论点“部分" R中的排序函数 [英] Argument "partial" of the sort function in R
问题描述
?sort指出partial参数可以是NULL或用于部分排序的索引向量.
?sort states that the partial argument may be NULL or a vector of indices for partial sorting.
我尝试过:
x <- c(1,3,5,2,4,6,7,9,8,10)
sort(x)
## [1] 1 2 3 4 5 6 7 8 9 10
sort(x, partial=5)
## [1] 1 3 4 2 5 6 7 9 8 10
sort(x, partial=2)
## [1] 1 2 5 3 4 6 7 9 8 10
sort(x, partial=4)
## [1] 1 2 3 4 5 6 7 9 8 10
我不确定partial在对向量进行排序时是什么意思.
I am not sure what partial means when sorting a vector.
推荐答案
如?sort所述，
如果partial不为NULL，则认为它包含结果元素的索引 通过部分排序将它们放置在已排序数组中的正确位置.
If partial is not NULL, it is taken to contain indices of elements of the result which are to be placed in their correct positions in the sorted array by partial sorting.
换句话说，以下断言始终为真:
In other words, the following assertion is always true:
stopifnot(sort(x, partial=pt_idx)[pt_idx] == sort(x)[pt_idx])
用于任何x和pt_idx，例如
x <- sample(100) # input vector
pt_idx <- sample(1:100, 5) # indices for partial arg
此行为与维基百科有关部分排序的文章中定义的行为不同.在R sort()的情况下，我们不一定要计算 k 个最小元素.
This behavior is different from the one defined in the Wikipedia article on partial sorting. In R sort()'s case we are not necessarily computing k smallest elements.
例如，如果
print(x)
## [1] 91 85 63 80 71 69 20 39 78 67 32 56 27 79 9 66 88 23 61 75 68 81 21 90 36 84 11 3 42 43
## [31] 17 97 57 76 55 62 24 82 28 72 25 60 14 93 2 100 98 51 29 5 59 87 44 37 16 34 48 4 49 77
## [61] 13 95 31 15 70 18 52 58 73 1 45 40 8 30 89 99 41 7 94 47 96 12 35 19 38 6 74 50 86 65
## [91] 54 46 33 22 26 92 53 10 64 83
和
pt_idx
## [1] 5 54 58 95 8
然后
sort(x, partial=pt_idx)
## [1] 1 3 2 4 5 6 7 8 11 12 9 10 13 15 14 16 17 18 23 30 31 27 21 32 36 34 35 19 20 37
## [31] 38 33 29 22 26 25 24 28 39 41 40 42 43 48 46 44 45 47 51 50 52 49 53 54 57 56 55 58 59 60
## [61] 62 64 63 61 65 66 70 72 73 69 68 71 67 79 78 82 75 81 80 77 76 74 89 85 88 87 83 84 86 90
## [91] 92 93 91 94 95 96 97 99 100 98
在这里x[5]，x[54]，...，x[8]放置在正确的位置-我们不能对其余元素说其他的话. HTH.
Here x[5], x[54], ..., x[8] are placed in their correct positions - and we cannot say anything else about the remaining elements. HTH.
编辑:部分排序可能会减少排序时间，当然，如果您对例如仅查找某些订单统计信息.
EDIT: Partial sorting may reduce the sorting time, of course if you are interested in e.g. finding only some of the order statistics.
require(microbenchmark)
x <- rnorm(100000)
microbenchmark(sort(x, partial=1:10)[1:10], sort(x)[1:10])
## Unit: milliseconds
## expr min lq median uq max neval
## sort(x, partial = 1:10)[1:10] 2.342806 2.366383 2.393426 3.631734 44.00128 100
## sort(x)[1:10] 16.556525 16.645339 16.745489 17.911789 18.13621 100
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