邮编距离 R

Question

我在 R 中使用邮政编码包，我想列出每个邮政编码半径 10、20 或 X 英里范围内的所有邮政编码。从那里我会将邮政编码数据汇总到 10、20 或 X 英里的总距离。我目前正在加入每个邮政编码，每个邮政编码（所以行数平方）。然后计算每个邮政编码之间的距离。然后消除大于 10,20, X 英里的距离。有没有更好的方法在 R 中做到这一点，所以我不必计算所有可能性？我是 R 的新手。谢谢！

Code is here:
#Bringing in Zipcode database. 
library(zipcode)
data(zipcode)

#Limiting to certain states that I want to include,
SEZips <- zipcode[zipcode$state %in% c("GA","AL", "SC", "NC"),]

#Duplicating the data set to join it together
SEZips2 <- SEZips

#To code in SQL
library(sqldf)

#Creating a common match so I can join all rows from both tables together
SEZips$Match <- 1
SEZips2$Match <- 1

#attaches every zip code to each zip
ZipList <- sqldf("
                 SELECT
                 A.zip as zip1,
                 A.longitude as lon1,
                 A.latitude as lat1,
                 B.zip as zip2,
                 B.longitude as lon2,
                 B.latitude as lat2
                 From SEZips A
                 Left Join SEZips2 B
                 on A.Match = B.Match
                 ")


#to get the distance calculation, use package geosphere, 
library(geosphere)

#radius of Earth in miles, adjust for km, etc.
r = 3959
#Creating Table of the coordinates. Makes it easy to calc distance
Points1 <- cbind(ZipList$lon1,ZipList$lat1)
Points2 <- cbind(ZipList$lon2,ZipList$lat2)
distance <- distHaversine(Points1,Points2,r)

#Adding distance back on to the original ZipList
ZipList$Distance <- distance

#To limit to a certain radius.E.g. 15 for 15 miles.
z = 15
#Eliminating matches > z 
ZipList2 <- ZipList[ZipList$Distance <= z,]

#Adding data to roll up, e.g. population
ZipPayroll <- read.csv("filepath/ZipPayroll.csv")

#Changin Zip to 5 character from integer. A little bit of pain
#Essentailly code says (add 5 0's, and then grab the right 5 characters)
ZipPayroll$Zip2 <- substr(paste("00000",ZipPayroll$zip,sep=""),nchar(paste("00000",ZipPayroll$zip,sep=""))-4,nchar(paste("00000",ZipPayroll$zip,sep="")))

#Joining Payroll info to SEZips dataframe
SEZips <- sqldf("
                SELECT
                A.*,
                B.Payroll, 
                B.Employees,
                B.Establishments
                From SEZips A
                Left Join ZipPayroll B
                on A.zip = B.Zip2
                ")

#Rolling up to 15 mile level
SEZips15 <- sqldf("
                  SELECT
                  A.zip1 as Zip, 
                  Sum(B.Payroll) as PayrollArea,
                  Sum(B.Employees) as EmployeesArea,
                  Sum(B.Establishments) as EstablishmentsArea
                  From ZipList2 A
                  Left Join SEZips B
                  on A.zip2 = B.zip
                  Group By A.zip1
                  ")

#Include the oringinal Zip data 
SEZips15 <- sqldf("
                  SELECT
                  A.*,
                  B.Payroll,
                  B.Employees,
                  B.Establishments as EstablishmentsArea
                  From SEZips15 A
                  Left Join SEZips B
                  on A.zip = B.zip
                  ")

#Calculate Average Pay for Zip and Area
SEZips15$AvgPayArea <- SEZips15$PayrollArea / SEZips15$EmployeesArea
SEZips15$AvgPay <- SEZips15$Payroll / SEZips15$Employees

Answer 1

我在下面添加了一个使用 spatialrisk 包的解决方案。此包中的关键函数是用 C++ (Rcpp) 编写的，因此速度非常快。

函数 spatialrisk::points_in_circle() 从中心点计算半径内的观测值。请注意，距离是使用 Haversine 公式计算的。

library(spatialrisk)
library(tidyverse)

zips_within_radius <- function(x,y,z) {
  points_in_circle(SEZips, x, y, lon = longitude, lat = latitude, radius = 10000) %>% 
    mutate(source_zip = z)
 }

由于输出的每个元素都是一个数据框，因此使用 purrr::map_dfr 将它们行绑定在一起：

pmap_dfr(list(SEZips$longitude, SEZips$latitude, SEZips$zip), zips_within_radius)