緯度經度坐標到 R 中的狀態代碼

Question

有沒有一種快速的方法可以將經緯度坐標轉換為 R 中的州代碼？ 我一直在使用 zipcode 包作為查找表，但是當我查詢大量經緯度值時它太慢了

如果不在 R 中，有沒有辦法使用谷歌地理編碼器或任何其他類型的快速查詢服務來做到這一點？

謝謝！

Answer 1

這里有兩種選擇，一種使用sf ，一種使用sp包函數。 sf是用於分析空間數據的更現代（並且在 2020 年推薦）包，但如果它仍然有用，我將保留我原來的 2012 年答案，展示如何使用sp相關函數執行此操作。

---

方法1（使用sf）：

library(sf)
library(spData)

## pointsDF: A data.frame whose first column contains longitudes and
##           whose second column contains latitudes.
##
## states:   An sf MULTIPOLYGON object with 50 states plus DC.
##
## name_col: Name of a column in `states` that supplies the states'
##           names.
lonlat_to_state <- function(pointsDF,
                            states = spData::us_states,
                            name_col = "NAME") {
    ## Convert points data.frame to an sf POINTS object
    pts <- st_as_sf(pointsDF, coords = 1:2, crs = 4326)

    ## Transform spatial data to some planar coordinate system
    ## (e.g. Web Mercator) as required for geometric operations
    states <- st_transform(states, crs = 3857)
    pts <- st_transform(pts, crs = 3857)

    ## Find names of state (if any) intersected by each point
    state_names <- states[[name_col]]
    ii <- as.integer(st_intersects(pts, states))
    state_names[ii]
}

## Test the function with points in Wisconsin, Oregon, and France
testPoints <- data.frame(x = c(-90, -120, 0), y = c(44, 44, 44))
lonlat_to_state(testPoints)
## [1] "Wisconsin" "Oregon"    NA

如果您需要更高分辨率的狀態邊界，請使用sf::st_read()或通過其他方式將您自己的矢量數據作為sf對象讀入。 一個不錯的選擇是安裝rnaturalearth包並使用它從rnaturalearthhires加載狀態向量層。 然后使用我們剛剛定義的lonlat_to_state()函數，如下所示：

library(rnaturalearth)
us_states_ne <- ne_states(country = "United States of America",
                          returnclass = "sf")
lonlat_to_state(testPoints, states = us_states_ne, name_col = "name")
## [1] "Wisconsin" "Oregon"    NA         

要獲得非常准確的結果，您可以從此頁面下載包含GADM 維護的美國行政邊界的地理包。 然后，加載狀態邊界數據並像這樣使用它們：

USA_gadm <- st_read(dsn = "gadm36_USA.gpkg", layer = "gadm36_USA_1")
lonlat_to_state(testPoints, states = USA_gadm, name_col = "NAME_1")
## [1] "Wisconsin" "Oregon"    NA        

---

方法2（使用sp）：

這是一個函數，它在較低的 48 個狀態中采用經緯度的 data.frame，並且對於每個點，返回它所在的狀態。

大多數函數只是簡單地准備了sp包中的over()函數所需的SpatialPoints和SpatialPolygons對象，它完成了計算點和多邊形的“交點”的真正繁重工作：

library(sp)
library(maps)
library(maptools)

# The single argument to this function, pointsDF, is a data.frame in which:
#   - column 1 contains the longitude in degrees (negative in the US)
#   - column 2 contains the latitude in degrees

lonlat_to_state_sp <- function(pointsDF) {
    # Prepare SpatialPolygons object with one SpatialPolygon
    # per state (plus DC, minus HI & AK)
    states <- map('state', fill=TRUE, col="transparent", plot=FALSE)
    IDs <- sapply(strsplit(states$names, ":"), function(x) x[1])
    states_sp <- map2SpatialPolygons(states, IDs=IDs,
                     proj4string=CRS("+proj=longlat +datum=WGS84"))

    # Convert pointsDF to a SpatialPoints object 
    pointsSP <- SpatialPoints(pointsDF, 
                    proj4string=CRS("+proj=longlat +datum=WGS84"))

    # Use 'over' to get _indices_ of the Polygons object containing each point 
        indices <- over(pointsSP, states_sp)

    # Return the state names of the Polygons object containing each point
    stateNames <- sapply(states_sp@polygons, function(x) x@ID)
    stateNames[indices]
}

# Test the function using points in Wisconsin and Oregon.
testPoints <- data.frame(x = c(-90, -120), y = c(44, 44))

lonlat_to_state_sp(testPoints)
[1] "wisconsin" "oregon" # IT WORKS

Answer 2

你可以用幾行 R 來完成。

library(sp)
library(rgdal)
#lat and long
Lat <- 57.25
Lon <- -9.41
#make a data frame
coords <- as.data.frame(cbind(Lon,Lat))
#and into Spatial
points <- SpatialPoints(coords)
#SpatialPolygonDataFrame - I'm using a shapefile of UK counties
counties <- readOGR(".", "uk_counties")
#assume same proj as shapefile!
proj4string(points) <- proj4string(counties)
#get county polygon point is in
result <- as.character(over(points, counties)$County_Name)

Answer 3

請參閱 sp 包中的?over 。

您需要將狀態邊界作為SpatialPolygonsDataFrame 。

Answer 4

示例數據（多邊形和點）

library(raster)
pols <- shapefile(system.file("external/lux.shp", package="raster"))
xy <- coordinates(p)

使用光柵::提取

extract(p, xy)

#   point.ID poly.ID ID_1       NAME_1 ID_2           NAME_2 AREA
#1         1       1    1     Diekirch    1         Clervaux  312
#2         2       2    1     Diekirch    2         Diekirch  218
#3         3       3    1     Diekirch    3          Redange  259
#4         4       4    1     Diekirch    4          Vianden   76
#5         5       5    1     Diekirch    5            Wiltz  263
#6         6       6    2 Grevenmacher    6       Echternach  188
#7         7       7    2 Grevenmacher    7           Remich  129
#8         8       8    2 Grevenmacher   12     Grevenmacher  210
#9         9       9    3   Luxembourg    8         Capellen  185
#10       10      10    3   Luxembourg    9 Esch-sur-Alzette  251
#11       11      11    3   Luxembourg   10       Luxembourg  237
#12       12      12    3   Luxembourg   11           Mersch  233

Answer 5

使用sf非常簡單：

library(maps)
library(sf)

## Get the states map, turn into sf object
US <- st_as_sf(map("state", plot = FALSE, fill = TRUE))

## Test the function using points in Wisconsin and Oregon
testPoints <- data.frame(x = c(-90, -120), y = c(44, 44))

# Make it a spatial dataframe, using the same coordinate system as the US spatial dataframe
testPoints <- st_as_sf(testPoints, coords = c("x", "y"), crs = st_crs(US))

#.. and perform a spatial join!
st_join(testPoints, US)


         ID        geometry
1 wisconsin  POINT (-90 44)
2    oregon POINT (-120 44)