Question

我有一个R data.frame，包含经度，纬度，横跨整个美国地图。当X个条目都在经度为几度的小地理区域内时。纬度为几度，我希望能够检测到这个，然后让我的程序返回地理边界框的坐标。是否有Python或R CRAN包已经这样做了？如果没有，我将如何确定这些信息？

Answer 1

我能够将Joran的回答与Dan H的评论结合起来。这是一个输出示例： cluster map

python代码为R：map（）和rect（）发出函数。这个美国示例地图创建于：

map('state', plot = TRUE, fill = FALSE, col = palette())

然后你可以在R GUI解释器中相应地应用rect（）（见下文）。

import math
from collections import defaultdict

to_rad = math.pi / 180.0   # convert lat or lng to radians
fname = "site.tsv"        # file format: LAT\tLONG
threshhold_dist=50         # adjust to your needs
threshhold_locations=15    # minimum # of locations needed in a cluster

def dist(lat1,lng1,lat2,lng2):
    global to_rad
    earth_radius_km = 6371

    dLat = (lat2-lat1) * to_rad
    dLon = (lng2-lng1) * to_rad
    lat1_rad = lat1 * to_rad
    lat2_rad = lat2 * to_rad

    a = math.sin(dLat/2) * math.sin(dLat/2) + math.sin(dLon/2) * math.sin(dLon/2) * math.cos(lat1_rad) * math.cos(lat2_rad)
    c = 2 * math.atan2(math.sqrt(a), math.sqrt(1-a)); 
    dist = earth_radius_km * c
    return dist

def bounding_box(src, neighbors):
    neighbors.append(src)
    # nw = NorthWest se=SouthEast
    nw_lat = -360
    nw_lng = 360
    se_lat = 360
    se_lng = -360

    for (y,x) in neighbors:
        if y > nw_lat: nw_lat = y
        if x > se_lng: se_lng = x

        if y < se_lat: se_lat = y
        if x < nw_lng: nw_lng = x

    # add some padding
    pad = 0.5
    nw_lat += pad
    nw_lng -= pad
    se_lat -= pad
    se_lng += pad

    # sutiable for r's map() function
    return (se_lat,nw_lat,nw_lng,se_lng)

def sitesDist(site1,site2): 
    #just a helper to shorted list comprehension below 
    return dist(site1[0],site1[1], site2[0], site2[1])

def load_site_data():
    global fname
    sites = defaultdict(tuple)

    data = open(fname,encoding="latin-1")
    data.readline() # skip header
    for line in data:
        line = line[:-1]
        slots = line.split("\t")
        lat = float(slots[0])
        lng = float(slots[1])
        lat_rad = lat * math.pi / 180.0
        lng_rad = lng * math.pi / 180.0
        sites[(lat,lng)] = (lat,lng) #(lat_rad,lng_rad)
    return sites

def main():
    sites_dict = {}
    sites = load_site_data()
    for site in sites: 
        #for each site put it in a dictionary with its value being an array of neighbors 
        sites_dict[site] = [x for x in sites if x != site and sitesDist(site,x) < threshhold_dist] 

    results = {}
    for site in sites: 
        j = len(sites_dict[site])
        if j >= threshhold_locations:
            coord = bounding_box( site, sites_dict[site] )
            results[coord] = coord

    for bbox in results:
        yx="ylim=c(%s,%s), xlim=c(%s,%s)" % (results[bbox]) #(se_lat,nw_lat,nw_lng,se_lng)
        print('map("county", plot=T, fill=T, col=palette(), %s)' % yx)
        rect='rect(%s,%s, %s,%s, col=c("red"))' % (results[bbox][2], results[bbox][0], results[bbox][3], results[bbox][2])
        print(rect)
        print("")

main()

这是一个示例TSV文件（site.tsv）

LAT     LONG
36.3312 -94.1334
36.6828 -121.791
37.2307 -121.96
37.3857 -122.026
37.3857 -122.026
37.3857 -122.026
37.3895 -97.644
37.3992 -122.139
37.3992 -122.139
37.402  -122.078
37.402  -122.078
37.402  -122.078
37.402  -122.078
37.402  -122.078
37.48   -122.144
37.48   -122.144
37.55   126.967

