如何在R中批量反转地理编码？

Question

我想在纽约地区获得lat-long坐标的zipcode

我尝试使用谷歌的反向地理编码器API，但每天限制为2500次点击，因此可以批量处理我的数据框。

接下来，我尝试使用带有数据集邮政编码的库（zipcode），但无法将纬度经度与列车数据集的坐标相匹配，因为所有纬度坐标都不在数据集中。

此外，虽然使用KNN来预测数据集的zipcode但无法获得正确的结果。

zipcode_latlon = zipcode[zipcode$state=="NY",c(1,4,5)]
train_latlon = train_data[,c("latitude","longitude")]
zip1 = rep(10007, nrow(train_latlon))
zip1 = as.character(zip1)
train_latlon = cbind(zip1, train_latlon)
colnames(train_latlon) = c("zip","latitude","longitude")
knn_fit = knn(zipcode_latlon, train_latlon,zipcode_latlon$zip, k=1)

需要知道如何从lat中批量获取zipcodes，任何方法在R中都会很好。

Answer 1

我认为你这是错误的做法。您可以找到没有地理编码器的纬度/经度坐标的邮政编码 - 您只需下载美国zipcodes shapefile here然后进行空间连接：

library(sp)
library(rgdal)

#import zips shapefile and transform CRS 
zips <- readOGR("cb_2015_us_zcta510_500k.shp")
zips <- spTransform(zips, CRS("+proj=longlat +datum=WGS84 +ellps=WGS84 +towgs84=0,0,0"))

#here is a sample with three cities in New York State and their coordinates      
df <- as.data.frame(matrix(nrow = 3, ncol =3))
colnames(df) <- c("lat", "lon", "city")

df$lon <- c(43.0481, 43.1610, 42.8864)
df$lat <- c(-76.1474, -77.6109,-78.8784)
df$city <- c("Syracuse", "Rochester", "Buffalo")

df
       lat     lon      city
1 -76.1474 43.0481  Syracuse
2 -77.6109 43.1610 Rochester
3 -78.8784 42.8864   Buffalo

#extract only the lon/lat                   
xy <- df[,c(1,2)]

#transform coordinates into a SpatialPointsDataFrame
spdf <- SpatialPointsDataFrame(coords = xy, data = df, proj4string = CRS("+proj=longlat +datum=WGS84 +ellps=WGS84 +towgs84=0,0,0"))

#subset only the zipcodes in which points are found
zips_subset <- zips[spdf, ]

#NOTE: the column in zips_subset containing zipcodes is ZCTA5CE10
#use over() to overlay points in polygons and then add that to the original dataframe

df$zip <- over(spdf, zips_subset[,"ZCTA5CE10"])

瞧！你有每个点的邮政编码

df
       lat     lon      city ZCTA5CE10
1 -76.1474 43.0481  Syracuse     13202
2 -77.6109 43.1610 Rochester     14604
3 -78.8784 42.8864   Buffalo     14202