将sf转换为未标记的ppp

Question

我希望将一个 sf 对象转换成一个未标记的 ppp 对象。（根据 这篇文章，现在支持从 sf 转换成 ppp。）

library(sf)

#Initialise sf object
pp <- structure(list(X = c(959207.877070254, 959660.734838225, 951483.685462513,  951527.767554883, 958310.673042469, 950492.05212104, 959660.734838225,  959207.877070254, 960500.020456073, 959660.734838225), Y = c(1944457.42827898,  1955543.76027363, 1939982.16629396, 1940216.55143212, 1954704.68186897,  1951434.68524296, 1955543.76027363, 1944457.42827898, 1955292.64874361,  1955543.76027363), geometry = structure(list(structure(c(959207.877070254,  1944457.42827898), class = c("XY", "POINT", "sfg")), structure(c(959660.734838225,  1955543.76027363), class = c("XY", "POINT", "sfg")), structure(c(951483.685462513,  1939982.16629396), class = c("XY", "POINT", "sfg")), structure(c(951527.767554883,  1940216.55143212), class = c("XY", "POINT", "sfg")), structure(c(958310.673042469,  1954704.68186897), class = c("XY", "POINT", "sfg")), structure(c(950492.05212104,  1951434.68524296), class = c("XY", "POINT", "sfg")), structure(c(959660.734838225,  1955543.76027363), class = c("XY", "POINT", "sfg")), structure(c(959207.877070254,  1944457.42827898), class = c("XY", "POINT", "sfg")), structure(c(960500.020456073,  1955292.64874361), class = c("XY", "POINT", "sfg")), structure(c(959660.734838225,  1955543.76027363), class = c("XY", "POINT", "sfg"))), class = c("sfc_POINT",  "sfc"), precision = 0, bbox = structure(c(xmin = 950492.05212104,  ymin = 1939982.16629396, xmax = 960500.020456073, ymax = 1955543.76027363 ), class = "bbox"), crs = structure(list(input = "EPSG:5179",      wkt = "PROJCRS[\"Korea 2000 / Unified CS\",\n    BASEGEOGCRS[\"Korea 2000\",\n        DATUM[\"Geocentric datum of Korea\",\n            ELLIPSOID[\"GRS 1980\",6378137,298.257222101,\n                LENGTHUNIT[\"metre\",1]]],\n        PRIMEM[\"Greenwich\",0,\n            ANGLEUNIT[\"degree\",0.0174532925199433]],\n        ID[\"EPSG\",4737]],\n    CONVERSION[\"Korea Unified Belt\",\n        METHOD[\"Transverse Mercator\",\n            ID[\"EPSG\",9807]],\n        PARAMETER[\"Latitude of natural origin\",38,\n            ANGLEUNIT[\"degree\",0.0174532925199433],\n            ID[\"EPSG\",8801]],\n        PARAMETER[\"Longitude of natural origin\",127.5,\n            ANGLEUNIT[\"degree\",0.0174532925199433],\n            ID[\"EPSG\",8802]],\n        PARAMETER[\"Scale factor at natural origin\",0.9996,\n            SCALEUNIT[\"unity\",1],\n            ID[\"EPSG\",8805]],\n        PARAMETER[\"False easting\",1000000,\n            LENGTHUNIT[\"metre\",1],\n            ID[\"EPSG\",8806]],\n        PARAMETER[\"False northing\",2000000,\n            LENGTHUNIT[\"metre\",1],\n            ID[\"EPSG\",8807]]],\n    CS[Cartesian,2],\n        AXIS[\"northing (X)\",north,\n            ORDER[1],\n            LENGTHUNIT[\"metre\",1]],\n        AXIS[\"easting (Y)\",east,\n            ORDER[2],\n            LENGTHUNIT[\"metre\",1]],\n    USAGE[\n        SCOPE[\"unknown\"],\n        AREA[\"Korea, Republic of (South Korea)\"],\n        BBOX[28.6,122.71,40.27,134.28]],\n    ID[\"EPSG\",5179]]"), class = "crs"), n_empty = 0L)), row.names = c(4177L,  15721L, 21365L, 21973L, 24836L, 59359L, 66313L, 70379L, 83277L,  90828L), class = c("sf", "data.frame"), sf_column = "geometry", agr = structure(c(X = NA_integer_,  Y = NA_integer_), .Label = c("constant", "aggregate", "identity" ), class = "factor"))

library(spatstat)

#Convert to ppp format: method 1
pp1 <- as.ppp(pp)
#Warning message:
#In as.ppp.sf(pp) : only first attribute column is used for marks
pp1 <- unmark(pp1)

#Convert to ppp format: method 2
pp2 <- as.ppp(st_coordinates(pp), st_convex_hull(st_union(pp)))
#Warning message:
#data contain duplicated points

identical(pp1, pp2)
[1] FALSE

第一种方法需要使用标记，我没有找到如何关闭它的方法（指定 marks=NULL 无法生效）。此后我会将标记移除，但这只是个权宜之计。

第二种方法可能有误，因为有警告出现，尽管我是依据这个解决方案来制作的。这两个输出并不相同。

这两种方法中是否存在正确的方法？为什么第二种方法会产生重复的点？有没有更直接的方法可以进行转化而不需要解决问题？

- syre

Answer 1

这两个命令应该返回相同的结果。不同的结果是由于最近对“spatstat”包进行更改所导致的临时问题。

（这些更改打破了一些由包“sf”、“maptools”和“sp”做出的假设，导致第二个命令无法找到正确的“as.owin”方法，因此它默认为矩形窗口）。

第一种方法没有问题。数据“pp”包含标记值，因此“as.ppp”处理标记，并警告只使用了第一列。如果您不想要标记，请在之后使用“unmark”。

第二种方法中收到的警告告诉您某些数据点位于同一位置。在第一种方法中，您不会收到此警告，因为位于相同位置的点具有不同的标记值。

警告不是错误。除了“as.owin”的分派的临时问题外，这两种方法都可以。

我建议您使用第一种方法。

如果您确实需要使用第二种方法，则需要更新所有这些包，并且在“spatstat”的情况下，安装可在github存储库中获得的开发版本。 如果仍然存在问题，请等待大约一周，直到所有四个包的完成包更新发布在CRAN上。

- Adrian Baddeley

---

我的错误，因为我还没有领悟到显而易见的事实，即几何列在 sf 中足以定位点，所以我无用地将坐标 X 和 Y 复制到单独的列中，然后 spatstat 将其解释为标记... - syre

Answer 2

我不确定以下回答是否适用于每个版本的 spatstat，但我认为如果您想生成一个未标记的 ppp 对象，您应该运行：

# packages
suppressPackageStartupMessages({
  library(sf)
  library(spatstat)
})

# Initialise sf object
pp <- structure(list(X = c(959207.877070254, 959660.734838225, 951483.685462513,  951527.767554883, 958310.673042469, 950492.05212104, 959660.734838225,  959207.877070254, 960500.020456073, 959660.734838225), Y = c(1944457.42827898,  1955543.76027363, 1939982.16629396, 1940216.55143212, 1954704.68186897,  1951434.68524296, 1955543.76027363, 1944457.42827898, 1955292.64874361,  1955543.76027363), geometry = structure(list(structure(c(959207.877070254,  1944457.42827898), class = c("XY", "POINT", "sfg")), structure(c(959660.734838225,  1955543.76027363), class = c("XY", "POINT", "sfg")), structure(c(951483.685462513,  1939982.16629396), class = c("XY", "POINT", "sfg")), structure(c(951527.767554883,  1940216.55143212), class = c("XY", "POINT", "sfg")), structure(c(958310.673042469,  1954704.68186897), class = c("XY", "POINT", "sfg")), structure(c(950492.05212104,  1951434.68524296), class = c("XY", "POINT", "sfg")), structure(c(959660.734838225,  1955543.76027363), class = c("XY", "POINT", "sfg")), structure(c(959207.877070254,  1944457.42827898), class = c("XY", "POINT", "sfg")), structure(c(960500.020456073,  1955292.64874361), class = c("XY", "POINT", "sfg")), structure(c(959660.734838225,  1955543.76027363), class = c("XY", "POINT", "sfg"))), class = c("sfc_POINT",  "sfc"), precision = 0, bbox = structure(c(xmin = 950492.05212104,  ymin = 1939982.16629396, xmax = 960500.020456073, ymax = 1955543.76027363 ), class = "bbox"), crs = structure(list(input = "EPSG:5179",      wkt = "PROJCRS[\"Korea 2000 / Unified CS\",\n    BASEGEOGCRS[\"Korea 2000\",\n        DATUM[\"Geocentric datum of Korea\",\n            ELLIPSOID[\"GRS 1980\",6378137,298.257222101,\n                LENGTHUNIT[\"metre\",1]]],\n        PRIMEM[\"Greenwich\",0,\n            ANGLEUNIT[\"degree\",0.0174532925199433]],\n        ID[\"EPSG\",4737]],\n    CONVERSION[\"Korea Unified Belt\",\n        METHOD[\"Transverse Mercator\",\n            ID[\"EPSG\",9807]],\n        PARAMETER[\"Latitude of natural origin\",38,\n            ANGLEUNIT[\"degree\",0.0174532925199433],\n            ID[\"EPSG\",8801]],\n        PARAMETER[\"Longitude of natural origin\",127.5,\n            ANGLEUNIT[\"degree\",0.0174532925199433],\n            ID[\"EPSG\",8802]],\n        PARAMETER[\"Scale factor at natural origin\",0.9996,\n            SCALEUNIT[\"unity\",1],\n            ID[\"EPSG\",8805]],\n        PARAMETER[\"False easting\",1000000,\n            LENGTHUNIT[\"metre\",1],\n            ID[\"EPSG\",8806]],\n        PARAMETER[\"False northing\",2000000,\n            LENGTHUNIT[\"metre\",1],\n            ID[\"EPSG\",8807]]],\n    CS[Cartesian,2],\n        AXIS[\"northing (X)\",north,\n            ORDER[1],\n            LENGTHUNIT[\"metre\",1]],\n        AXIS[\"easting (Y)\",east,\n            ORDER[2],\n            LENGTHUNIT[\"metre\",1]],\n    USAGE[\n        SCOPE[\"unknown\"],\n        AREA[\"Korea, Republic of (South Korea)\"],\n        BBOX[28.6,122.71,40.27,134.28]],\n    ID[\"EPSG\",5179]]"), class = "crs"), n_empty = 0L)), row.names = c(4177L,  15721L, 21365L, 21973L, 24836L, 59359L, 66313L, 70379L, 83277L,  90828L), class = c("sf", "data.frame"), sf_column = "geometry", agr = structure(c(X = NA_integer_,  Y = NA_integer_), .Label = c("constant", "aggregate", "identity" ), class = "factor"))

# Convert to ppp format: method 1
(pp1 <- as.ppp(st_geometry(pp)))
#> Planar point pattern: 10 points
#> window: rectangle = [950492.1, 960500] x [1939982.2, 1955543.8] units

^{本文内容于2021-03-01由reprex package (v0.3.0)创建}

你提供的两种方法并不等价，因为sf包中定义的as.ppp.sfc函数使用了一个矩形的owin对象：

# packages
suppressPackageStartupMessages({
  library(sf)
  library(spatstat)
})

#Initialise sf object
pp <- structure(list(X = c(959207.877070254, 959660.734838225, 951483.685462513,  951527.767554883, 958310.673042469, 950492.05212104, 959660.734838225,  959207.877070254, 960500.020456073, 959660.734838225), Y = c(1944457.42827898,  1955543.76027363, 1939982.16629396, 1940216.55143212, 1954704.68186897,  1951434.68524296, 1955543.76027363, 1944457.42827898, 1955292.64874361,  1955543.76027363), geometry = structure(list(structure(c(959207.877070254,  1944457.42827898), class = c("XY", "POINT", "sfg")), structure(c(959660.734838225,  1955543.76027363), class = c("XY", "POINT", "sfg")), structure(c(951483.685462513,  1939982.16629396), class = c("XY", "POINT", "sfg")), structure(c(951527.767554883,  1940216.55143212), class = c("XY", "POINT", "sfg")), structure(c(958310.673042469,  1954704.68186897), class = c("XY", "POINT", "sfg")), structure(c(950492.05212104,  1951434.68524296), class = c("XY", "POINT", "sfg")), structure(c(959660.734838225,  1955543.76027363), class = c("XY", "POINT", "sfg")), structure(c(959207.877070254,  1944457.42827898), class = c("XY", "POINT", "sfg")), structure(c(960500.020456073,  1955292.64874361), class = c("XY", "POINT", "sfg")), structure(c(959660.734838225,  1955543.76027363), class = c("XY", "POINT", "sfg"))), class = c("sfc_POINT",  "sfc"), precision = 0, bbox = structure(c(xmin = 950492.05212104,  ymin = 1939982.16629396, xmax = 960500.020456073, ymax = 1955543.76027363 ), class = "bbox"), crs = structure(list(input = "EPSG:5179",      wkt = "PROJCRS[\"Korea 2000 / Unified CS\",\n    BASEGEOGCRS[\"Korea 2000\",\n        DATUM[\"Geocentric datum of Korea\",\n            ELLIPSOID[\"GRS 1980\",6378137,298.257222101,\n                LENGTHUNIT[\"metre\",1]]],\n        PRIMEM[\"Greenwich\",0,\n            ANGLEUNIT[\"degree\",0.0174532925199433]],\n        ID[\"EPSG\",4737]],\n    CONVERSION[\"Korea Unified Belt\",\n        METHOD[\"Transverse Mercator\",\n            ID[\"EPSG\",9807]],\n        PARAMETER[\"Latitude of natural origin\",38,\n            ANGLEUNIT[\"degree\",0.0174532925199433],\n            ID[\"EPSG\",8801]],\n        PARAMETER[\"Longitude of natural origin\",127.5,\n            ANGLEUNIT[\"degree\",0.0174532925199433],\n            ID[\"EPSG\",8802]],\n        PARAMETER[\"Scale factor at natural origin\",0.9996,\n            SCALEUNIT[\"unity\",1],\n            ID[\"EPSG\",8805]],\n        PARAMETER[\"False easting\",1000000,\n            LENGTHUNIT[\"metre\",1],\n            ID[\"EPSG\",8806]],\n        PARAMETER[\"False northing\",2000000,\n            LENGTHUNIT[\"metre\",1],\n            ID[\"EPSG\",8807]]],\n    CS[Cartesian,2],\n        AXIS[\"northing (X)\",north,\n            ORDER[1],\n            LENGTHUNIT[\"metre\",1]],\n        AXIS[\"easting (Y)\",east,\n            ORDER[2],\n            LENGTHUNIT[\"metre\",1]],\n    USAGE[\n        SCOPE[\"unknown\"],\n        AREA[\"Korea, Republic of (South Korea)\"],\n        BBOX[28.6,122.71,40.27,134.28]],\n    ID[\"EPSG\",5179]]"), class = "crs"), n_empty = 0L)), row.names = c(4177L,  15721L, 21365L, 21973L, 24836L, 59359L, 66313L, 70379L, 83277L,  90828L), class = c("sf", "data.frame"), sf_column = "geometry", agr = structure(c(X = NA_integer_,  Y = NA_integer_), .Label = c("constant", "aggregate", "identity" ), class = "factor"))

# Convert to ppp format: method 1
(pp1 <- as.ppp(st_geometry(pp)))
#> Planar point pattern: 10 points
#> window: rectangle = [950492.1, 960500] x [1939982.2, 1955543.8] units

# Method 2
(pp2 <- as.ppp(
  X = st_coordinates(pp), 
  W = as.owin(st_bbox(pp))
))
#> Warning: data contain duplicated points
#> Planar point pattern: 10 points
#> window: rectangle = [950492.1, 960500] x [1939982.2, 1955543.8] units

identical(pp1, pp2)
#> [1] TRUE

此外，第一种方法返回没有警告信息，因为作者设置了check = FALSE。

anyDuplicated(pp1)
#> [1] TRUE

^{本文于2021年03月01日使用 reprex包 (v0.3.0) 创建。}

更多详情请参见此处。

- agila