從2個向量創建矩陣，其矩陣值取決于向量元素-有解無憂

我有兩個向量

seqX <- c("seq1","seq2","seq3")
seqY <- c("seqA","seqB")

并希望創建一個seqX x seqY矩陣，其中每個單元格包含由用戶函式計算的值，該值依賴于特定單元格的每個seqX x seqY組合的值。例如像這樣的用戶函式

my.function.example <- function (seqX,seqY) {
   return(paste(seqX,"-",seqY))
}

應該產生如下所示的結果：

     [,1]      [,2]      [,3]
[1,] seq1-seqA seq2-seqA seq3-seqA
[2,] seq1-seqB seq2-seqB seq3-seqB

我迷失了tapply，sapply等，因此我尋求您的幫助。我的兩個問題是：

如何創建矩陣并應用依賴于兩個向量的函式？
如果 seqX <- seqY （這意味著所有組合將出現兩次，我將不得不冗余計算它們），我怎樣才能最有效地做到這一點

非常感謝您的幫助 - 提前感謝您！

親切的問候，馬丁

uj5u.com熱心網友回復：

你騙用outer.

t(outer(seqX, seqY, my.function.example))
#     [,1]          [,2]          [,3]         
#[1,] "seq1 - seqA" "seq2 - seqA" "seq3 - seqA"
#[2,] "seq1 - seqB" "seq2 - seqB" "seq3 - seqB"

t(outer(seqX, seqY, paste, sep="-"))
#     [,1]        [,2]        [,3]       
#[1,] "seq1-seqA" "seq2-seqA" "seq3-seqA"
#[2,] "seq1-seqB" "seq2-seqB" "seq3-seqB"

基準

bench::mark(check=FALSE,
         GKi = t(outer(seqX, seqY, my.function.example)),
         Shree = sapply(seqX, function(x) my.function.example(x, seqY)), #Adapted
         Peter = with(expand.grid(a=seqX, b=seqY), matrix(my.function.example(a, b), length(seqY), byrow=TRUE)) #Adapted
         )

結果

  expression      min  median itr/s…1 mem_a…2 gc/se…3 n_itr  n_gc total…? result
  <bch:expr> <bch:tm> <bch:t>   <dbl> <bch:b>   <dbl> <int> <dbl> <bch:t> <list>
1 GKi          7.47μs  11.1μs  73248.      0B    51.3  9993     7   136ms <NULL>
2 Shree       22.09μs    30μs  28128.  4.95KB    62.0  9978    22   355ms <NULL>
3 Peter        46.1μs    76μs  11531.      0B    70.4  4750    29   412ms <NULL>

GKi 比 Shree 快 2.5 倍，比 Peter 快 6 倍。

uj5u.com熱心網友回復：

seqX <- c("seq1","seq2","seq3")
seqY <- c("seqA","seqB")

xy <- expand.grid(x = seqX, y = seqY)

matrix(paste(xy$x, xy$y, sep = "-"), nrow = 2, byrow = TRUE)
#>      [,1]        [,2]        [,3]       
#> [1,] "seq1-seqA" "seq2-seqA" "seq3-seqA"
#> [2,] "seq1-seqB" "seq2-seqB" "seq3-seqB"


seqY <- seqX

xy <- expand.grid(x = seqX, y = seqY)

mx <- matrix(paste(xy$x, xy$y, sep = "-"), nrow = 3, byrow = TRUE)

mx[upper.tri(mx)] <- NA

mx

#>      [,1]        [,2]        [,3]       
#> [1,] "seq1-seq1" NA          NA         
#> [2,] "seq1-seq2" "seq2-seq2" NA         
#> [3,] "seq1-seq3" "seq2-seq3" "seq3-seq3"

^{使用reprex v2.0.2創建于 2022-10-16}

uj5u.com熱心網友回復：

這是一種方法sapply-

sapply(seqX, function(x) paste(x, "-", seqY))

     seq1          seq2          seq3         
[1,] "seq1 - seqA" "seq2 - seqA" "seq3 - seqA"
[2,] "seq1 - seqB" "seq2 - seqB" "seq3 - seqB"

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標籤：r

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