- sample(1:1000)
rsum.rollapply <- function(x, n = 3L) rollapply(x, n, sum)
rsum.sapply <- function(x, n = 3L) sapply(1:(length(x)-n+1),function(i){sum(x[i:(i+n-1)])})
rsum.filter <- function(x, n = 3L) filter(x, rep(1, n))[-c(1, length(x))]
rsum.cumsum <- function(x, n = 3L) tail(cumsum(x) - cumsum(c(rep(0, n), head(x, -n))), -n + 1)
rsum.outer <- function(x, n = 3L) rowSums(outer(1:(length(x)-n+1),1:n,FUN=function(i,j){x[(j - 1) + i]}))
library(microbenchmark)
microbenchmark(
rsum.rollapply(x),
rsum.sapply(x),
rsum.filter(x),
rsum.cumsum(x),
rsum.outer(x)
)
# Unit: microseconds
# expr min lq median uq max neval
# rsum.rollapply(x) 9464.495 9929.4480 10223.2040 10752.7960 11808.779 100
# rsum.sapply(x) 3013.394 3251.1510 3466.9875 4031.6195 7029.333 100
# rsum.filter(x) 161.278 178.7185 229.7575 242.2375 359.676 100
# rsum.cumsum(x) 65.280 70.0800 88.1600 95.1995 181.758 100
# rsum.outer(x) 66.880 73.7600 82.8795 87.0400 131.519 100
于 2013-10-05T17:58:57.980 回答
1
如果您需要真正的速度，请尝试
rsum.cumdiff <- function(x, n = 3L) (cs <- cumsum(x))[-(1:(n-1))] - c(0,cs[1:(length(x)-n)])
一切都在 base R 中，更新 flodel 的微基准测试不言而喻
x <- sample(1:1000)
rsum.rollapply <- function(x, n = 3L) rollapply(x, n, sum)
rsum.sapply <- function(x, n = 3L) sapply(1:(length(x)-n+1),function(i){sum(x[i:(i+n-1)])})
rsum.filter <- function(x, n = 3L) filter(x, rep(1, n))[-c(1, length(x))]
rsum.cumsum <- function(x, n = 3L) tail(cumsum(x) - cumsum(c(rep(0, n), head(x, -n))), -n + 1)
rsum.outer <- function(x, n = 3L) rowSums(outer(1:(length(x)-n+1),1:n,FUN=function(i,j){x[(j - 1) + i]}))
rsum.cumdiff <- function(x, n = 3L) (cs <- cumsum(x))[-(1:(n-1))] - c(0, cs[1:(length(x)-n)])
all.equal(rsum.rollapply(x), rsum.sapply(x))
# [1] TRUE
all.equal(rsum.sapply(x), rsum.filter(x))
# [1] TRUE
all.equal(rsum.filter(x), rsum.outer(x))
# [1] TRUE
all.equal(rsum.outer(x), rsum.cumsum(x))
# [1] TRUE
all.equal(rsum.cumsum(x), rsum.cumdiff(x))
# [1] TRUE
library(microbenchmark)
microbenchmark(
rsum.rollapply(x),
rsum.sapply(x),
rsum.filter(x),
rsum.cumsum(x),
rsum.outer(x),
rsum.cumdiff(x)
)
# Unit: microseconds
# expr min lq mean median uq max neval
# rsum.rollapply(x) 3369.211 4104.2415 4630.89799 4391.7560 4767.2710 12002.904 100
# rsum.sapply(x) 850.425 999.2730 1355.56383 1086.0610 1246.5450 6915.877 100
# rsum.filter(x) 48.970 67.1525 97.28568 96.2430 113.6975 248.728 100
# rsum.cumsum(x) 47.515 62.7885 89.12085 82.1825 106.6675 230.303 100
# rsum.outer(x) 69.819 85.3340 160.30133 92.6070 109.0920 5740.119 100
# rsum.cumdiff(x) 9.698 12.6070 70.01785 14.3040 17.4555 5346.423 100
## R version 3.5.1 "Feather Spray"
## zoo and microbenchmark compiled under R 3.5.3
奇怪的是，第二次通过微基准测试一切都更快了：
microbenchmark(
rsum.rollapply(x),
rsum.sapply(x),
rsum.filter(x),
rsum.cumsum(x),
rsum.outer(x),
rsum.cumdiff(x)
)
# Unit: microseconds
# expr min lq mean median uq max neval
# rsum.rollapply(x) 3127.272 3477.5750 3869.38566 3593.4540 3858.9080 7836.603 100
# rsum.sapply(x) 844.122 914.4245 1059.89841 965.3335 1032.2425 5184.968 100
# rsum.filter(x) 47.031 60.8490 80.53420 74.1830 90.9100 260.365 100
# rsum.cumsum(x) 45.092 55.2740 69.90630 64.4855 81.4555 122.668 100
# rsum.outer(x) 68.850 76.6070 88.49533 82.1825 91.8800 166.304 100
# rsum.cumdiff(x) 9.213 11.1520 13.18387 12.1225 13.5770 49.456 100
于 2020-02-03T17:35:16.897 回答
1
runner也可以使用库
x <- c(1, 2, 3, 10, 20, 30)
runner::sum_run(x, k=3, na_pad = T)
#> [1] NA NA 6 15 33 60
或者slider也有用
x <- c(1, 2, 3, 10, 20, 30)
slider::slide_sum(x, before = 2, complete = T)
#> [1] NA NA 6 15 33 60
由reprex 包（v2.0.0）于 2021-06-14 创建
于 2021-06-14T12:01:40.083 回答