数据：

Question

我旨在基于两个数据集中的坐标来确定数据集中2中与数据集中1中每个条目最近的条目。数据集1包含180,000行（仅1,800个唯一坐标），数据集2包含4,500行（完整的4,500个唯一坐标）。

我试图从stackoverflow上复制类似问题的答案。例如：

R - Finding closest neighboring point and number of neighbors within a given radius, coordinates lat-long

Calculating the distance between points in different data frames

但是这些并不能解决我想要的问题（它们要么加入数据框，要么检查单个数据框内的距离）。

到目前为止，我在Find the nearest X,Y coordinate using R和related posts中找到的解决方案最接近。

我在这篇文章中遇到的问题是，它可以计算出单个数据框内坐标之间的距离，而我无法理解在RANN::nn2中要更改哪些参数才能在两个数据框内做到这一点。

建议的无效代码：

library(RANN)
dataset1[,4]<- nn2(data=dataset1, query=dataset2, k=2)

注释/问题：

1）应该向查询提供哪个数据集，以找到数据集2中与数据集1中给定值最接近的值？

2）有什么方法可以避免数据集似乎需要相同的宽度（列数）的问题？

3）如何将输出（SRD_ID和distance）添加到数据集1中的相关条目？

4）eps函数中的RANN::nn2参数有什么用途？

目的是使用数据集中2中最接近的站点ID以及数据集中1中的条目与数据集中2中的最近条目之间的距离填充数据集1中的SRC_ID和distance列。 / p>

下面的表格最小化了预期的结果。 注意：SRC_ID和distance值是我手动添加的示例值，几乎可以肯定是不正确的，并且可能不会被代码复制。

       id HIGH_PRCN_LAT HIGH_PRCN_LON SRC_ID distance
1 3797987      52.88121     -2.873734     55      350 
2 3798045      53.80945     -2.439163     76     2100

数据：

_{r详细信息}

platform        x86_64-w64-mingw32
version.string  R version 3.5.3 (2019-03-11)

_{数据集1输入（不缩小为唯一坐标）}

structure(list(id = c(1L, 2L, 4L, 5L, 
6L, 7L, 8L, 9, 10L, 3L), 
    HIGH_PRCN_LAT = c(52.881442267773, 57.8094538200198, 34.0233529, 
    63.8087900198, 53.6888144440184, 63.4462810678651, 21.6075544376207, 
    78.324442654172, 66.85532539759495, 51.623544596), HIGH_PRCN_LON = c(-2.87377812157822, 
    -2.23454414781635, -3.0984448341, -2.439163178635, -7.396111601421454, 
    -5.162345043546359, -8.63311254098095, 3.813289888829932, 
    -3.994325961186105, -8.9065532453272409), SRC_ID = c(NA, NA, 
    NA, NA, NA, NA, NA, NA, NA, NA), distance = c(NA, NA, 
    NA, NA, NA, NA, NA, NA, NA, NA)), row.names = c(NA, 10L), class = "data.frame")

_{数据集2输入}

structure(list(SRC_ID = c(55L, 54L, 23L, 11L, 44L, 21L, 76L, 
5688L, 440L, 61114L), HIGH_PRCN_LAT = c(68.46506, 50.34127, 61.16432, 
42.57807, 52.29879, 68.52132, 87.83912, 55.67825, 29.74444, 34.33228
), HIGH_PRCN_LON = c(-5.0584, -5.95506, -5.75546, -5.47801, -3.42062, 
-6.99441, -2.63457, -2.63057, -7.52216, -1.65532)), row.names = c(NA, 
10L), class = "data.frame")

Answer 1

我写了一个参考此thread的答案。对该功能进行了修改，以确保报告距离并避免硬编码。请注意，它计算欧几里得距离。

library(data.table)
#Euclidean distance 
mydist <- function(a, b, df1, x, y){

          dt <- data.table(sqrt((df1[[x]]-a)^2 + (df1[[y]]-b)^2))

          return(data.table(Closest.V1  = which.min(dt$V1),
                            Distance    = dt[which.min(dt$V1)]))
           }

setDT(df1)[, j = mydist(HIGH_PRCN_LAT, HIGH_PRCN_LON, setDT(df2), 
                        "HIGH_PRCN_LAT", "HIGH_PRCN_LON"), 
                         by = list(id, HIGH_PRCN_LAT, HIGH_PRCN_LON)]

