我试着用R来计算矩阵中一系列值的移动平均值。R中似乎没有一个内置函数可以让我计算移动平均线。有任何软件包提供这样的服务吗?还是需要我自己写?
当前回答
下面是一个简单的带有过滤器的函数,演示了一种方法来处理带有填充的开始和结束NAs,并使用自定义权重计算加权平均值(由过滤器支持):
wma <- function(x) {
wts <- c(seq(0.5, 4, 0.5), seq(3.5, 0.5, -0.5))
nside <- (length(wts)-1)/2
# pad x with begin and end values for filter to avoid NAs
xp <- c(rep(first(x), nside), x, rep(last(x), nside))
z <- stats::filter(xp, wts/sum(wts), sides = 2) %>% as.vector
z[(nside+1):(nside+length(x))]
}
其他回答
编辑:非常喜欢添加侧参数,例如,一个日期向量的过去7天的移动平均值(或总和,或…)。
对于那些只想自己计算的人来说,它无非是:
# x = vector with numeric data
# w = window length
y <- numeric(length = length(x))
for (i in seq_len(length(x))) {
ind <- c((i - floor(w / 2)):(i + floor(w / 2)))
ind <- ind[ind %in% seq_len(length(x))]
y[i] <- mean(x[ind])
}
y
但是让它独立于mean()会很有趣,所以你可以计算任何“移动”函数!
# our working horse:
moving_fn <- function(x, w, fun, ...) {
# x = vector with numeric data
# w = window length
# fun = function to apply
# side = side to take, (c)entre, (l)eft or (r)ight
# ... = parameters passed on to 'fun'
y <- numeric(length(x))
for (i in seq_len(length(x))) {
if (side %in% c("c", "centre", "center")) {
ind <- c((i - floor(w / 2)):(i + floor(w / 2)))
} else if (side %in% c("l", "left")) {
ind <- c((i - floor(w) + 1):i)
} else if (side %in% c("r", "right")) {
ind <- c(i:(i + floor(w) - 1))
} else {
stop("'side' must be one of 'centre', 'left', 'right'", call. = FALSE)
}
ind <- ind[ind %in% seq_len(length(x))]
y[i] <- fun(x[ind], ...)
}
y
}
# and now any variation you can think of!
moving_average <- function(x, w = 5, side = "centre", na.rm = FALSE) {
moving_fn(x = x, w = w, fun = mean, side = side, na.rm = na.rm)
}
moving_sum <- function(x, w = 5, side = "centre", na.rm = FALSE) {
moving_fn(x = x, w = w, fun = sum, side = side, na.rm = na.rm)
}
moving_maximum <- function(x, w = 5, side = "centre", na.rm = FALSE) {
moving_fn(x = x, w = w, fun = max, side = side, na.rm = na.rm)
}
moving_median <- function(x, w = 5, side = "centre", na.rm = FALSE) {
moving_fn(x = x, w = w, fun = median, side = side, na.rm = na.rm)
}
moving_Q1 <- function(x, w = 5, side = "centre", na.rm = FALSE) {
moving_fn(x = x, w = w, fun = quantile, side = side, na.rm = na.rm, 0.25)
}
moving_Q3 <- function(x, w = 5, side = "centre", na.rm = FALSE) {
moving_fn(x = x, w = w, fun = quantile, side = side, na.rm = na.rm, 0.75)
}
可以使用runner包来移动函数。在本例中是mean_run函数。cummean的问题是它不处理NA值,但mean_run可以。Runner包还支持不规则时间序列,Windows可以依赖于日期:
library(runner)
set.seed(11)
x1 <- rnorm(15)
x2 <- sample(c(rep(NA,5), rnorm(15)), 15, replace = TRUE)
date <- Sys.Date() + cumsum(sample(1:3, 15, replace = TRUE))
mean_run(x1)
#> [1] -0.5910311 -0.2822184 -0.6936633 -0.8609108 -0.4530308 -0.5332176
#> [7] -0.2679571 -0.1563477 -0.1440561 -0.2300625 -0.2844599 -0.2897842
#> [13] -0.3858234 -0.3765192 -0.4280809
mean_run(x2, na_rm = TRUE)
#> [1] -0.18760011 -0.09022066 -0.06543317 0.03906450 -0.12188853 -0.13873536
#> [7] -0.13873536 -0.14571604 -0.12596067 -0.11116961 -0.09881996 -0.08871569
#> [13] -0.05194292 -0.04699909 -0.05704202
mean_run(x2, na_rm = FALSE )
#> [1] -0.18760011 -0.09022066 -0.06543317 0.03906450 -0.12188853 -0.13873536
#> [7] NA NA NA NA NA NA
#> [13] NA NA NA
mean_run(x2, na_rm = TRUE, k = 4)
#> [1] -0.18760011 -0.09022066 -0.06543317 0.03906450 -0.10546063 -0.16299272
#> [7] -0.21203756 -0.39209010 -0.13274756 -0.05603811 -0.03894684 0.01103493
#> [13] 0.09609256 0.09738460 0.04740283
mean_run(x2, na_rm = TRUE, k = 4, idx = date)
#> [1] -0.187600111 -0.090220655 -0.004349696 0.168349653 -0.206571573 -0.494335093
#> [7] -0.222969541 -0.187600111 -0.087636571 0.009742884 0.009742884 0.012326968
#> [13] 0.182442234 0.125737145 0.059094786
还可以指定其他选项,如延迟和仅在特定索引处滚动。更多内容在包和函数文档中。
caTools包具有非常快速的滚动mean/min/max/sd和其他一些功能。我只使用过runmean和runsd,它们是迄今为止提到的其他包中最快的。
您可以使用RcppRoll来实现用c++编写的快速移动平均线。只需调用roll_mean函数。文档可以在这里找到。
否则,这个(较慢的)for循环应该可以做到:
ma <- function(arr, n=15){
res = arr
for(i in n:length(arr)){
res[i] = mean(arr[(i-n):i])
}
res
}
事实上,RcppRoll非常好。
cantdutchthis发布的代码必须在窗口的第四行进行修正:
ma <- function(arr, n=15){
res = arr
for(i in n:length(arr)){
res[i] = mean(arr[(i-n+1):i])
}
res
}
这里给出了另一种处理缺失的方法。
第三种方法,改进cantdutch这段代码来计算部分平均与否,如下:
ma <- function(x, n=2,parcial=TRUE){
res = x #set the first values
if (parcial==TRUE){
for(i in 1:length(x)){
t<-max(i-n+1,1)
res[i] = mean(x[t:i])
}
res
}else{
for(i in 1:length(x)){
t<-max(i-n+1,1)
res[i] = mean(x[t:i])
}
res[-c(seq(1,n-1,1))] #remove the n-1 first,i.e., res[c(-3,-4,...)]
}
}