r - 向量化函数中 sapply() 的意外行为

Question

我想尝试一下模块化算法并编写了一些看似无辜的函数......但对以下意想不到的行为感到非常惊讶:

crt <- function(x, mods = c(5, 7)) {
  sapply(mods, \(y) x %% y)
}
crt <- Vectorize(crt)

crt(20)
##      [,1]
## [1,]    0
## [2,]    6

crt(55)
##      [,1]
## [1,]    0
## [2,]    6

crt(1:100)
##      [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13] [,14]
## [1,]    1    2    3    4    0    1    2    3    4     0     1     2     3     4
## [2,]    1    2    3    4    5    6    0    1    2     3     4     5     6     0
##      [,15] [,16] [,17] [,18] [,19] [,20] [,21] [,22] [,23] [,24] [,25] [,26]
## [1,]     0     1     2     3     4     0     1     2     3     4     0     1
## [2,]     1     2     3     4     5     6     0     1     2     3     4     5
##      [,27] [,28] [,29] [,30] [,31] [,32] [,33] [,34] [,35] [,36] [,37] [,38]
## [1,]     2     3     4     0     1     2     3     4     0     1     2     3
## [2,]     6     0     1     2     3     4     5     6     0     1     2     3
##      [,39] [,40] [,41] [,42] [,43] [,44] [,45] [,46] [,47] [,48] [,49] [,50]
## [1,]     4     0     1     2     3     4     0     1     2     3     4     0
## [2,]     4     5     6     0     1     2     3     4     5     6     0     1
##      [,51] [,52] [,53] [,54] [,55] [,56] [,57] [,58] [,59] [,60] [,61] [,62]
## [1,]     1     2     3     4     0     1     2     3     4     0     1     2
## [2,]     2     3     4     5     6     0     1     2     3     4     5     6
##      [,63] [,64] [,65] [,66] [,67] [,68] [,69] [,70] [,71] [,72] [,73] [,74]
## [1,]     3     4     0     1     2     3     4     0     1     2     3     4
## [2,]     0     1     2     3     4     5     6     0     1     2     3     4
##      [,75] [,76] [,77] [,78] [,79] [,80] [,81] [,82] [,83] [,84] [,85] [,86]
## [1,]     0     1     2     3     4     0     1     2     3     4     0     1
## [2,]     5     6     0     1     2     3     4     5     6     0     1     2
##      [,87] [,88] [,89] [,90] [,91] [,92] [,93] [,94] [,95] [,96] [,97] [,98]
## [1,]     2     3     4     0     1     2     3     4     0     1     2     3
## [2,]     3     4     5     6     0     1     2     3     4     5     6     0
##      [,99] [,100]
## [1,]     4      0
## [2,]     1      2

crt(x = 1:100, mods = c(12, 60))
##   [1]  1  2  3  4  5  6  7  8  9 10 11 12  1 14  3 16  5 18  7 20  9 22 11 24  1
##  [26] 26  3 28  5 30  7 32  9 34 11 36  1 38  3 40  5 42  7 44  9 46 11 48  1 50
##  [51]  3 52  5 54  7 56  9 58 11  0  1  2  3  4  5  6  7  8  9 10 11 12  1 14  3
##  [76] 16  5 18  7 20  9 22 11 24  1 26  3 28  5 30  7 32  9 34 11 36  1 38  3 40

为什么最后一个函数调用 crt(x = 1:100, mods = c(12, 60)) 给出了完全不同的输出？第一个向量化输出 crt(1:100) 是我想要和期望的，最后一个在结构上似乎没有什么不同，但结果是......为什么？我该如何解决这个问题以获得与第一个相同的输出？

Answer 1

根据?Vectorize

The arguments named in the vectorize.args argument to Vectorize are the arguments passed in the ... list to mapply. Only those that are actually passed will be vectorized; default values will not.

这里，在 OP 的函数中，'mods' 有一个默认值。如果我们删除它

crt <- function(x, mods) {
   sapply(mods, \(y) x %% y)
 }
crt <- Vectorize(crt)

-测试

> crt(1:100, mods = c(5, 7))
  [1] 1 2 3 4 0 6 2 1 4 3 1 5 3 0 0 2 2 4 4 6 1 1 3 3 0 5 2 0 4 2 1 4 3 6 0 1 2 3 4 5 1 0 3 2 0 4 2 6 4 1 1 3 3 5 0 0 2 2 4 4 1 6 3 1 0 3 2 5 4 0 1 2 3 4
 [75] 0 6 2 1 4 3 1 5 3 0 0 2 2 4 4 6 1 1 3 3 0 5 2 0 4 2
> crt(1:100, mods = c(12, 60))
  [1]  1  2  3  4  5  6  7  8  9 10 11 12  1 14  3 16  5 18  7 20  9 22 11 24  1 26  3 28  5 30  7 32  9 34 11 36  1 38  3 40  5 42  7 44  9 46 11 48  1
 [50] 50  3 52  5 54  7 56  9 58 11  0  1  2  3  4  5  6  7  8  9 10 11 12  1 14  3 16  5 18  7 20  9 22 11 24  1 26  3 28  5 30  7 32  9 34 11 36  1 38
 [99]  3 40

这里的输出格式在两个级别上确定 - 1) sapply 默认使用 simplify = TRUE 和 Vectorize 默认情况下有 SIMPLIFY = TRUE

此外，根据定义的函数，Vectorize 并不是内部真正需要的，它使用 *apply 函数进行循环，我们已经有了 crt 用 sapply 定义，它循环遍历“mods”。在这些参数上应用 %% 的函数是 %%，默认情况下它已经是矢量化函数。