  #     id HIGH_PRCN_LAT HIGH_PRCN_LON Closest.V1 Distance.V1
  # 1:   1      52.88144     -2.873778          5   0.7990743
  # 2:   2      57.80945     -2.234544          8   2.1676868
  # 3:   4      34.02335     -3.098445         10   1.4758202
  # 4:   5      63.80879     -2.439163          3   4.2415854
  # 5:   6      53.68881     -7.396112          2   3.6445416
  # 6:   7      63.44628     -5.162345          3   2.3577811
  # 7:   8      21.60755     -8.633113          9   8.2123762
  # 8:   9      78.32444      3.813290          7  11.4936496
  # 9:  10      66.85533     -3.994326          1   1.9296370
  # 10:  3      51.62354     -8.906553          2   3.2180026

您可以使用RANN::nn2，但需要确保使用正确的语法。正在关注作品！

as.data.frame(RANN::nn2(df2[,c(2,3)],df1[,c(2,3)],k=1))

#    nn.idx   nn.dists
# 1       5  0.7990743
# 2       8  2.1676868
# 3      10  1.4758202
# 4       3  4.2415854
# 5       2  3.6445416
# 6       3  2.3577811
# 7       9  8.2123762
# 8       7 11.4936496
# 9       1  1.9296370
# 10      2  3.2180026

Answer 2

数据

x = structure(list(id = c(1L, 2L, 4L, 5L, 
6L, 7L, 8L, 9, 10L, 3L), 
    HIGH_PRCN_LAT = c(52.881442267773, 57.8094538200198, 34.0233529, 
    63.8087900198, 53.6888144440184, 63.4462810678651, 21.6075544376207, 
    78.324442654172, 66.85532539759495, 51.623544596), HIGH_PRCN_LON = c(-2.87377812157822, 
    -2.23454414781635, -3.0984448341, -2.439163178635, -7.396111601421454, 
    -5.162345043546359, -8.63311254098095, 3.813289888829932, 
    -3.994325961186105, -8.9065532453272409), SRC_ID = c(NA, NA, 
    NA, NA, NA, NA, NA, NA, NA, NA), distance = c(NA, NA, 
    NA, NA, NA, NA, NA, NA, NA, NA)), row.names = c(NA, 10L), class = "data.frame")

y = structure(list(SRC_ID = c(55L, 54L, 23L, 11L, 44L, 21L, 76L, 
 5688L, 440L, 61114L), HIGH_PRCN_LAT = c(68.46506, 50.34127, 61.16432, 
 42.57807, 52.29879, 68.52132, 87.83912, 55.67825, 29.74444, 34.33228
 ), HIGH_PRCN_LON = c(-5.0584, -5.95506, -5.75546, -5.47801, -3.42062, 
 -6.99441, -2.63457, -2.63057, -7.52216, -1.65532)), row.names = c(NA, 
 10L), class = "data.frame")

解决方案。注意“ 3：2”按此顺序获得“经度/纬度”。

library(raster)

d <- pointDistance(x[,3:2], y[,3:2], lonlat=TRUE, allpairs=T) 
i <- apply(d, 1, which.min)

x$SRC_ID = y$SRC_ID[i]
x$distance = d[cbind(1:nrow(d), i)]
x

#   id HIGH_PRCN_LAT HIGH_PRCN_LON SRC_ID   distance
#1   1      52.88144     -2.873778     44   74680.48
#2   2      57.80945     -2.234544   5688  238553.51
#3   4      34.02335     -3.098445  61114  137385.18
#4   5      63.80879     -2.439163     23  340642.70
#5   6      53.68881     -7.396112     44  308458.73
#6   7      63.44628     -5.162345     23  256176.88
#7   8      21.60755     -8.633113    440  908292.28
#8   9      78.32444      3.813290     76 1064419.47
#9  10      66.85533     -3.994326     55  185119.29
#10  3      51.62354     -8.906553     54  251580.45

说明

plot(x[,3:2], ylim=c(0,90), col="blue", pch=20)
points(y[,3:2], col="red", pch=20)
for (i in 1:nrow(x)) {
    j <- y$SRC_ID==x$SRC_ID[i]
    arrows(x[i,3], x[i,2], y[j,3], y[j,2],length=.1) 
}

text(x[,3:2], labels=x$id, pos=1, cex=.75)
text(y[,3:2], labels=y$SRC_ID, pos=3, cex=.75)

r-查找两个大型数据集之间的最近坐标

数据：

2 个答案: