r替换低于矩阵中每列平均值的值

我有一个巨大的矩阵，我想用一个NA来替换低于每列平均值或中位数的值。 例如，使用此矩阵：

我想得到：

for(i in 1:ncol(ex)){
  ex[, i][ex[, i] < colMeans(ex)[i]] <- NA
}
ex
     [,1] [,2] [,3]
[1,]   NA   NA  0.6
[2,]   NA  0.9   NA
[3,]  0.6  0.9   NA
[4,]  0.9  0.7   NA

上面的代码使用for循环，我希望有一个更快的矢量化版本。

我们可以使用sweep

---

这是另一个基本的解决方案

ex <- replace(ex, ex < t(replicate(nrow(ex),colMeans(ex))),NA)

---

将@Ronak Shah和@ThomasIsCoding提出的解决方案与较大矩阵上的微基准进行比较，得出以下结果：

# Generate matrix
set.seed(1)
ex <- matrix(data = round(runif(100000), 1), nrow = 1000, ncol = 100)
ex
colMeans(ex)

# for-loop solution
ex2 <- ex
for(i in 1:ncol(ex2)){
  ex2[, i][ex2[, i] < colMeans(ex2)[i]] <- NA
}
ex2

# Solution with sweep
ex3 <- ex
ex3[sweep(ex3, 2, colMeans(ex3), "<")] <- NA
ex3

# Solution with replace
ex4 <- ex
ex4 <- replace(ex4, ex4 < t(replicate(nrow(ex4), colMeans(ex4))), NA)
ex4

# Transposing solution
ex5 <- ex
ex5[t(t(ex5) < colMeans(ex5))] <- NA
ex5

# Apply solution
ex6 <- ex
apply(ex6, 2, function(x) replace(x, x < mean(x), NA))
ex6

# Identical
all.equal(ex2, ex3, ex4, ex5, ex6)

# Microbenchmark
library(microbenchmark)
comp <- microbenchmark(
  for_loop = {
    ex2 <- ex
    for(i in 1:ncol(ex2)){
      ex2[, i][ex2[, i] < colMeans(ex2)[i]] <- NA
    }},

  sweep = {
    ex3 <- ex
    ex3[sweep(ex3, 2, colMeans(ex3), "<")] <- NA
  },

  replace = {
    ex4 <- ex
    ex4 <- replace(ex4, ex4 < t(replicate(nrow(ex4), colMeans(ex4))), NA)
  },

  transpose = {
    ex5 <- ex
    ex5[t(t(ex5) < colMeans(ex5))] <- NA
  },

  apply = {
    ex6 <- ex
    apply(ex6, 2, function(x) replace(x, x < mean(x), NA))
  }
)

library(ggplot2)
autoplot(comp)

它们给出了相同的结果，但扫描法似乎是最快的

ex[t(t(ex) < colMeans(ex))] <- NA

apply(ex, 2, function(x) replace(x, x < mean(x), NA))

ex <- replace(ex, ex < t(replicate(nrow(ex),colMeans(ex))),NA)

# Generate matrix
set.seed(1)
ex <- matrix(data = round(runif(100000), 1), nrow = 1000, ncol = 100)
ex
colMeans(ex)

# for-loop solution
ex2 <- ex
for(i in 1:ncol(ex2)){
  ex2[, i][ex2[, i] < colMeans(ex2)[i]] <- NA
}
ex2

# Solution with sweep
ex3 <- ex
ex3[sweep(ex3, 2, colMeans(ex3), "<")] <- NA
ex3

# Solution with replace
ex4 <- ex
ex4 <- replace(ex4, ex4 < t(replicate(nrow(ex4), colMeans(ex4))), NA)
ex4

# Transposing solution
ex5 <- ex
ex5[t(t(ex5) < colMeans(ex5))] <- NA
ex5

# Apply solution
ex6 <- ex
apply(ex6, 2, function(x) replace(x, x < mean(x), NA))
ex6

# Identical
all.equal(ex2, ex3, ex4, ex5, ex6)

# Microbenchmark
library(microbenchmark)
comp <- microbenchmark(
  for_loop = {
    ex2 <- ex
    for(i in 1:ncol(ex2)){
      ex2[, i][ex2[, i] < colMeans(ex2)[i]] <- NA
    }},

  sweep = {
    ex3 <- ex
    ex3[sweep(ex3, 2, colMeans(ex3), "<")] <- NA
  },

  replace = {
    ex4 <- ex
    ex4 <- replace(ex4, ex4 < t(replicate(nrow(ex4), colMeans(ex4))), NA)
  },

  transpose = {
    ex5 <- ex
    ex5[t(t(ex5) < colMeans(ex5))] <- NA
  },

  apply = {
    ex6 <- ex
    apply(ex6, 2, function(x) replace(x, x < mean(x), NA))
  }
)

library(ggplot2)
autoplot(comp)