r - 计算向量中每 n 个值的平均值

Question

所以让我们说我有一个向量

a <- rnorm(6000)

我想计算第 1 个值到第 60 个值的平均值，然后再次计算第 61 个值到第 120 个值的平均值，以此类推。所以基本上我想计算每 60 个值的平均值，从该向量给我 100 个平均值。我知道我可以做一个 for 循环，但我想知道是否有更好的方法来做到这一点？

Answer 1

我会用

 colMeans(matrix(a, 60))
.colMeans(a, 60, length(a) / 60)  # more efficient (without reshaping to matrix)

用户增强 adunaic的请求

This only works if there are 60x100 data points. If you have an incomplete 60 at the end then this errors. It would be good to have a general solution for others looking at this problem for ideas.

BinMean <- function (vec, every, na.rm = FALSE) {
  n <- length(vec)
  x <- .colMeans(vec, every, n %/% every, na.rm)
  r <- n %% every
  if (r) x <- c(x, mean.default(vec[(n - r + 1):n], na.rm = na.rm))
  x
  }

a <- 1:103
BinMean(a, every = 10)
# [1]   5.5  15.5  25.5  35.5  45.5  55.5  65.5  75.5  85.5  95.5 102.0

分组操作的替代解决方案(效率较低)

BinMean2 <- function (vec, every, na.rm = FALSE) {
  grp <- as.integer(ceiling(seq_along(vec) / every))
  grp <- structure(grp, class = "factor",
                   levels = as.character(seq_len(grp[length(grp)])) )
  lst <- .Internal(split(vec, grp))
  unlist(lapply(lst, mean.default, na.rm = na.rm), use.names = FALSE)
  }

速度

library(microbenchmark)
a <- runif(1e+4)
microbenchmark(BinMean(a, 100), BinMean2(a, 100))
#Unit: microseconds
#             expr      min        lq       mean    median        uq       max
#  BinMean(a, 100)   40.400   42.1095   54.21286   48.3915   57.6555   205.702
# BinMean2(a, 100) 1216.823 1335.7920 1758.90267 1434.9090 1563.1535 21467.542