使用我的数据集，我的python脚本的输出，显示在美国地图上。为了清晰起见，我改变了颜色。

rect(-74.989,39.7667, -73.0419,41.5209, col=c("red"))
rect(-123.005,36.8144, -121.392,38.3672, col=c("green"))
rect(-78.2422,38.2474, -76.3,39.9282, col=c("blue"))

2013-05-01为Yacob添加

这两行为你提供了全部目标......

map("county", plot=T )
rect(-122.644,36.7307, -121.46,37.98, col=c("red"))

如果您想缩小地图的一部分，可以使用ylim和xlim

map("county", plot=T, ylim=c(36.7307,37.98), xlim=c(-122.644,-121.46))
# or for more coloring, but choose one or the other map("country") commands
map("county", plot=T, fill=T, col=palette(), ylim=c(36.7307,37.98), xlim=c(-122.644,-121.46))
rect(-122.644,36.7307, -121.46,37.98, col=c("red"))

您需要使用“世界”地图......

map("world", plot=T )

自从我使用下面发布的这个python代码已经很长时间了，所以我会尽力帮助你。

threshhold_dist is the size of the bounding box, ie: the geographical area
theshhold_location is the number of lat/lng points needed with in
    the bounding box in order for it to be considered a cluster.

这是一个完整的例子。 TSV文件位于pastebin.com上。我还包括一个从R生成的图像，它包含所有rect（）命令的输出。

# pyclusters.py
# May-02-2013
# -John Taylor

# latlng.tsv is located at http://pastebin.com/cyvEdx3V
# use the "RAW Paste Data" to preserve the tab characters

import math
from collections import defaultdict

# See also: http://www.geomidpoint.com/example.html
# See also: http://www.movable-type.co.uk/scripts/latlong.html

to_rad = math.pi / 180.0  # convert lat or lng to radians
fname = "latlng.tsv"      # file format: LAT\tLONG
threshhold_dist=20        # adjust to your needs
threshhold_locations=20   # minimum # of locations needed in a cluster
earth_radius_km = 6371

def coord2cart(lat,lng):
    x = math.cos(lat) * math.cos(lng)
    y = math.cos(lat) * math.sin(lng)
    z = math.sin(lat)
    return (x,y,z)

def cart2corrd(x,y,z):
    lon = math.atan2(y,x)
    hyp = math.sqrt(x*x + y*y)
    lat = math.atan2(z,hyp)
    return(lat,lng)

def dist(lat1,lng1,lat2,lng2):
    global to_rad, earth_radius_km

    dLat = (lat2-lat1) * to_rad
    dLon = (lng2-lng1) * to_rad
    lat1_rad = lat1 * to_rad
    lat2_rad = lat2 * to_rad

    a = math.sin(dLat/2) * math.sin(dLat/2) + math.sin(dLon/2) * math.sin(dLon/2) * math.cos(lat1_rad) * math.cos(lat2_rad)
    c = 2 * math.atan2(math.sqrt(a), math.sqrt(1-a)); 
    dist = earth_radius_km * c
    return dist

def bounding_box(src, neighbors):
    neighbors.append(src)
    # nw = NorthWest se=SouthEast
    nw_lat = -360
    nw_lng = 360
    se_lat = 360
    se_lng = -360

    for (y,x) in neighbors:
        if y > nw_lat: nw_lat = y
        if x > se_lng: se_lng = x

        if y < se_lat: se_lat = y
        if x < nw_lng: nw_lng = x

    # add some padding
    pad = 0.5
    nw_lat += pad
    nw_lng -= pad
    se_lat -= pad
    se_lng += pad

    #print("answer:")
    #print("nw lat,lng : %s %s" % (nw_lat,nw_lng))
    #print("se lat,lng : %s %s" % (se_lat,se_lng))

    # sutiable for r's map() function
    return (se_lat,nw_lat,nw_lng,se_lng)

def sitesDist(site1,site2): 
    # just a helper to shorted list comprehensioin below 
    return dist(site1[0],site1[1], site2[0], site2[1])

def load_site_data():
    global fname
    sites = defaultdict(tuple)

    data = open(fname,encoding="latin-1")
    data.readline() # skip header
    for line in data:
        line = line[:-1]
        slots = line.split("\t")
        lat = float(slots[0])
        lng = float(slots[1])
        lat_rad = lat * math.pi / 180.0
        lng_rad = lng * math.pi / 180.0
        sites[(lat,lng)] = (lat,lng) #(lat_rad,lng_rad)
    return sites

