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如何将夏威夷和阿拉斯加添加到以下代码（取自 Josh O'Brien 的回答：Latitude Longitude Coordinates to State Code in R）？

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

latlong2state <- 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 Alaska (ak) and Hawaii (hi)

ak <- data.frame(lon = c(-151.0074), lat = c(63.0694))
hi <- data.frame(lon = c(-157.8583), lat = c(21.30694))
nc <- data.frame(lon = c(-77.335), lat = c(34.671))


latlong2state(ak)
latlong2state(hi)
latlong2state(nc)

latlong2state(ak)and代码返回 NA ，latlong2state(hi)但如果代码修改正确，阿拉斯加和夏威夷将作为结果返回。

任何帮助表示赞赏！

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2 回答 2

5

您需要使用包含 50 个州的地图，而您正在使用的地图states <- map('state', fill=TRUE, col="transparent", plot=FALSE)没有夏威夷和阿拉斯加。

例如，您可以从这里下载 20m 美国地图，并将其解压缩到您的当前目录中。然后，您应该cb_2013_us_state_5m在 R 当前目录中有一个名为的文件夹。

我已经对您发布的代码进行了一些调整，为夏威夷和阿尔萨卡工作，没有尝试过其他状态。

library(sp)
library(rgeos)
library(rgdal)

# 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

latlong2state <- function(pointsDF) {
  states <-readOGR(dsn='cb_2013_us_state_5m',layer='cb_2013_us_state_5m')
  states <- spTransform(states, CRS("+proj=longlat"))

  pointsSP <- SpatialPoints(pointsDF,proj4string=CRS("+proj=longlat"))

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

# Test the function using points in Alaska (ak) and Hawaii (hi)

ak <- data.frame(lon = c(-151.0074), lat = c(63.0694))
hi <- data.frame(lon = c(-157.8583), lat = c(21.30694))

latlong2state(ak)
latlong2state(hi)
于 2015-02-10T00:46:09.513 回答
2

那是基于包中的一个数据集maps，它只包括较低的 48 个。对于你的任务，它需要一个包含所有状态的 shapefile。Census.gov 网站始终是找到这些的好地方。我对您发布的函数进行了一些更改，以便它可以与这个新的 shapefile 一起使用。

#download a shapefile with ALL states
tmp_dl <- tempfile()
download.file("http://www2.census.gov/geo/tiger/GENZ2013/cb_2013_us_state_20m.zip", tmp_dl)
unzip(tmp_dl, exdir=tempdir())
ST <- readOGR(tempdir(), "cb_2013_us_state_20m")

latlong2state <- function(pointsDF) {
    # Just copied the earlier code with some key changes
    states <- ST

    # Convert pointsDF to a SpatialPoints object 
    # USING THE CRS THAT MATCHES THE SHAPEFILE
    pointsCRS <- "+proj=longlat +datum=NAD83 +no_defs +ellps=GRS80 +towgs84=0,0,0"
    pointsSP <- SpatialPoints(pointsDF, proj4string=CRS(pointsCRS))

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

    # Return the state names of the Polygons object containing each point
    as.vector(indices$NAME)
}

让我们测试一下！

ak <- data.frame(lon = c(-151.0074), lat = c(63.0694))
hi <- data.frame(lon = c(-157.8583), lat = c(21.30694))
nc <- data.frame(lon = c(-77.335), lat = c(34.671))

latlong2state(ak)
[1] "Alaska"

latlong2state(hi)
[1] "Hawaii"

latlong2state(nc)
[1] "North Carolina"
于 2015-02-10T01:08:22.980 回答