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
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,...)]
}
}
其他回答
为了配合坎迪奇西斯和罗德里戈·雷麦黛奥的回答;
moving_fun <- function(x, w, FUN, ...) {
# x: a double vector
# w: the length of the window, i.e., the section of the vector selected to apply FUN
# FUN: a function that takes a vector and return a summarize value, e.g., mean, sum, etc.
# Given a double type vector apply a FUN over a moving window from left to the right,
# when a window boundary is not a legal section, i.e. lower_bound and i (upper bound)
# are not contained in the length of the vector, return a NA_real_
if (w < 1) {
stop("The length of the window 'w' must be greater than 0")
}
output <- x
for (i in 1:length(x)) {
# plus 1 because the index is inclusive with the upper_bound 'i'
lower_bound <- i - w + 1
if (lower_bound < 1) {
output[i] <- NA_real_
} else {
output[i] <- FUN(x[lower_bound:i, ...])
}
}
output
}
# example
v <- seq(1:10)
# compute a MA(2)
moving_fun(v, 2, mean)
# compute moving sum of two periods
moving_fun(v, 2, sum)
动物园包中的滚动平均值/最大值/中位数(rollmean) TTR中的移动平均线 马云在预测
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))]
}
你可以通过以下方法计算窗口宽度为k的向量x的移动平均值:
apply(embed(x, k), 1, mean)
vector_avg <- function(x){
sum_x = 0
for(i in 1:length(x)){
if(!is.na(x[i]))
sum_x = sum_x + x[i]
}
return(sum_x/length(x))
}