在R程序中，想要使用特定值从向量值填充矩阵的特定列的各行。

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英文:

In R Program, want to fill a matrix from vector values across rows at a a specific column with specific value

问题

  1. 在R程序中，我有一个包含0和1的矩阵。见下面：

0 1 0 1 0 0
0 1 1 0 0 0
0 0 0 1 0 1
0 1 1 0 1 0
0 0 0 0 0 1

  2. 我想按行填充矩阵，使用列出的值（c("J" "J" "A" "A" "A" "A" "...一直到矩阵结束"）但从每行的第一个1开始。见下面：

0 J J A A A
0 J J A A A
0 0 0 J J A
0 J J A A A
0 0 0 0 0 J

  2. 截至目前，我已经创建了一个值列表和一个确定第一个1所在位置的函数。我不知道如何将其应用于获得我想要的矩阵。
  4. `pattern<- c("A","A","A","A","A")`
  5. `pattern <- c("J","J",  rep(pattern, length.out = ncol(Matrix)-2))`
  6. `indices<- apply(Matrix, 1, function(row) min(which(row == 1)))`
英文:

In R Program, I have a matrix containing 0 and 1's. See Below

  1. 0 1 0 1 0 0 
  2. 0 1 1 0 0 0 
  3. 0 0 0 1 0 1
  4. 0 1 1 0 1 0 
  5. 0 0 0 0 0 1

I want to fill the matrix byrow from values listed (c("J" "J" "A" "A" "A" "A" "... continue "A" until end of matrix") but begin at the first 1 in each row. See below:

  1. 0 J J A A A
  2. 0 J J A A A 
  3. 0 0 0 J J A
  4. 0 J J A A A
  5. 0 0 0 0 0 J

As of now, I have created a values list and a function to determine where the first 1 is. I'm lost on how to apply this to get the matrix I want.

pattern&lt;- c(&quot;A&quot;,&quot;A&quot;,&quot;A&quot;,&quot;A&quot;,&quot;A&quot;)
pattern &lt;- c(&quot;J&quot;,&quot;J&quot;, rep(pattern, length.out = ncol(Matrix)-2))
indices&lt;- apply(Matrix, 1, function(row) min(which(row == 1)))

答案1

得分: 0

获取零的第一部分，然后填充“J”，最后添加“A”。

  1. t(apply(mat, 1, function(x){
  2.   res <- which(x == 1)[1] - 1
  3.   res <- replace(res, is.na(res), length(x))
  4.   c(x[0:res], 
  5.     rep("J", min(c(2, (length(x) - res)))),
  6.     rep("A", max(c(0, (length(x) - res) - 2))))}))
  7.      [,1] [,2] [,3] [,4] [,5] [,6]
  8. [1,] "J"  "J"  "A"  "A"  "A"  "A" 
  9. [2,] "0"  "J"  "J"  "A"  "A"  "A" 
  10. [3,] "0"  "J"  "J"  "A"  "A"  "A" 
  11. [4,] "0"  "0"  "0"  "J"  "J"  "A" 
  12. [5,] "0"  "J"  "J"  "A"  "A"  "A" 
  13. [6,] "0"  "0"  "0"  "0"  "0"  "J" 
  14. [7,] "0"  "0"  "0"  "0"  "0"  "0"

数据

  1. mat <- structure(c(1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0, 0, 1, 
  2. 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 
  3. 1, 0, 1, 0), dim = 7:6)
英文:

Get the first part of zeros, then fill the "J"s, finally add the "A"s.

