Question

我正试图想出一种方法将一些数据加在一起。我想带一个候选人的载体加入（可以排除其中一些）与另一个候选人的载体（所有应该使用，或至少，应该指出未找到匹配）。我希望加入是1比1。在我卷起袖子并为此自己制定算法之前，我想我会问我所寻找的机器是否已经存在。如果没有工作R代码，也许一些完善的算法？理想情况下，我也希望解决方案能够使用日期。

以下是我认为重要的一些案例：

#' # case: all y in xs
#' c(1, 2, 3, 4, 5, 6) %>% closest_to(c(4, 5, 6)) # => c(4, 5, 6)
#' # case: all xs in y
#' c(3, 5) %>% closest_to(c(1, 2, 3, 4, 5)) # => c(NA, NA, 3, NA, 5)
#' c(2, 5) %>% closest_to(c(2, 4, 5, 6)) # => c(2, NA, 5, NA)
#' # case: no xs in y
#' c(1, 2, 3) %>% closest_to(c(4, 5, 6)) # => c(1, 2, 3)
#' c(1, 2) %>% closest_to(c(4, 5, 6)) # => c(NA, 1, 2)
#' c(5, 6) %>% closest_to(c(1, 2)) # => c(1, 2)
#' # case: xs is empty
#' c() %>% closest_to(c(1, 2)) # => c(NA, NA)
#' # case: xs is y
#' c(1, 2, 3) %>% closest_to(c(1, 2, 3)) # => c(1, 2, 3)
#' # case: some values shared (revert to minimum distance of maximum length)
#' c(1, 2, 5) %>% closest_to(c(2, 4, 5, 6)) # => c(1, 2, 5, NA)
#' c(2, 5) %>% closest_to(c(1, 3, 6)) # => c(NA, 2, 5) (choose some nice way to handle ties)

Answer 1

在这里，我找到了一些可以与您的问题相匹配的答案，但最后一个答案。

#' # case: all y in xs
#' c(1, 2, 3, 4, 5, 6) %>% closest_to(c(4, 5, 6)) # => c(4, 5, 6)
vector1<-c(1, 2, 3, 4, 5, 6)
closest_to<-c(4,5,6)
intersect(vector1, closest_to)
#[1] 4 5 6

#' # case: all xs in y
#' c(3, 5) %>% closest_to(c(1, 2, 3, 4, 5)) # => c(NA, NA, 3, NA, 5)
#' c(2, 5) %>% closest_to(c(2, 4, 5, 6)) # => c(2, NA, 5, NA)
#' # case: no xs in y
#' c(1, 2, 3) %>% closest_to(c(4, 5, 6)) # => c(1, 2, 3)
#' c(1, 2) %>% closest_to(c(4, 5, 6)) # => c(NA, 1, 2)
#' c(5, 6) %>% closest_to(c(1, 2)) # => c(1, 2)

vector2<-c(3, 5)
closest_to2<-c(1, 2, 3, 4, 5, 6)
closest_to2 %in% vector2
#[1] FALSE FALSE  TRUE FALSE  TRUE FALSE
which(closest_to2 %in% vector2)
#[1] 3 5




#' # case: xs is empty
#' c() %>% closest_to(c(1, 2)) # => c(NA, NA)
c() %in% c(1, 2)
#logical(0)


#' # case: xs is y
#' c(1, 2, 3) %>% closest_to(c(1, 2, 3)) # => c(1, 2, 3)
vector1<-c(1,2,3)
vector2<-c(1,2,3)
setequal(vector1, vector2)
#[1] TRUE


#' # case: some values shared (revert to minimum distance of maximum length)
#' c(1, 2, 5) %>% closest_to(c(2, 4, 5, 6)) # => c(1, 2, 5, NA)
#' c(2, 5) %>% closest_to(c(1, 3, 6)) # => c(NA, 2, 5) (choose some nice way to handle ties)

# No idea

干杯！，

通过寻找最接近的值来匹配向量

1 个答案: