我遇到了将县分配到某些城市的问题。通过acs包
> geo.lookup(state = "NY", place = "New York")
state state.name county.name place place.name
1 36 New York <NA> NA <NA>
2 36 New York Bronx County, Kings County, New York County, Queens County, Richmond County 51000 New York city
3 36 New York Oneida County 51011 New York Mills village
，你可以看到，“纽约”，例如，有一堆县。洛杉矶，波特兰，俄克拉荷马，哥伦布等也是如此。如何将这些数据分配给“县”？
以下代码目前用于将“county.name”与相应的县FIPS代码进行匹配。不幸的是，它仅适用于查询中只有一个县名输出的情况。
dat <- c("New York, NY","Boston, MA","Los Angeles, CA","Dallas, TX","Palo Alto, CA")
dat <- strsplit(dat, ",")
dat
library(tigris)
library(acs)
data(fips_codes) # FIPS codes with state, code, county information
GeoLookup <- lapply(dat,function(x) {
geo.lookup(state = trimws(x[2]), place = trimws(x[1]))[2,]
})
df <- bind_rows(GeoLookup)
#Rename cols to match
colnames(fips_codes) = c("state.abb", "statefips", "state.name", "countyfips", "county.name")
# Here is a problem, because it works with one item in "county.name" but not more than one (see output below).
df <- df %>% left_join(fips_codes, by = c("state.name", "county.name"))
df
返回：
state state.name county.name place place.name state.abb statefips countyfips
1 36 New York Bronx County, Kings County, New York County, Queens County, Richmond County 51000 New York city <NA> <NA> <NA>
2 25 Massachusetts Suffolk County 7000 Boston city MA 25 025
3 6 California Los Angeles County 20802 East Los Angeles CDP CA 06 037
4 48 Texas Collin County, Dallas County, Denton County, Kaufman County, Rockwall County 19000 Dallas city <NA> <NA> <NA>
5 6 California San Mateo County 20956 East Palo Alto city CA 06 081
为了保留数据，最好将 left_join 匹配为“查找包含county.name的{{1}}（不附加xy city 在名称中），或默认选择第一项。很高兴看到如何做到这一点。
总的来说：我认为，没有比这种方法更好的方法了吗？
感谢您的帮助！
答案 0 :(得分：2)
下面的代码如何为加入创建“长”数据框。我们使用tidyverse管道运算符来链接操作。 strsplit返回一个列表，我们unnest将列表值（与state.name和place.name的每个组合一起使用的县名称）堆叠到一个长数据框中county.name现在有了自己的行。
library(tigris)
library(acs)
library(tidyverse)
dat = geo.lookup(state = "NY", place = "New York")
state state.name county.name place place.name 1 36 New York <NA> NA <NA> 2 36 New York Bronx County, Kings County, New York County, Queens County, Richmond County 51000 New York city 3 36 New York Oneida County 51011 New York Mills village
dat = dat %>%
group_by(state.name, place.name) %>%
mutate(county.name = strsplit(county.name, ", ")) %>%
unnest
state state.name place place.name county.name <chr> <chr> <int> <chr> <chr> 1 36 New York NA <NA> <NA> 2 36 New York 51000 New York city Bronx County 3 36 New York 51000 New York city Kings County 4 36 New York 51000 New York city New York County 5 36 New York 51000 New York city Queens County 6 36 New York 51000 New York city Richmond County 7 36 New York 51011 New York Mills village Oneida County
更新：关于评论中的第二个问题，假设您已经拥有都市区的向量，那么：
dat <- c("New York, NY","Boston, MA","Los Angeles, CA","Dallas, TX","Palo Alto, CA")
df <- map_df(strsplit(dat, ", "), function(x) {
geo.lookup(state = x[2], place = x[1])[-1, ] %>%
group_by(state.name, place.name) %>%
mutate(county.name = strsplit(county.name, ", ")) %>%
unnest
})
df
state state.name place place.name county.name 1 36 New York 51000 New York city Bronx County 2 36 New York 51000 New York city Kings County 3 36 New York 51000 New York city New York County 4 36 New York 51000 New York city Queens County 5 36 New York 51000 New York city Richmond County 6 36 New York 51011 New York Mills village Oneida County 7 25 Massachusetts 7000 Boston city Suffolk County 8 25 Massachusetts 7000 Boston city Suffolk County 9 6 California 20802 East Los Angeles CDP Los Angeles County 10 6 California 39612 Lake Los Angeles CDP Los Angeles County 11 6 California 44000 Los Angeles city Los Angeles County 12 48 Texas 19000 Dallas city Collin County 13 48 Texas 19000 Dallas city Dallas County 14 48 Texas 19000 Dallas city Denton County 15 48 Texas 19000 Dallas city Kaufman County 16 48 Texas 19000 Dallas city Rockwall County 17 48 Texas 40516 Lake Dallas city Denton County 18 6 California 20956 East Palo Alto city San Mateo County 19 6 California 55282 Palo Alto city Santa Clara County
更新2：如果我理解你的评论，对于有多个县的城市（实际上是地名），我们只想要包含与城市名称相同的县（对于例如，在纽约市的纽约县），或列表中的第一个县。以下代码选择与城市同名的县，如果没有，则选择该城市的第一个县。您可能需要稍微调整一下以使其适用于整个美国。例如，要使其适用于路易斯安那州，您可能需要gsub(" County| Parish"...而不是gsub(" County"...。
map_df(strsplit(dat, ", "), function(x) {
geo.lookup(state = x[2], place = x[1])[-1, ] %>%
group_by(state.name, place.name) %>%
mutate(county.name = strsplit(county.name, ", ")) %>%
unnest %>%
slice(max(1, which(grepl(sub(" [A-Za-z]*$","", place.name), gsub(" County", "", county.name))), na.rm=TRUE))
})
state state.name place place.name county.name <chr> <chr> <int> <chr> <chr> 1 36 New York 51000 New York city New York County 2 36 New York 51011 New York Mills village Oneida County 3 25 Massachusetts 7000 Boston city Suffolk County 4 6 California 20802 East Los Angeles CDP Los Angeles County 5 6 California 39612 Lake Los Angeles CDP Los Angeles County 6 6 California 44000 Los Angeles city Los Angeles County 7 48 Texas 19000 Dallas city Dallas County 8 48 Texas 40516 Lake Dallas city Denton County 9 6 California 20956 East Palo Alto city San Mateo County 10 6 California 55282 Palo Alto city Santa Clara County
答案 1 :(得分：1)
您可以使用类似下面代码的方式准备数据吗？
{{1}}
它有点乱，你需要plyr和stringr包。准备好数据后，您应该能够加入数据