  1. t(apply(mat, 1, function(x){
  2.   res &lt;- which(x == 1)[1] - 1
  3.   res &lt;- replace(res, is.na(res), length(x))
  4.   c(x[0:res], 
  5.     rep(&quot;J&quot;, min(c(2, (length(x) - res)))),
  6.     rep(&quot;A&quot;, max(c(0, (length(x) - res) - 2))))}))
  7.      [,1] [,2] [,3] [,4] [,5] [,6]
  8. [1,] &quot;J&quot;  &quot;J&quot;  &quot;A&quot;  &quot;A&quot;  &quot;A&quot;  &quot;A&quot; 
  9. [2,] &quot;0&quot;  &quot;J&quot;  &quot;J&quot;  &quot;A&quot;  &quot;A&quot;  &quot;A&quot; 
  10. [3,] &quot;0&quot;  &quot;J&quot;  &quot;J&quot;  &quot;A&quot;  &quot;A&quot;  &quot;A&quot; 
  11. [4,] &quot;0&quot;  &quot;0&quot;  &quot;0&quot;  &quot;J&quot;  &quot;J&quot;  &quot;A&quot; 
  12. [5,] &quot;0&quot;  &quot;J&quot;  &quot;J&quot;  &quot;A&quot;  &quot;A&quot;  &quot;A&quot; 
  13. [6,] &quot;0&quot;  &quot;0&quot;  &quot;0&quot;  &quot;0&quot;  &quot;0&quot;  &quot;J&quot; 
  14. [7,] &quot;0&quot;  &quot;0&quot;  &quot;0&quot;  &quot;0&quot;  &quot;0&quot;  &quot;0&quot;

Data

  1. mat &lt;- structure(c(1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0, 0, 1, 
  2. 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 
  3. 1, 0, 1, 0), dim = 7:6)

答案2

得分: 0

  1. vec = c("J", "J", rep("A", ncol(m) - 2))
  2. m |>
  3.   apply(MARGIN = 1, \(x) {
  4.     x = cumsum(x) > 0
  5.     y = rep("0", length(x))
  6.     y[x] = vec[seq_len(sum(x))]
  7.     y
  8.   }) |>
  9.   t()
  10. #      [,1] [,2] [,3] [,4] [,5] [,6]
  11. # [1,] "0"  "J"  "J"  "A"  "A"  "A" 
  12. # [2,] "0"  "J"  "J"  "A"  "A"  "A" 
  13. # [3,] "0"  "0"  "0"  "J"  "J"  "A" 
  14. # [4,] "0"  "J"  "J"  "A"  "A"  "A" 
  15. # [5,] "0"  "0"  "0"  "0"  "0"  "J"

Using this sample data:

  1. m = read.table(text = '0 1 0 1 0 0 
  2. 0 1 1 0 0 0 
  3. 0 0 0 1 0 1
  4. 0 1 1 0 1 0 
  5. 0 0 0 0 0 1') |>
  6. as.matrix()
  2. <details>
  3. <summary>英文:</summary>

vec = c("J", "J", rep("A", ncol(m) - 2))
m |>
apply(MARGIN = 1, (x) {
x = cumsum(x) > 0
y = rep("0", length(x))
y[x] = vec[seq_len(sum(x))]
y
}) |>
t()

[,1] [,2] [,3] [,4] [,5] [,6]

[1,] "0" "J" "J" "A" "A" "A"

[2,] "0" "J" "J" "A" "A" "A"

[3,] "0" "0" "0" "J" "J" "A"

[4,] "0" "J" "J" "A" "A" "A"

[5,] "0" "0" "0" "0" "0" "J"

  2. ---
  4. Using this sample data:

m = read.table(text = '0 1 0 1 0 0
0 1 1 0 0 0
0 0 0 1 0 1
0 1 1 0 1 0
0 0 0 0 0 1') |> as.matrix()

