Question

如何使用经度和纬度找到最近的机场站？

例如我在我的数据库中有这个json数据存储，

 "location" : {
    "long" : "Devon, 8 Market Road, Plympton, Plymouth PL7 1QW, United Kingdom",
    "street_number" : "",
    "route" : "Market Road",
    "locality" : "Plymouth",
    "administrative_area_level_1" : "England",
    "country" : "United Kingdom",
    "postal_code" : "PL7 1QW",
    "lat" : "50.38693379999999",
    "lng" : "-4.0598999999999705"
}

我知道我的地点是Plymouth，因此我将通过以下网址向 Weather Underground 请求电台数据：

http://api.wunderground.com/api/[MY-API-CODE]/geolookup/conditions/q/UK/Plymouth.json

我是这样做的：

locality <- 'Plymouth'

pullUrl <- paste(apiUrl, 'geolookup/conditions/q/UK/', locality, '.json', sep='')

# Reading in as raw lines from the web service.
conn <- url(pullUrl)
rawData <- readLines(conn, n=-1L, ok=TRUE)

# Convert to a JSON.
geoData <- fromJSON(paste(rawData, collapse=""))

# Get the station data in location only.
# Turn the result into a data frame.
stationsDF <- as.data.frame(geoData$location$nearby_weather_stations$airport$station)

所以我得到3个站点：

      city state        country icao         lat         lon
1 Plymouth       United Kingdom EGDB 50.35491562 -4.12105608
2   Exeter                   UK EGTE 50.73714066 -3.40577006
3 Culdrose                   UK EGDR 50.08427429 -5.25711393

但我的问题是如何确保我获得EGDB 而不是 EGTE或EGDR - 因为Plympton 更接近到普利茅斯？

那么我可以在我的数据库中使用下面的lat和lng来确定哪个电台是最近的吗？

"lat" : "50.38693379999999",
"lng" : "-4.0598999999999705"

那我怎么知道上面的lat和lng 去 EGDB 50.35491562 -4.12105608？

有什么想法吗？

修改

stationsDF <- as.data.frame(geoData$location$nearby_weather_stations$airport$station, stringsAsFactors=FALSE)
df <- setDT(stationsDF)
loc <- c(lat = "50.38693379999999", lng = "-4.0598999999999705")
dists <- geosphere::distHaversine(as.numeric(loc[c('lng', 'lat')]), df[, c('lon', 'lat')])

错误：

Error in .pointsToMatrix(p2) * toRad : 
  non-numeric argument to binary operator
In addition: Warning message:
In .pointsToMatrix(p2) : NAs introduced by coercion

编辑2：

stationsDF <- as.data.frame(geoData$location$nearby_weather_stations$airport$station, stringsAsFactors=FALSE)
dput(stationsDF)

输出：

structure(list(city = c("Plymouth", "Exeter", "Culdrose"), state = c("", 
"", ""), country = c("United Kingdom", "UK", "UK"), icao = c("EGDB", 
"EGTE", "EGDR"), lat = c("50.35491562", "50.73714066", "50.08427429"
), lon = c("-4.12105608", "-3.40577006", "-5.25711393")), .Names = c("city", 
"state", "country", "icao", "lat", "lon"), class = "data.frame", row.names = c(NA, 
-3L))

编辑3：

虽然：

str(stationsDF)

输出：

'data.frame':   3 obs. of  6 variables:
 $ city   : chr  "Plymouth" "Exeter" "Culdrose"
 $ state  : chr  "" "" ""
 $ country: chr  "United Kingdom" "UK" "UK"
 $ icao   : chr  "EGDB" "EGTE" "EGDR"
 $ lat    : chr  "50.35491562" "50.73714066" "50.08427429"
 $ lon    : chr  "-4.12105608" "-3.40577006" "-5.25711393"

Answer 1

如果您已有数据，请说

df <- read.table(text = 'city state        country icao         lat         lon
                   1 Plymouth     "United Kingdom" EGDB 50.35491562 -4.12105608
                   2   Exeter                   UK EGTE 50.73714066 -3.40577006
                   3 Culdrose                   UK EGDR 50.08427429 -5.25711393', head = T)

loc <- c(lat = "50.38693379999999", lng = "-4.0598999999999705")

然后，您可以使用geosphere::distHaversine来计算loc与df的每次观察之间的距离（默认为米）：

dists <- geosphere::distHaversine(as.numeric(loc[c('lng', 'lat')]), df[, c('lon', 'lat')])

dists
## [1]  5617.667 60493.398 91661.079

使用which.min，您可以为df编制索引以获得结果：

df[which.min(dists), ]
##   city    state        country icao      lat       lon
## 1    1 Plymouth United Kingdom EGDB 50.35492 -4.121056

R + Weather Underground - 如何使用经度和纬度来获得最近的机场站？

1 个答案: