我有一个脚本,可使用tidycensus软件包提取GeoID。这些数据非常有价值,但我也想参考经纬度(即使它们是近似值)。问题是,我不知道怎么做或是否有可能。或者,是否有任何方法可以利用另一个空间包来合并此类数据。请参见下面的代码示例。
>>> arr = [*range(5)]
>>> print(*arr)
0 1 2 3 4
谢谢!
答案 0 :(得分:0)
如果您在geometry = TRUE
调用中指定get_decennial()
,则tidycensus
将返回多边形几何以及普查估计数。然后,您可以使用sf::st_centroid()
获取每个多边形的质心(中间)。
在此示例中,我获得了人口普查数据和几何图形,然后使用每个县多边形的形心创建第二个sf
对象。最后,创建一个快速地图,以便您可以看到多边形及其质心。
library(tidyverse)
library(tidycensus)
library(sf)
#> Linking to GEOS 3.6.1, GDAL 2.1.3, PROJ 4.9.3
county <- get_decennial(
geography = "county",
state = "SC",
variables = c(county = "COUNTY", state = "STATE"),
output = "wide",
geometry = TRUE
)
#> Getting data from the 2010 decennial Census
count_centroid <- county %>%
st_transform(2273) %>% # convert to projected coord system for better centroid
st_centroid()
#> Warning in st_centroid.sf(.): st_centroid assumes attributes are constant
#> over geometries of x
head(count_centroid)
#> Simple feature collection with 6 features and 4 fields
#> geometry type: POINT
#> dimension: XY
#> bbox: xmin: 1506653 ymin: 420377.9 xmax: 1983425 ymax: 1170306
#> epsg (SRID): 2273
#> proj4string: +proj=lcc +lat_1=34.83333333333334 +lat_2=32.5 +lat_0=31.83333333333333 +lon_0=-81 +x_0=609600 +y_0=0 +ellps=GRS80 +towgs84=0,0,0,0,0,0,0 +units=ft +no_defs
#> # A tibble: 6 x 5
#> GEOID NAME county state geometry
#> <chr> <chr> <dbl> <dbl> <POINT [ft]>
#> 1 45009 Bamberg County, South Carolina 9 45 (1983425 502618.9)
#> 2 45001 Abbeville County, South Carolina 1 45 (1559068 872366.5)
#> 3 45005 Allendale County, South Carolina 5 45 (1890285 420377.9)
#> 4 45007 Anderson County, South Carolina 7 45 (1506653 981064.2)
#> 5 45021 Cherokee County, South Carolina 21 45 (1814307 1170306)
#> 6 45011 Barnwell County, South Carolina 11 45 (1866998 521545.1)
ggplot(county) +
geom_sf() +
geom_sf(data = count_centroid) +
theme_void()
由reprex package(v0.3.0)于2019-06-16创建