  2. </details>
  6. # 答案3
  7. **得分**: 0
  9. n <- ncol(m)
  10. vec <- c("J", "J", rep("A", n - 2))
  11. t(apply(mat, 1, \(x)c(numeric(which(x>0)[1]-1), vec)[seq(n)]))
  13.    [,1] [,2] [,3] [,4] [,5] [,6]
  14. [1,] "0"  "J"  "J"  "A"  "A"  "A" 
  15. [2,] "0"  "J"  "J"  "A"  "A"  "A" 
  16. [3,] "0"  "0"  "0"  "J"  "J"  "A" 
  17. [4,] "0"  "J"  "J"  "A"  "A"  "A" 
  18. [5,] "0"  "0"  "0"  "0"  "0"  "J" 
  20. <details>
  21. <summary>英文:</summary>
  23. feels easier to just do:
  25.     n &lt;- ncol(m)
  26.     vec &lt;- c(&quot;J&quot;, &quot;J&quot;, rep(&quot;A&quot;, n - 2))
  27.     t(apply(mat, 1, \(x)c(numeric(which(x&gt;0)[1]-1), vec)[seq(n)]))
  28.     
  29.        [,1] [,2] [,3] [,4] [,5] [,6]
  30.     [1,] &quot;0&quot;  &quot;J&quot;  &quot;J&quot;  &quot;A&quot;  &quot;A&quot;  &quot;A&quot; 
  31.     [2,] &quot;0&quot;  &quot;J&quot;  &quot;J&quot;  &quot;A&quot;  &quot;A&quot;  &quot;A&quot; 
  32.     [3,] &quot;0&quot;  &quot;0&quot;  &quot;0&quot;  &quot;J&quot;  &quot;J&quot;  &quot;A&quot; 
  33.     [4,] &quot;0&quot;  &quot;J&quot;  &quot;J&quot;  &quot;A&quot;  &quot;A&quot;  &quot;A&quot; 
  34.     [5,] &quot;0&quot;  &quot;0&quot;  &quot;0&quot;  &quot;0&quot;  &quot;0&quot;  &quot;J&quot; 
  36. </details>
  40. # 答案4
  41. **得分**: 0
  43. 以下是已翻译的内容：
  45. A few vectorized options:
  47. ```R
  48. m <- max.col(cbind(mat, 1L), "f")
  49. m <- rbind(m - 1L, pmin(2L, ncol(mat) - m + 1L), pmax(0L, ncol(mat) - m - 1L))
  50. matrix(c("0", "J", "A")[rep.int(row(m), m)], nrow(mat), ncol(mat), 1)
  51. #>      [,1] [,2] [,3] [,4] [,5] [,6]
  52. #> [1,] "0"  "J"  "J"  "A"  "A"  "A" 
  53. #> [2,] "0"  "J"  "J"  "A"  "A"  "A" 
  54. #> [3,] "0"  "0"  "0"  "J"  "J"  "A" 
  55. #> [4,] "0"  "J"  "J"  "A"  "A"  "A" 
  56. #> [5,] "0"  "0"  "0"  "0"  "0"  "J"

或者

  1. m <- max.col(cbind(mat, 1L), "f")
  2. array(c("0", "J", "A")[(col(mat) >= m) + (col(mat) > m + 1L) + 1L], dim(mat))
  3. #>      [,1] [,2] [,3] [,4] [,5] [,6]
  4. #> [1,] "0"  "J"  "J"  "A"  "A"  "A" 
  5. #> [2,] "0"  "J"  "J"  "A"  "A"  "A" 
  6. #> [3,] "0"  "0"  "0"  "J"  "J"  "A" 
  7. #> [4,] "0"  "J"  "J"  "A"  "A"  "A" 
  8. #> [5,] "0"  "0"  "0"  "0"  "0"  "J"

或者

  1. matrix(
  2.   rep.int(c("0", "J", "A"), c(ncol(mat), 2L, ncol(mat) - 2L))[
  3.     sequence(rep(ncol(mat), nrow(mat)), ncol(mat) - max.col(cbind(mat, 1L), "f") + 2L)
  4.   ], nrow(mat), ncol(mat), 1
  5. )
  6. #>      [,1] [,2] [,3] [,4] [,5] [,6]
  7. #> [1,] "0"  "J"  "J"  "A"  "A"  "A" 
  8. #> [2,] "0"  "J"  "J"  "A"  "A"  "A" 
  9. #> [3,] "0"  "0"  "0"  "J"  "J"  "A" 
  10. #> [4,] "0"  "J"  "J"  "A"  "A"  "A" 
  11. #> [5,] "0"  "0"  "0"  "0"  "0"  "J"

Benchmarking shows the advantage of using a vectorized approach. Define various approaches as functions:

  1. f1 <- function(mat) {
  2.   m <- max.col(cbind(mat, 1L), "f")
  3.   m <- rbind(m - 1L, pmin(2L, ncol(mat) - m + 1L), pmax(0L, ncol(mat) - m - 1L))
  4.   matrix(c("0", "J", "A")[rep.int(row(m), m)], nrow(mat), ncol(mat), 1)
  5. }
  7. f2 <- function(mat) {
  8.   m <- max.col(cbind(mat, 1L), "f")
  9.   array(c("0", "J", "A")[(col(mat) >= m) + (col(mat) > m + 1L) + 1L], dim(mat))
  10. }
  12. f3 <- function(mat) {
  13.   d <- dim(mat)
  14.   matrix(
  15.     rep.int(c("0", "J", "A"), c(d[2], 2L, d[2] - 2L))[
  16.       sequence(rep(d[2], d[1]), d[2] - max.col(cbind(mat, 1L), "f") + 2L)
  17.     ], d[1], d[2], 1
  18.   )
  19. }
  21. Andre <- function(mat) {
  22.   # from Andre Wildberg
  23.   t(apply(mat, 1, function(x){
  24.     res <- which(x == 1)[1] - 1
  25.     res <- replace(res, is.na(res), length(x))
  26.     c(x[0:res], 
  27.       rep("J", min(c(2, (length(x) - res)))),
  28.       rep("A", max(c(0, (length(x) - res) - 2))))}))
  29. }

Benchmark on a large-ish matrix.

