我有以下数据集 w
和关键变量 x
对于两种情况。
Case 1:
x = 4
w = c(1,2,4,4,4,4,6,7,8,9,10,11,12,14,15)
Case2:
x = 12
w = c(1,2,4,4,4,4,6,7,8,9,10,11,12,14,15)
我想创建一个函数来搜索
x
通过数据集w
并将根据 x
将原始数据集子集到较小的数据集的位置在 w
.输出将是具有与搜索关键字相同的上限值的较小尺寸的数据集。下面是我试图用 R 编写的函数:create_chunk <- function(val, tab, L=1L, H=length(tab))
{
if(H >= L)
{
mid = L + ((H-L)/2)
## If the element is present within middle length
if(tab[mid] > val)
{
## subset the original data in reduced size and again do mid position value checking
## then subset the data
} else
{
mid = mid + (mid/2)
## Increase the mid position to go for right side checking
}
}
}
在我正在寻找的输出中:
Output for Case 1:
Dataset containing: 1,2,4,4,4,4
Output for Case 2:
Dataset containing: 1,2,4,4,4,4,6,7,8,9,10,11,12
Please note:
1. Dataset may contain duplicate values for search key and
all the duplicate values are expected in the output dataset.
2. I have huge size datasets (say around 2M rows) from
where I am trying to subset smaller dataset as per my requirement of search key.
新更新:案例 3
输入数据:
date value size stockName
1 2016-08-12 12:44:43 10093.40 4 HWA IS Equity
2 2016-08-12 12:44:38 10093.35 2 HWA IS Equity
3 2016-08-12 12:44:47 10088.00 2 HWA IS Equity
4 2016-08-12 12:44:52 10089.95 1 HWA IS Equity
5 2016-08-12 12:44:53 10089.95 1 HWA IS Equity
6 2016-08-12 12:44:54 10088.95 1 HWA IS Equity
搜索关键字是:
10089.95
在值列中。预期输出为:
date value size stockName
1 2016-08-12 12:44:47 10088.00 2 HWA IS Equity
2 2016-08-12 12:44:54 10088.95 1 HWA IS Equity
3 2016-08-12 12:44:52 10089.95 1 HWA IS Equity
4 2016-08-12 12:44:53 10089.95 1 HWA IS Equity
最佳答案
您可以执行此操作来处理重复值。如果有重复,将返回其最高位置。请注意 A
应该是非递减的。
binSearch <- function(A, value, left=1, right=length(A)){
if (left > right)
return(-1)
middle <- (left + right) %/% 2
if (A[middle] == value){
while (A[middle] == value)
middle<-middle+1
return(middle-1)
}
else {
if (A[middle] > value)
return(binSearch(A, value, left, middle - 1))
else
return(binSearch(A, value, middle + 1, right))
}
}
w[1:binSearch(w,x1)]
# [1] 1 2 4 4 4 4
w[1:binSearch(w,x2)]
# [1] 1 2 4 4 4 4 6 7 8 9 10 11 12
但是,正如评论中提到的,您可以简单地使用
findInterval
达到同样的目的:w[1:findInterval(x1,w)]
如您所知,二分查找的顺序为
log(n)
但正如 ?findInterval
中所述,它还受益于 log(n)
因为第一个参数的长度是一个:The function findInterval finds the index of one vector x in another, vec, where the latter must be non-decreasing. Where this is trivial, equivalent to apply( outer(x, vec, ">="), 1, sum), as a matter of fact, the internal algorithm uses interval search ensuring O(n * log(N)) complexity where n <- length(x) (and N <- length(vec)). For (almost) sorted x, it will be even faster, basically O(n).
编辑
根据您的编辑和新设置,您可以这样做(假设您的数据在
df
中):o <- order(df$value)
rows <- o[1:findInterval(key, df$value[o])]
df[rows,]
或者等效地,使用建议的
binSearch
功能:o <- order(df$value)
rows <- o[1:binSearch(df$value[o], key)]
df[rows,]
数据
x1 <- 4
x2 <- 12
w <- c(1,2,4,4,4,4,6,7,8,9,10,11,12,14,15)
key <- 10089.95
关于r - 在 R 中创建子集数据的二分搜索类似概念,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/39632507/