如何使用sf :: st_centroid计算多边形的质心? [英] How to calculate centroid of polygon using sf::st_centroid?

问题描述

我正在尝试使用新的"sf"软件包来处理R中的某些巴西人口普查数据.我可以导入数据，但是在尝试创建原始多边形的质心时会出现错误

I am trying to manipulate some Brazilian Census data in R using the new "sf" package. I am able to import the data, but I get an error when I try to create the centroids of the original polygons

library(sf)

#Donwload data  
filepath <- 'ftp://geoftp.ibge.gov.br/organizacao_do_territorio/malhas_territoriais/malhas_de_setores_censitarios__divisoes_intramunicipais/censo_2010/setores_censitarios_shp/ac/ac_setores_censitarios.zip'
download.file(filepath,'ac_setores_censitarios.zip')
unzip('ac_setores_censitarios.zip')
d <- st_read('12SEE250GC_SIR.shp',stringsAsFactors = F) 

现在，我尝试创建一个新的几何列，其中包含列"geometry"的质心，但是会出现错误:

Now I try to create a new geometry column containing the centroid of column "geometry", but get an error:

d$centroid <- st_centroid(d$geometry)
Warning message:
In st_centroid.sfc(d$geometry) :
  st_centroid does not give correct centroids for longitude/latitude data

我该如何解决?

推荐答案

sf下的所有GEOS函数都需要投影坐标才能正常工作，因此应在适当投影的数据上运行st_centroid.我对巴西可用的CRS知之甚少，但EPSG:29101似乎运行良好:

All the GEOS functions underlying sf need projected coordinates to work properly, so you should run st_centroid on appropriately projected data. I don't know much about Brazil's available CRS's, but EPSG:29101 appears to work fine:

library(tidyverse)

d$centroids <- st_transform(d, 29101) %>% 
  st_centroid() %>% 
  # this is the crs from d, which has no EPSG code:
  st_transform(., '+proj=longlat +ellps=GRS80 +no_defs') %>%
  # since you want the centroids in a second geometry col:
  st_geometry()

# check with
plot(st_geometry(d))
plot(d[, 'centroids'], add = T, col = 'red', pch = 19)

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