从其他数据框中查找最近的点 [英] Finding closest point from other data frame

问题描述

我有两个数据框，一个是具有80万行的X和Y坐标，另一个数据框是70000行，具有X和Y坐标.我想知道R中的逻辑和代码，我想将第1帧中的数据点与第2帧中的最近点相关联.是否有任何标准包装?

I have two data frames , one is with 0.8 million rows with x and Y coordinates, another data frame is of 70000 rows with X and Y coordinates. I want to know logic and code in R where I want to associate data point from frame 1 to closest point in data frame 2. Is there any standard package to do so ?

我正在嵌套运行循环.但这非常慢，因为它要进行80万次* 70000次的迭代，这非常耗时.

I am running nested for loop. But this is very slow as it is getting iterated for 0.8 million * 70000 times which is very time consuming.

推荐答案

我找到了使用data.table库获得预期结果的更快方法:

I found a faster way to get the expected result using the data.table library:

library(data.table)

time0 <- Sys.time()

以下是一些随机数据:

df1 <- data.table(x = runif(8e5), y = runif(8e5))
df2 <- data.table(x = runif(7e4), y = runif(7e4))

假设(x，y)是正交法坐标系中的坐标，则可以如下计算距离的平方:

Assuming (x,y) are the coordinates in an orthonormal coordinate system, you can compute the square of the distance as follow:

dist <- function(a, b){
                dt <- data.table((df2$x-a)^2+(df2$y-b)^2)
                return(which.min(dt$V1))}

现在您可以将此功能应用于数据以获得预期的结果:

And now you can applied this function to your data to get the expected result:

results <- df1[, j = list(Closest =  dist(x, y)), by = 1:nrow(df1)]

time1 <- Sys.time()
print(time1 - time0)

我花了大约30分钟的时间才能在慢速的计算机上获得结果.

It tooked me around 30 minutes to get the result on a slow computer.

根据要求，我尝试使用sapply或plyr软件包中的adply尝试其他几种解决方案.我已经在较小的数据帧上测试了这些解决方案，以使其更快.

As asked, I have tried severals other solutions using sapply or using adply from the plyr package. I have tested these solutions on smaller data frames to make it faster.

library(data.table)
library(plyr)
library(microbenchmark)

########################
## Test 1: data.table ##
########################

dt1 <- data.table(x = runif(1e4), y = runif(1e4))
dt2 <- data.table(x = runif(5e3), y = runif(5e3))

dist1 <- function(a, b){
                dt <- data.table((dt2$x-a)^2+(dt2$y-b)^2)
                return(which.min(dt$V1))}

results1 <- function() return(dt1[, j = list(Closest =  dist1(x, y)), by = 1:nrow(dt1)])

###################
## Test 2: adply ##
###################

df1 <- data.frame(x = runif(1e4), y = runif(1e4))
df2 <- data.frame(x = runif(5e3), y = runif(5e3))

dist2 <- function(df){
                dt <- data.table((df2$x-df$x)^2+(df2$y-df$y)^2)
                return(which.min(dt$V1))}

results2 <- function() return(adply(.data = df1, .margins = 1, .fun = dist2))

####################
## Test 3: sapply ##
####################

df1 <- data.frame(x = runif(1e4), y = runif(1e4))
df2 <- data.frame(x = runif(5e3), y = runif(5e3))

dist2 <- function(df){
                dt <- data.table((df2$x-df$x)^2+(df2$y-df$y)^2)
                return(which.min(dt$V1))}

results3 <- function() return(sapply(1:nrow(df1), function(x) return(dist2(df1[x,]))))

microbenchmark(results1(), results2(), results3(), times = 20)

#Unit: seconds
#       expr      min       lq     mean   median       uq      max neval
# results1() 4.046063 4.117177 4.401397 4.218234 4.538186 5.724824    20
# results2() 5.503518 5.679844 5.992497 5.886135 6.041192 7.283477    20
# results3() 4.718865 4.883286 5.131345 4.949300 5.231807 6.262914    20

第一个解决方案似乎比其他两个解决方案要快得多.对于更大的数据集，情况更是如此.

The first solution seems to be significantly faster than the 2 other. This is even more true for a larger dataset.

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