用两个数据集求最近邻的欧几里德距离

编程问答 2022-04-24

问题描述

我有两组 first_data。第一个是 100 (x_0,y_0) 个数据：

x_0 <- seq(1,10,by=1)
y_0 <- seq(1,by=1)
data <- expand.grid(x_0,y_0)

第二个是 5 个 (x,y) 数据，称为 second_data：

x <- c(2,4,6,8,10)
y <- c(3,5,7,9,11)
color <- c("green","green","red","red")
second_data<- data.frame(x,y,color)

我需要为 3NN 应用欧几里得距离公式，以根据欧几里得距离确定第一个数据集中的每个点是绿色还是红色。基本上，我需要找到每100对点的距离，5次，然后使用下面的代码选择距离最小的3个。

我想我需要一个循环，但我没有正确理解：

out <- rep(NA,nrow(first_data))
K=3

for(k in 1:nrow(first_data)){
green <- mutate(second_data,distance = sqrt(x - first_data[k]^2)+(y-first_data[k]^2)) %>%
  slice_min(distance,n=K) %>% filter(color=='green') %>% nrow()
  out[k] <- ifelse(new_blue >= (K+1)/2,'green','red')
}

解决方法

如果我理解正确的话，get.knn 包中的 FNN 函数会很容易做到这一点：

library(FNN)    
neighbors3 <- get.knnx(second_data[,-3],data,k=3)
str(neighbors3)
# List of 2
#  $ nn.index: int [1:100,1:3] 1 1 1 1 1 1 2 2 2 3 ...
#  $ nn.dist : num [1:100,1:3] 2.24 2 2.24 2.83 3.61 ...
head(neighbors3$nn.index)
#      [,1] [,2] [,3]
# [1,]    1    2    3
# [2,]    1    2    3
# [3,]    1    2    3
# [4,]    1    2    3
# [5,]    1    2    3
# [6,]    1    2    3

列表元素 neighbors3$nn.index 为 second_data 中的每一行提供 data 中的三个最近邻。现在获取邻居的颜色：

result <- matrix(color[neighbors3$nn.index],100,3)
head(result); cat("\n"); tail(result)
#      [,1]    [,2]    [,3] 
# [1,] "green" "green" "red"
# [2,] "green" "green" "red"
# [3,] "green" "green" "red"
# [4,] "green" "green" "red"
# [5,] "green" "green" "red"
# [6,] "green" "green" "red"
# 
#        [,1]  [,2]  [,3]   
#  [95,] "red" "red" "green"
#  [96,] "red" "red" "red"  
#  [97,] "red" "red" "red"  
#  [98,] "red" "red" "red"  
#  [99,] "red" "red" "red"  
# [100,] "red" "red" "red"

如果需要，您可以将所有内容与原始数据合并：

results <- cbind(data,neighbors3$nn.index,result,neighbors3$nn.dist)
colnames(results) <- c("x0","y_0","nn1","nn2","nn3","col1","col2","col3","dist1","dist2","dist3")
head(results)
#   x0 y_0 nn1 nn2 nn3  col1  col2 col3    dist1    dist2    dist3
# 1  1   1   1   2   3 green green  red 2.236068 5.000000 7.810250
# 2  2   1   1   2   3 green green  red 2.000000 4.472136 7.211103
# 3  3   1   1   2   3 green green  red 2.236068 4.123106 6.708204
# 4  4   1   1   2   3 green green  red 2.828427 4.000000 6.324555
# 5  5   1   1   2   3 green green  red 3.605551 4.123106 6.082763
# 6  6   1   1   2   3 green green  red 4.472136 4.472136 6.000000
,

如果我正确理解，这将产生正确的分类

data$color <- NA
k <- 3
for (i in 1:nrow(data)){
  d <- data.frame()
  dat_aux <- second_data
  for (j in 1:k){
    d_j <- which.min((dat_aux$x - data$Var1[i])^2 + (dat_aux$y - data$Var2[i])^2)
    d <- c(d,dat_aux$color[d_j])
    dat_aux[d_j,] <- NA
  }
  data$color[i] <- names(sort(table(unlist(d)),decreasing = T))[1]
}
data
loopsnearest-neighborrrstatistics