如何使用st_join()与sf包进行空间连接

Ben*_*Ben 20 r spatial r-sf

这是我一直在努力的玩具示例

# Make points
point1 <- c(.5, .5)
point2 <- c(.6, .6)
point3 <- c(3, 3)
mpt <- st_multipoint(rbind(point1, point2, point3))  # create multipoint

# Make polygons
square1 <- rbind(c(0, 0), c(1, 0), c(1,1), c(0, 1), c(0, 0))
square2 <- rbind(c(0, 0), c(2, 0), c(2,2), c(0, 2), c(0, 0))
square3 <- rbind(c(0, 0), c(-1, 0), c(-1,-1), c(0, -1), c(0, 0))
mpol <- st_multipolygon(list(list(square1), list(square2), list(square2)))  # create multipolygon

# Convert to class' sf'
pts <- st_sf(st_sfc(mpt))
polys <- st_sf(st_sfc(mpol))

# Determine which points fall inside which polygons
st_join(pts, polys, join = st_contains)
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最后一行产生

Error in as.data.frame.default(x[[i]], optional = TRUE, stringsAsFactors = stringsAsFactors) : 
  cannot coerce class "c("sfc_MULTIPOINT", "sfc")" to a data.frame
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如何进行空间连接以确定哪些点落在哪个多边形内?

lbu*_*ett 14

我也在努力解决sf软件包的功能,如果这不正确或有更好的方法,请道歉.我认为这里的一个问题是,如果像你的例子那样构建几何图形,你就不会得到你的想法:

> pts
Simple feature collection with 1 feature and 0 fields
geometry type:  MULTIPOINT
dimension:      XY
bbox:           xmin: 0.5 ymin: 0.5 xmax: 3 ymax: 3
epsg (SRID):    NA
proj4string:    NA
                     st_sfc.mpt.
1 MULTIPOINT(0.5 0.5, 0.6 0.6...

> polys
Simple feature collection with 1 feature and 0 fields
geometry type:  MULTIPOLYGON
dimension:      XY
bbox:           xmin: 0 ymin: 0 xmax: 2 ymax: 2
epsg (SRID):    NA
proj4string:    NA
                    st_sfc.mpol.
1 MULTIPOLYGON(((0 0, 1 0, 1 ...
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你可以看到,你只有一个"功能"无论是在ptspolys.这意味着您正在构建一个"多面"特征(即由3个部分构成的多边形),而不是三个不同的多边形.这些点也是如此.

在挖掘了一下之后,我发现使用WKT表示法构建几何图形的方式不同(在我看来更容易):

polys <- st_as_sfc(c("POLYGON((0 0 , 0 1 , 1 1 , 1 0, 0 0))",
                     "POLYGON((0 0 , 0 2 , 2 2 , 2 0, 0 0 ))", 
                     "POLYGON((0 0 , 0 -1 , -1 -1 , -1 0, 0 0))")) %>% 
  st_sf(ID = paste0("poly", 1:3))    

pts <- st_as_sfc(c("POINT(0.5 0.5)",
                   "POINT(0.6 0.6)",
                   "POINT(3 3)")) %>%
  st_sf(ID = paste0("point", 1:3))

> polys
Simple feature collection with 3 features and 1 field
geometry type:  POLYGON
dimension:      XY
bbox:           xmin: -1 ymin: -1 xmax: 2 ymax: 2
epsg (SRID):    NA
proj4string:    NA
     ID                              .
1 poly1 POLYGON((0 0, 0 1, 1 1, 1 0...
2 poly2 POLYGON((0 0, 0 2, 2 2, 2 0...
3 poly3 POLYGON((0 0, 0 -1, -1 -1, ...

> pts
Simple feature collection with 3 features and 1 field
geometry type:  POINT
dimension:      XY
bbox:           xmin: 0.5 ymin: 0.5 xmax: 3 ymax: 3
epsg (SRID):    NA
proj4string:    NA
      ID              .
1 point1 POINT(0.5 0.5)
2 point2 POINT(0.6 0.6)
3 point3     POINT(3 3)
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你现在都可以看到,polyspts有三个特点.

我们现在可以使用以下方法找到"交叉矩阵":

# Determine which points fall inside which polygons
pi <- st_contains(polys,pts, sparse = F) %>% 
  as.data.frame() %>% 
  mutate(polys = polys$ID) %>% 
  select(dim(pi)[2],1:dim(pi)[1])
colnames(pi)[2:dim(pi)[2]] = levels(pts$ID)

> pi
  polys point1 point2 point3
1 poly1   TRUE   TRUE  FALSE
2 poly2   TRUE   TRUE  FALSE
3 poly3  FALSE  FALSE  FALSE
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意义(在评论中指出@symbolixau)多边形1和2包含点1和2,而多边形3不包含任何点.相反,点3不包含在任何多边形中.

HTH.

  • 我认为**结果是说点1和2都在多边形1和2中(它们重叠),而点3不在任何多边形中,并且多边形3不包含任何点.(我使用除了`sf`之外的其他方法来尝试同样的问题来帮助解释结果) (4认同)
  • 很有意思.希望其他人可以加入这个问题.我希望`sf`文档有更好的`st_join`示例,并从头开始构建简单的功能. (2认同)
  • 如果您正在寻找有关空间功能的更好文档,可以在这里查看https://postgis.net/docs/reference.html (2认同)

Edz*_*sma 5

我看到不同的输出:

> # Determine which points fall inside which polygons
> st_join(pts, polys, join = st_contains)
Simple feature collection with 1 feature and 0 fields
geometry type:  MULTIPOINT
dimension:      XY
bbox:           xmin: 0.5 ymin: 0.5 xmax: 3 ymax: 3
epsg (SRID):    NA
proj4string:    NA
                        geometry
1 MULTIPOINT(0.5 0.5, 0.6 0.6...
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这是最新的 CRAN 版本吗sf