从lat / lon coordinates找到5个最近的地铁站

时间:2015-11-08 23:04:28

标签: r sp

我正在尝试查找从一个数据集(set1)到另一个数据集(set2)的5个最近的站点。 This帖子是我正在使用的基础,找到最接近的单个似乎很简单,但我正在编写for循环来处理它并且效率不高。此外,我得到了错误,不明白为什么它不起作用。理想情况下,我想使用set1查找set2中最近的电台,查找5个最近的电台,并为set1的每个唯一ID添加每个电台的列。

编辑:这个问题与How to assign a name to lat-long observations based on shortest distance不同,因为我试图找到5个最近的站点,而不仅仅是一个距离。此外,该方法对于找到最小值是不同的。请重新打开这个问题。

dput:

set1 <- structure(list(id = c(5984, 7495, 4752, 2654, 4578, 9865, 3265, 
1252, 4679, 1346), lat = c(48.39167, 48.148056, 48.721111, 47.189167, 
47.054443, 47.129166, 47.306667, 47.84, 47.304167, 48.109444), 
    lon = c(13.671114, 12.866947, 15.94223, 11.099736, 12.958342, 
    14.203892, 11.86389, 16.526674, 16.193064, 17.071392)), row.names = c(NA, 
10L), class = "data.frame", .Names = c("id", "lat", "lon"))

set2 <- structure(list(id = 1:10, lat = structure(c(35.8499984741211, 
34.75, 70.9329986572266, 78.25, 69.6829986572266, 74.515998840332, 
70.3659973144531, 67.265998840332, 63.6990013122559, 60.1990013122559
), .Dim = 10L), lon = structure(c(14.4829998016357, 32.4000015258789, 
-8.66600036621094, 15.4670000076294, 18.9160003662109, 19.0160007476807, 
31.0990009307861, 14.3660001754761, 9.59899997711182, 11.0830001831055
), .Dim = 10L)), row.names = c(NA, 10L), class = "data.frame", .Names = c("id", 
"lat", "lon"))

代码:

library(rgeos)
library(sp)


set1sp <- SpatialPoints(set1)
set2sp <- SpatialPoints(set2)
for (i in length(set1$id)){
  for (j in 4:9){
    if(i == 1) {
      sub <- set2
      set1[i,j] <- apply(gDistance(set1sp, set2sp, byid=TRUE), 1, which.min)
      sub <- filter(sub, id != set1[i,j])}
    else{
      set1[i,j] <- apply(gDistance(set1sp, set2sp, byid=TRUE), 1, which.min)
      sub <- filter(sub, id != set1[i,j])}
  }
}

输出错误:

 Error in `[<-.data.frame`(`*tmp*`, i, j, value = c(8L, 8L, 8L, 8L, 8L,  : 
  replacement has 10 rows, data has 1 

2 个答案:

答案 0 :(得分:1)

我必须设置投影系统以及scheduler.every '5m' do # Some Fancy Code Logic That Runs Every 5 Minutes end set1sp的坐标才能使set2sp正常工作。我假设WGS84。

gDistance

这个循环将根据你的for循环的一般结构返回你想要的输出。

dummyset1= set1
dummyset2= set2
coordinates(set1) = c('lon', 'lat')
coordinates(set2) = c('lon', 'lat')
proj4string(set1) = "+proj=longlat +datum=WGS84"
proj4string(set2) = "+proj=longlat +datum=WGS84"
set1sp = set1
set2sp = set2
set1 = dummyset1
set2 = dummyset2

结果输出为:

for (i in 1:length(set1$id)){
    #Store the projected data in a dummy variable sub
    sub <- set2sp
    for (j in 4:8){
        if (j == 4){
           set1[i,j] <- apply(gDistance(set2sp['id'], set1sp['id'][i,], byid=TRUE), 1, which.min)
           #Remove the index of the closest point from sub.
           sub <- sub[which(sub$id != set1[i,j]), ]
        }
        else {
           #Note that sub is now being checked instead of set2sp. This is because sub has had the index of the closest point removed.
           set1[i,j] <- apply(gDistance(sub['id'], set1sp['id'][i,], byid=TRUE), 1, which.min)
           sub <- sub[which(sub$id != set1[i,j]), ]
        }
    }
}

答案 1 :(得分:1)

下面计算第2组中所有点的大圆距离。然后在第1组中取最小值,然后对它们进行排序;然后是情节。

library(sp)
coordinates(set1) = c('lon', 'lat')
coordinates(set2) = c('lon', 'lat')
proj4string(set1) = "+proj=longlat +datum=WGS84"
proj4string(set2) = "+proj=longlat +datum=WGS84"
d = apply(spDists(set1,set2),2,min)
order(d)[1:5]
# [1]  1 10  9  2  8
plot(set2, pch=2, axes=TRUE)
points(set1)
o = order(d)[1:5]
points(set2[o,], col = 'red', pch=16)

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