Question

我试图用R反转地理编码。我首先使用ggmap，但无法使用我的API密钥。现在我用googleway尝试它。

newframe[,c("Front.lat","Front.long")]

  Front.lat Front.long
1 -37.82681   144.9592
2 -37.82681   145.9592

newframe$address <- apply(newframe, 1, function(x){
  google_reverse_geocode(location = as.numeric(c(x["Front.lat"], 
x["Front.long"])),
                         key = "xxxx")
})

这将变量作为列表提取，但我无法弄清楚结构。

我正在努力弄清楚如何将下面列出的地址组件提取为newframe中的变量

postal_code，administrative_area_level_1，administrative_area_level_2，locality，route，street_number

我希望每个地址组件都是一个单独的变量。

Answer 1

Google的API以JSON格式返回响应。其中，当翻译成R时自然形成嵌套列表。在googleway内部，这是通过jsonlite::fromJSON()

完成的

在googleway中，通过使用simplify参数，我已经选择返回原始JSON或列表。

我故意从Google的回复中返回所有数据，并将其留给用户通过常规的列表子集操作提取他们感兴趣的元素。

说了这么多，在googleway的开发版本中，我写了一些函数来帮助访问各种API调用的元素。以下是其中三个可能对您有用的内容

## Install the development version
# devtools::install_github("SymbolixAU/googleway")

res <- google_reverse_geocode(
  location = c(df[1, 'Front.lat'], df[1, 'Front.long']), 
  key = apiKey
  )

geocode_address(res)
# [1] "45 Clarke St, Southbank VIC 3006, Australia"                    
# [2] "Bank Apartments, 275-283 City Rd, Southbank VIC 3006, Australia"
# [3] "Southbank VIC 3006, Australia"                                  
# [4] "Melbourne VIC, Australia"                                       
# [5] "South Wharf VIC 3006, Australia"                                
# [6] "Melbourne, VIC, Australia"                                      
# [7] "CBD & South Melbourne, VIC, Australia"                          
# [8] "Melbourne Metropolitan Area, VIC, Australia"                    
# [9] "Victoria, Australia"                                            
# [10] "Australia"

geocode_address_components(res)
#        long_name short_name                                  types
# 1             45         45                          street_number
# 2  Clarke Street  Clarke St                                  route
# 3      Southbank  Southbank                    locality, political
# 4 Melbourne City  Melbourne administrative_area_level_2, political
# 5       Victoria        VIC administrative_area_level_1, political
# 6      Australia         AU                     country, political
# 7           3006       3006                            postal_code

geocode_type(res)
# [[1]]
# [1] "street_address"
# 
# [[2]]
# [1] "establishment"      "general_contractor" "point_of_interest" 
# 
# [[3]]
# [1] "locality"  "political"
# 
# [[4]]
# [1] "colloquial_area" "locality"        "political"

Answer 2

在对newframe $地址进行反向地理编码后，可以进一步提取地址组件，如下所示：

# Make a boolean array of the valid ("OK" status) responses (other statuses may be "NO_RESULTS", "REQUEST_DENIED" etc).
sel <- sapply(c(1: nrow(newframe)), function(x){
  newframe$address[[x]]$status == 'OK'
})

# Get the address_components of the first result (i.e. best match) returned per geocoded coordinate.
address.components <- sapply(c(1: nrow(newframe[sel,])), function(x){
  newframe$address[[x]]$results[1,]$address_components
})

# Get all possible component types.
all.types <- unique(unlist(sapply(c(1: length(address.components)), function(x){
  unlist(lapply(address.components[[x]]$types, function(l) l[[1]]))
})))

# Get "long_name" values of the address_components for each type present (the other option is "short_name").
all.values <- lapply(c(1: length(address.components)), function(x){
  types <- unlist(lapply(address.components[[x]]$types, function(l) l[[1]]))
  matches <- match(all.types, types)
  values <- address.components[[x]]$long_name[matches]
})

# Bind results into a dataframe.
all.values <- do.call("rbind", all.values)
all.values <- as.data.frame(all.values)
names(all.values) <- all.types

# Add columns and update original data frame.
newframe[, all.types] <- NA
newframe[sel,][, all.types] <- all.values

请注意，我只保留了每个组件给出的第一种类型，有效地跳过了'政治＆＃34;键入，因为它出现在多个组件中，可能是多余的，例如＆＃34; administrative_area_level_1，政治＆＃34;。

Answer 3

您可以轻松使用ggmap:revgeocode;看下面：

library(ggmap)
df <- cbind(df,do.call(rbind,
        lapply(1:nrow(df),
          function(i) 
            revgeocode(as.numeric(
              df[i,2:1]), output = "more")      
                [c("administrative_area_level_1","locality","postal_code","address")])))

#output:
df
#   Front.lat Front.long administrative_area_level_1  locality
#   1 -37.82681   144.9592                    Victoria Southbank
#   2 -37.82681   145.9592                    Victoria    Noojee
#     postal_code                                     address
#   1        3006 45 Clarke St, Southbank VIC 3006, Australia
#   2        3833 Cec Dunns Track, Noojee VIC 3833, Australia

您可以将"route"和"street_number"添加到要提取的变量中，但是您可以看到第二个地址没有街道号，这会导致错误。

注意： 您也可以使用sub并从地址中提取信息。

<强> 数据：的

df <- structure(list(Front.lat = c(-37.82681, -37.82681), Front.long = 
      c(144.9592, 145.9592)), .Names = c("Front.lat", "Front.long"), class = "data.frame", 
      row.names = c(NA, -2L))

反向地理编码以提取地址组件

3 个答案: