这是最有效的解决方案，我发现迄今：
Manipulating and mapping US Census data in R using the acs, tigris and leaflet packages
library(tigris)
library(acs)
library(stringr) #to pad fips codes
library(gdtools)
#grab the spatial data (tigris)
#note that you can use the county names inthe tigris package but not in the acs.fetch function from the ACS pacakge so I'm using FIPS numbers here.
#Grab the spatial data
counties<-c(5,47,61,81,85)
#solve the 'an error occurred in the secure channel support'
#firewall issue? #nope.
#https://www2.census.gov/geo/tiger/GENZ2015/shp/
#download via chrome works fine.
library(gdtools) #did not fix it.
#libcurl may fix it
#https://stackoverflow.com/questions/29688026/vb6-winhhtp-error-occurred-in-the-secure-channel-support
library(curl)
tracts<-tracts(state='NY', county = c(5,47,61,81,85), cb=TRUE)
#It does!
##----------------get the tabular data--------------------
#zevross.com/blog
#get the tabular data
#in order to do this, you will need an API key from the US Census.
#Go to https://api.census.gov/data/key_signup.html
#to request one (takes a minute or two) and then
#use the api.key.install function in the `acs` package to use the key.
api.key.install(key="GETYOUROWNKEEY")
#make a geographic set to grab tabular data (acs)
geo<-geo.make(state=c("NY"), county = c(5,47,61,81,85), tract = "*")
#package not updated to 2013 data, so 2012 used as terminal year
income<-acs.fetch(endyear=2012, span=5, geography=geo, table.number="B19001", col.names ="pretty")
#pretty gives fully column names, not census abbreviation.
#B19001_001 and *.017 are total income and income over $200k
#what results is not data, but a list of what is available.
names(attributes(income)) #shows what's available
attr(income, "acs.colnames")
#convert to data frame for merging.
income_df<-data.frame(paste0(
str_pad([email protected]$state,2,"left", pad="0"),
str_pad([email protected]$county,3,"left", pad = "0"),
str_pad([email protected]$tract,6,"left", pad="0")),
[email protected][,c(
"Household Income: Total:",
"Household Income: $200,000 or more")],
stringsAsFactors=FALSE)
#that worked, 12/18/2017
library(dplyr) #required for select
income_df<-select(income_df, 1:3)
rownames(income_df)<-1:nrow(income_df)
names(income_df)<-c("GEOID","total","over_200")
income_df$percent <-100*(income_df$over_200/income_df$total)
#works!
我不知道这会有所帮助，但2005 - 2009年B07010可以通过优秀的'ACS访问'包。 – Tiernan