  1. mat <- matrix(sample(0:1, 1e5, 1, c(0.75, 0.25)), 1e4)
  3. microbenchmark::microbenchmark(
  4.   f1 = f1(mat),
  5.   f2 = f2(mat),
  6.   f3 = f3(mat),
  7.   Andre = Andre(mat),
  8.   check = "equal"
  9. )
  10. #> Unit: milliseconds
  11. #>   expr       min        lq      mean    median        uq        max neval
  12. #>     f1  1.391700  1.894001  2.127106  1.967701  2.097001   7.966101   100
  13. #>     f2  1.616000  2.240750  2.691387  2.361451  2.590051   7.142301   100
  14. #>     f3  1.118401  1.570600  1.745991  1.619800  1.739251   5.924802   100
  15. #>  Andre 68.022601 70.696101 73.181934 72.200000 73.931750 117.784401   100
英文:

A few vectorized options:

  1. m &lt;- max.col(cbind(mat, 1L), &quot;f&quot;)
  2. m &lt;- rbind(m - 1L, pmin(2L, ncol(mat) - m + 1L), pmax(0L, ncol(mat) - m - 1L))
  3. matrix(c(&quot;0&quot;, &quot;J&quot;, &quot;A&quot;)[rep.int(row(m), m)], nrow(mat), ncol(mat), 1)
  4. #&gt;      [,1] [,2] [,3] [,4] [,5] [,6]
  5. #&gt; [1,] &quot;0&quot;  &quot;J&quot;  &quot;J&quot;  &quot;A&quot;  &quot;A&quot;  &quot;A&quot; 
  6. #&gt; [2,] &quot;0&quot;  &quot;J&quot;  &quot;J&quot;  &quot;A&quot;  &quot;A&quot;  &quot;A&quot; 
  7. #&gt; [3,] &quot;0&quot;  &quot;0&quot;  &quot;0&quot;  &quot;J&quot;  &quot;J&quot;  &quot;A&quot; 
  8. #&gt; [4,] &quot;0&quot;  &quot;J&quot;  &quot;J&quot;  &quot;A&quot;  &quot;A&quot;  &quot;A&quot; 
  9. #&gt; [5,] &quot;0&quot;  &quot;0&quot;  &quot;0&quot;  &quot;0&quot;  &quot;0&quot;  &quot;J&quot;

Or

  1. m &lt;- max.col(cbind(mat, 1L), &quot;f&quot;)
  2. array(c(&quot;0&quot;, &quot;J&quot;, &quot;A&quot;)[(col(mat) &gt;= m) + (col(mat) &gt; m + 1L) + 1L], dim(mat))
  3. #&gt;      [,1] [,2] [,3] [,4] [,5] [,6]
  4. #&gt; [1,] &quot;0&quot;  &quot;J&quot;  &quot;J&quot;  &quot;A&quot;  &quot;A&quot;  &quot;A&quot; 
  5. #&gt; [2,] &quot;0&quot;  &quot;J&quot;  &quot;J&quot;  &quot;A&quot;  &quot;A&quot;  &quot;A&quot; 
  6. #&gt; [3,] &quot;0&quot;  &quot;0&quot;  &quot;0&quot;  &quot;J&quot;  &quot;J&quot;  &quot;A&quot; 
  7. #&gt; [4,] &quot;0&quot;  &quot;J&quot;  &quot;J&quot;  &quot;A&quot;  &quot;A&quot;  &quot;A&quot; 
  8. #&gt; [5,] &quot;0&quot;  &quot;0&quot;  &quot;0&quot;  &quot;0&quot;  &quot;0&quot;  &quot;J&quot;