def main():
    color_list = ( "red", "blue", "green", "yellow", "orange", "brown", "pink", "purple" )
    color_idx = 0
    sites_dict = {}
    sites = load_site_data()
    for site in sites: 
        #for each site put it in a dictionarry with its value being an array of neighbors 
        sites_dict[site] = [x for x in sites if x != site and sitesDist(site,x) < threshhold_dist] 

    print("")
    print('map("state", plot=T)') # or use: county instead of state
    print("")


    results = {}
    for site in sites: 
        j = len(sites_dict[site])
        if j >= threshhold_locations:
            coord = bounding_box( site, sites_dict[site] )
            results[coord] = coord

    for bbox in results:
        yx="ylim=c(%s,%s), xlim=c(%s,%s)" % (results[bbox]) #(se_lat,nw_lat,nw_lng,se_lng)

        # important!
        # if you want an individual map for each cluster, uncomment this line
        #print('map("county", plot=T, fill=T, col=palette(), %s)' % yx)
        if len(color_list) == color_idx:
            color_idx = 0
        rect='rect(%s,%s, %s,%s, col=c("%s"))' % (results[bbox][2], results[bbox][0], results[bbox][3], results[bbox][1], color_list[color_idx])
        color_idx += 1
        print(rect)
    print("")


main()

pyclusters.py / R image result

Answer 2

我通过首先创建距离矩阵然后在其上运行聚类来定期执行此操作。这是我的代码。

library(geosphere)
library(cluster)
clusteramounts <- 10
distance.matrix <- (distm(points.to.group[,c("lon","lat")]))
clustersx <- as.hclust(agnes(distance.matrix, diss = T))
points.to.group$group <- cutree(clustersx, k=clusteramounts)

我不确定它是否完全解决了你的问题。您可能希望使用不同的k进行测试，也可能需要对某些第一个集群进行第二次集群，以防它们太大，例如明尼苏达州有一个点，加利福尼亚州有一个点。当你有points.to.group $ group时，你可以通过找到每组的最大和最小纬度来获得边界框。

如果你希望X为20，而你在纽约有18分，在达拉斯有22分，你必须决定是否需要一个小盒子和一个非常大的盒子（每个20分），如果你有更好的话达拉斯队包括22分，或者如果你想将达拉斯的22分分成两组。在某些情况下，基于距离的聚类可能是好的。但它当然取决于你为什么要分组。

/克里斯

Answer 3

一些想法：

Ad-hoc＆amp;近似值：“二维直方图”。创建任意程度宽度的任意“矩形”箱，为每个箱分配一个ID。在bin中放置一个点意味着“将该点与bin的ID相关联”。每次添加到垃圾箱后，询问垃圾箱它有多少积分。缺点：没有正确地“看到”一个跨越边界的点集群;和：当你向北移动时，“恒定纵向宽度”的箱子实际上（空间上）较小。
使用Python的“Shapely”库。按照它的“缓冲点”的库存示例，并执行缓冲区的级联联合。寻找某个区域上的球体，或“包含”一定数量的原始点。请注意，Shapely本质上并不是“geo-savy”，因此如果需要，您必须添加更正。
使用具有空间处理功能的真正数据库。 MySQL，Oracle，Postgres（使用PostGIS），MSSQL（我认为）都有“Geometry”和“Geography”数据类型，你可以对它们进行空间查询（来自你的Python脚本）。

每种都有不同的美元和时间成本（在学习曲线中）......以及不同程度的地理空间精度。您必须选择适合您预算和/或要求的内容。

Answer 4

如果你使用整齐，你可以扩展我的cluster_points function 通过形状几何的.bounds属性返回集群的边界框，例如：

clusterlist.append(cluster, (poly.buffer(-b)).bounds)

Answer 5

可能像

def dist(lat1,lon1,lat2,lon2):
    #just return normal x,y dist
    return sqrt((lat1-lat2)**2+(lon1-lon2)**2)

def sitesDist(site1,site2):
    #just a helper to shorted list comprehensioin below
    return dist(site1.lat,site1.lon,site2.lat,site2.lon)
sites_dict = {}
threshhold_dist=5 #example dist
for site in sites:
    #for each site put it in a dictionarry with its value being an array of neighbors
    sites_dict[site] = [x for x in sites if x != site and sitesDist(site,x) < threshhold_dist]
print "\n".join(sites_dict)

检测地理集群

5 个答案: