我有一个sf
对象,其中包含通过.shp
文件获得的都市区域的多边形信息(区域)。对于给定的纬度/经度对,我想确定它属于哪个区域。我想我可以利用sf::st_contains()
,但是我无法以正确的格式获得lat / lon。
这可以是“矢量化”的。这是一个例子:
library(sf)
library(tidyverse)
新加坡shapefile:
singapore <- st_read("~/data/master-plan-2014-subzone-boundary-no-sea-shp/MP14_SUBZONE_NO_SEA_PL.shp", quiet=TRUE, stringsAsFactors=FALSE)
singapore <- st_transform(singapore, 4326)
回收中心的CSV:
centers <- read_csv("~/data/recycl.csv")
glimpse(centers)
## Observations: 407
## Variables: 10
## $ lng <dbl> 104.0055, 103.7677, 103.7456, 103.7361, 103.8106, 103.962...
## $ lat <dbl> 1.316764, 1.296245, 1.319204, 1.380412, 1.286512, 1.33355...
## $ inc_crc <chr> "F8907D68D7EB64A1", "ED1F74DC805CEC8B", "F48D575631DCFECB...
## $ name <chr> "RENEW (Recycling Nation's Electronic Waste)", "RENEW (Re...
## $ block_house_num <chr> "10", "84", "698", "3", "2", "1", "1", "1", "357", "50", ...
## $ bldg_name <chr> "Changi Water Reclamation Plant", "Clementi Woods", "Comm...
## $ floor <chr> NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, N...
## $ post_code <int> 498785, 126811, 608784, 689814, 159047, 486036, 39393, 55...
## $ street <chr> "Changi East Close", "West Coast Road , Clementi Woods Co...
## $ unit <chr> "(Lobby)", "#B1-01 (Management Office)", "(School foyer)"...
将^^转变为一个简单的特征对象:
map2(centers$lng, centers$lat, ~st_point(c(.x, .y))) %>%
st_sfc(crs = 4326) %>%
st_sf(centers[,-(1:2)], .) -> centers_sf
这可能比行方式更快,但我会让其他人玩得开心:
bind_cols(
centers,
singapore[as.numeric(st_within(centers_sf, singapore)),]
) %>%
select(lng, lat, inc_crc, subzone_name=SUBZONE_N) %>%
mutate(subzone_name = str_to_title(subzone_name))
## # A tibble: 407 x 4
## lng lat inc_crc subzone_name
## <dbl> <dbl> <chr> <chr>
## 1 104.0055 1.316764 F8907D68D7EB64A1 Changi Airport
## 2 103.7677 1.296245 ED1F74DC805CEC8B Clementi Woods
## 3 103.7456 1.319204 F48D575631DCFECB Teban Gardens
## 4 103.7361 1.380412 1F910E0086FD4798 Peng Siang
## 5 103.8106 1.286512 55A0B9E7CBD34AFE Alexandra Hill
## 6 103.9624 1.333555 C664D09D9CD5325F Xilin
## 7 103.8542 1.292778 411F79EAAECFE609 City Hall
## 8 103.8712 1.375876 F4516742CFD4228E Serangoon North Ind Estate
## 9 103.8175 1.293319 B05B32DF52D922E7 Alexandra North
## 10 103.9199 1.335878 58E9EAF06206C772 Bedok Reservoir
## # ... with 397 more rows
迟到的回应,我自己正在寻找答案。
结束了这个:
library(sf)
library(tidyverse)
nc = st_read(system.file("shape/nc.shp", package="sf"),
stringsAsFactors = FALSE)
d <-
data_frame(lon = runif(1e3, -84.5, -75.5),
lat = runif(1e3, 34, 36.6),
somevariable = rnorm(1e3, 1000, 100))
geo_inside <- function(lon, lat, map, variable) {
variable <- enquo(variable)
# slow if lots of lons and lats or big sf - needs improvement
pt <-
tibble::data_frame(x = lon,
y = lat) %>%
st_as_sf(coords = c("x", "y"), crs = st_crs(map))
pt %>% st_join(map) %>% pull(!!variable)
}
d <-
d %>%
mutate(county = geo_inside(lon, lat, nc, NAME))
glimpse(d)
Observations: 1,000
Variables: 4
$ lon <dbl> -79.68728, -79.06104, -83.92082, -76.36866, -75.8635...
$ lat <dbl> 36.11349, 35.67239, 35.08802, 35.78083, 36.55786, 34...
$ somevariable <dbl> 910.9803, 1010.6816, 919.3937, 924.0845, 1154.0975, ...
$ county <chr> "Guilford", "Chatham", "Cherokee", "Tyrrell", NA, NA...
d %>%
ggplot() +
geom_sf(data = nc) +
geom_point(aes(lon, lat, colour = county)) +
theme(legend.position = "none")
虽然对速度不满意,但似乎做到了这一点。
埃纳尔
在lon / lat上使用st_point()
然后它可以与其他sf
函数一起使用。
例:
find_precinct <- function(precincts, point) {
precincts %>%
filter(st_contains(geometry, point) == 1) %>%
`[[`("WARDS_PREC")
}
ggmap::geocode("nicollet mall, st paul") %>%
rowwise() %>%
mutate(point = c(lon, lat) %>%
st_point() %>%
list(),
precinct = find_precinct(msvc_precincts, point)
)
如果你有一个坐标数据框架(mydf
),将它们转换为sf对象,然后与sf map
多边形相交:
mydf_sf <- sf::st_as_sf(mydf, coords=c("lon","lat"), crs=4326)
int <- sf::st_intersects(mydf_sf , map)
mydf$country <- map$country_name[unlist(int)]
在https://gis.stackexchange.com/a/318629/36710有一个完整的工作示例