Or

  1. matrix(
  2.   rep.int(c(&quot;0&quot;, &quot;J&quot;, &quot;A&quot;), c(ncol(mat), 2L, ncol(mat) - 2L))[
  3.     sequence(rep(ncol(mat), nrow(mat)), ncol(mat) - max.col(cbind(mat, 1L), &quot;f&quot;) + 2L)
  4.   ], nrow(mat), ncol(mat), 1
  5. )
  6. #&gt;      [,1] [,2] [,3] [,4] [,5] [,6]
  7. #&gt; [1,] &quot;0&quot;  &quot;J&quot;  &quot;J&quot;  &quot;A&quot;  &quot;A&quot;  &quot;A&quot; 
  8. #&gt; [2,] &quot;0&quot;  &quot;J&quot;  &quot;J&quot;  &quot;A&quot;  &quot;A&quot;  &quot;A&quot; 
  9. #&gt; [3,] &quot;0&quot;  &quot;0&quot;  &quot;0&quot;  &quot;J&quot;  &quot;J&quot;  &quot;A&quot; 
  10. #&gt; [4,] &quot;0&quot;  &quot;J&quot;  &quot;J&quot;  &quot;A&quot;  &quot;A&quot;  &quot;A&quot; 
  11. #&gt; [5,] &quot;0&quot;  &quot;0&quot;  &quot;0&quot;  &quot;0&quot;  &quot;0&quot;  &quot;J&quot;

Benchmarking shows the advantage of using a vectorized approach. Define various approaches as functions:

  1. f1 &lt;- function(mat) {
  2.   m &lt;- max.col(cbind(mat, 1L), &quot;f&quot;)
  3.   m &lt;- rbind(m - 1L, pmin(2L, ncol(mat) - m + 1L), pmax(0L, ncol(mat) - m - 1L))
  4.   matrix(c(&quot;0&quot;, &quot;J&quot;, &quot;A&quot;)[rep.int(row(m), m)], nrow(mat), ncol(mat), 1)
  5. }
  7. f2 &lt;- function(mat) {
  8.   m &lt;- max.col(cbind(mat, 1L), &quot;f&quot;)
  9.   array(c(&quot;0&quot;, &quot;J&quot;, &quot;A&quot;)[(col(mat) &gt;= m) + (col(mat) &gt; m + 1L) + 1L], dim(mat))
  10. }
  12. f3 &lt;- function(mat) {
  13.   d &lt;- dim(mat)
  14.   matrix(
  15.     rep.int(c(&quot;0&quot;, &quot;J&quot;, &quot;A&quot;), c(d[2], 2L, d[2] - 2L))[
  16.       sequence(rep(d[2], d[1]), d[2] - max.col(cbind(mat, 1L), &quot;f&quot;) + 2L)
  17.     ], d[1], d[2], 1
  18.   )
  19. }
  21. Andre &lt;- function(mat) {
  22.   # from Andre Wildberg
  23.   t(apply(mat, 1, function(x){
  24.     res &lt;- which(x == 1)[1] - 1
  25.     res &lt;- replace(res, is.na(res), length(x))
  26.     c(x[0:res], 
  27.       rep(&quot;J&quot;, min(c(2, (length(x) - res)))),
  28.       rep(&quot;A&quot;, max(c(0, (length(x) - res) - 2))))}))
  29. }

Benchmark on a large-ish matrix.

  1. mat &lt;- matrix(sample(0:1, 1e5, 1, c(0.75, 0.25)), 1e4)
  3. microbenchmark::microbenchmark(
  4.   f1 = f1(mat),
  5.   f2 = f2(mat),
  6.   f3 = f3(mat),
  7.   Andre = Andre(mat),
  8.   check = &quot;equal&quot;
  9. )
  10. #&gt; Unit: milliseconds
  11. #&gt;   expr       min        lq      mean    median        uq        max neval
  12. #&gt;     f1  1.391700  1.894001  2.127106  1.967701  2.097001   7.966101   100
  13. #&gt;     f2  1.616000  2.240750  2.691387  2.361451  2.590051   7.142301   100
  14. #&gt;     f3  1.118401  1.570600  1.745991  1.619800  1.739251   5.924802   100
  15. #&gt;  Andre 68.022601 70.696101 73.181934 72.200000 73.931750 117.784401   100

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  • 本文由 发表于 2023年6月8日 00:50:31
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