如何有效地将经/纬对映射到邮政编码（无API） [英] How to efficiently map lat/long pairs to zipcodes (without an API)

问题描述

我正在将大量纬度/经度坐标对映射到相关的邮政编码。由于通话限制，记录数量太大，无法使用Google Maps或地名之类的API。

I'm mapping a large number of lat/long coordinate pairs to associated zip codes. The number of records is too large to use an API like Google Maps or geonames because of call limits.

我有一个查找表，其中包含邮政编码和每个邮政编码的纬度/经度中心。您可以在此处获取查询表：

I have a lookup table which contains zip codes and the lat/long centroid of each zip code. You can get the lookup table here:

# zipcode data with lat/lon coordinates
url <- "http://www.boutell.com/zipcodes/zipcode.zip"
fil <- "ziplatlong.zip"

# download an unzip
if (!file.exists(fil)) { download.file(url, fil) }
unzip(fil, exdir="zips")

library(readr)
ziplkp<-read_csv("zips/zipcode.csv")

对于我数据中的每个经纬度，通过查找经纬度对和查找表中每个质心之间的最小绝对差值，将其与最接近的邮政编码质心相匹配。

For each lat/long pair in my data, I'd like to match it to the nearest zip code centroid by finding the minimum absolute difference between that lat/long pair and each centroid in my lookup table.

将这种查找功能逐行应用于大量记录的最有效方法是什么？

样本数据：经纬度列表：

latlongdata <- 
  structure(list(dropoff_longitude = c(-73.981705, -73.993553, 
-73.973305, -73.988823, -73.938484, -74.015503, -73.95472, -73.9571, 
-73.971298, -73.99794), dropoff_latitude = c(40.760559, 40.756348, 
40.762646, 40.777504, 40.684692, 40.709881, 40.783371, 40.776657, 
40.752148, 40.720535)), row.names = c(8807218L, 9760613L, 3175671L, 
10878727L, 2025038L, 5345659L, 14474481L, 1650223L, 684883L, 
9129975L), class = "data.frame", .Names = c("dropoff_longitude", 
"dropoff_latitude"))

    print(latlongdata)
         dropoff_longitude dropoff_latitude
8807218          -73.98171         40.76056
9760613          -73.99355         40.75635
3175671          -73.97330         40.76265
10878727         -73.98882         40.77750
2025038          -73.93848         40.68469
5345659          -74.01550         40.70988
14474481         -73.95472         40.78337
1650223          -73.95710         40.77666
684883           -73.97130         40.75215
9129975          -73.99794         40.72053

** ZipLooker函数：查找距输入的最小绝对距离协调对到最近的邮政编码质心，并返回该邮政编码

**ZipLooker function: Finds the minimum absolute distance from an input coordinate pair to the nearest zip code centroid and returns that zip

library(dplyr)
ZipLooker<-function(dropoff_longitude,dropoff_latitude){
  if(is.na(dropoff_longitude)|is.na(dropoff_latitude)){
    z<-NA_character_
  } else {

    tryCatch({
      x<-ziplkp1 
      x$latdiff=abs(dropoff_latitude-x$Latitude)
      x$londiff=abs(dropoff_longitude-x$Longitude)
      x$totdiff=x$latdiff+x$londiff
      z<-head(top_n(x,1,-totdiff),n=1)$Postal
      return(z)
    }, error=function(e) NA)
  }
}

使用dplyr的rowwsie（）函数应用Ziplooker函数

  latlongdata %>% 
  rowwise() %>% 
  mutate(zipcode=ZipLooker(dropoff_longitude,dropoff_latitude)
         )

推荐答案

为您提供完整的解决方案：

Here's a complete solution for you:

library(sp)
library(maptools)
library(zipcode)

# grab the zip code boundaries
url <- "http://www2.census.gov/geo/tiger/GENZ2014/shp/cb_2014_us_zcta510_500k.zip"
fil <- "ztca.zip"

# don't waste bandwidth
if (!file.exists(fil)) { download.file(url, fil) }
unzip(fil, exdir="ztca")

# read them in (this takes a bit)
ztca <- readShapePoly("ztca/cb_2014_us_zcta510_500k.shp", verbose=TRUE)

# extract NY
ny <- ztca[as.character(ztca$ZCTA5CE10) %in% as.character(zipcode[zipcode$state=="NY",]$zip),]

# your points
latlongdata <- 
  structure(list(dropoff_longitude = c(-73.981705, -73.993553, 
-73.973305, -73.988823, -73.938484, -74.015503, -73.95472, -73.9571, 
-73.971298, -73.99794), dropoff_latitude = c(40.760559, 40.756348, 
40.762646, 40.777504, 40.684692, 40.709881, 40.783371, 40.776657, 
40.752148, 40.720535)), row.names = c(8807218L, 9760613L, 3175671L, 
10878727L, 2025038L, 5345659L, 14474481L, 1650223L, 684883L, 
9129975L), class = "data.frame", .Names = c("dropoff_longitude", 
"dropoff_latitude"))

# make them all super spatial-like (must be in lon,lat format)
pts <- SpatialPoints(as.matrix(latlongdata[,1:2]))

# figure out where they are (this can take a bit)
dat <- pts %over% ny

# merge your data back in (there are many ways to do this)
dat$lon <- latlongdata$dropoff_longitude
dat$lat <- latlongdata$dropoff_latitude
rownames(dat) <- rownames(latlongdata)

# boom
dat
##          ZCTA5CE10     AFFGEOID10 GEOID10 ALAND10 AWATER10       lon      lat
## 8807218      10019 8600000US10019   10019 1780742        0 -73.98171 40.76056
## 9760613      10018 8600000US10018   10018  836253        0 -73.99355 40.75635
## 3175671      10022 8600000US10022   10022 1107169        0 -73.97330 40.76265
## 10878727     10069 8600000US10069   10069  249044        0 -73.98882 40.77750
## 2025038      11221 8600000US11221   11221 3582803        0 -73.93848 40.68469
## 5345659      10280 8600000US10280   10280  300652    38759 -74.01550 40.70988
## 14474481     10128 8600000US10128   10128 1206195        0 -73.95472 40.78337
## 1650223      10028 8600000US10028   10028  811363        0 -73.95710 40.77666
## 684883       10017 8600000US10017   10017  820953        0 -73.97130 40.75215
## 9129975      10013 8600000US10013   10013 1425085        0 -73.99794 40.72053

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