r 替換低於矩陣中每一列均值的值

Question

我有一個巨大的matrix ，我想在其中用NA替換低於每列均值（或中值）的值。 例如，使用這個matrix ：

set.seed(1)
ex <- matrix(data = round(runif(12), 1), nrow = 4, ncol = 3)
ex
     [,1] [,2] [,3]
[1,]  0.3  0.2  0.6
[2,]  0.4  0.9  0.1
[3,]  0.6  0.9  0.2
[4,]  0.9  0.7  0.2

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

上面的代碼使用 for 循環，我想要一個更快的矢量化版本。

Answer 1

該解決方案與建議@Ronak Shah和@ThomasIsCoding比較microbenchmark上一個更大的matrix提供了以下結果：

# 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)

它們給出相同的結果，但sweep方法似乎是最快的。

Answer 2

我們可以使用sweep 。

ex[sweep(ex, 2, colMeans(ex), `<`)] <- 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

或者有一些移調

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

---

由於它是一個矩陣，我們也可以使用apply columnwise

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

Answer 3

這是另一個基本的 R 解決方案

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