我需要根据与第一个链接在一起的其他两个数据框中的值来更新数据框。
目标df t_offices在此有4个感兴趣的字段：
administrative_area_level_1 administrative_area_level_2 country locality
1 Arizona Maricopa County United States Phoenix
2 District of Columbia <NA> United States Washington
3 <NA> <NA> India <NA>
4 New York Albany County United States Albany
5 Utrecht Nieuwegein Netherlands Nieuwegein
6 Connecticut Fairfield County United States Stamford
707 Illinois <NA> United States <NA>
4241 Illinois <NA> United States West Chicago
999998 Alabama <NA> United States Altoona
999999 Pennsylvania <NA> United States Washington
我需要更新administrative_area_level_2中包含美国记录的县的NA值。值以df t_places：
state_ab place_name county_name place_nameshort
1 AL Abanda CDP Chambers County Abanda
2 AL Abbeville city Henry County Abbeville
3 AL Adamsville city Jefferson County Adamsville
4 AL Addison town Winston County Addison
5 AL Akron town Hale County Akron
6 AL Alabaster city Shelby County Alabaster
12 AL Altoona town Blount County, Etowah County Altoona
4298 DC Washington city District of Columbia Washington
7527 IL West Chicago city DuPage County Washington
32611 PA Washington township Armstrong County West Chicago
32612 PA Washington township Berks County Washington
place_nameshort是place_name的截断版本，没有指定（例如“城市”，“城镇”等）
我在州和地方加入t_offices和t_places以获得正确的县。这可能会返回多个县1），因为county_name可以包含由逗号分隔的多个县，以及2）因为截断的place_nameshort可以在同一状态内返回同义词。我需要只是那些县是明确的（单县回归）。
由于t_places仅包含state_ab，我需要r_states的第三个数据框state_name：
state_ab state_name
1 AL Alabama
2 AK Alaska
3 AZ Arizona
4 AR Arkansas
5 CA California
6 CO Colorado
9 DC District of Columbia
17 IL Illinois
42 PA Pennsylvania
通过t_places与r_states加入state_ab，我可以state_name与t_offices$administrative_area_level_1匹配。
这是我的尝试，这是不完整的，因为由于状态内同义词而无法控制多个县，并且无论如何都不起作用。
no_county <- (!is.na(t_offices$country)
& t_offices$country == "United States"
& !is.na(t_offices$administrative_area_level_1)
& is.na(t_offices$administrative_area_level_2)
& !is.na(t_offices$locality))
t_offices$administrative_area_level_2[no_county] <-
t_places$county_name[!grepl(",", t_places$county_name)
& match(t_places$place_nameshort, t_offices$locality[no_county])
& match(t_places$state_ab,
r_states$state_ab[match(r_states$state_name,
t_offices$administrative_area_level_1[no_county])])]
编辑：按照@ r2evans的建议，这是我的新编码尝试，但仍然不起作用：
# split multiple counties into columns
library(splitstackshape)
t_places <- cSplit(t_places, "county_name", sep = ", ", drop = F, type.convert = F)
# merge state names into places
places_statename <- merge(t_places, r_states[,2:3])
# define condition to select t_offices records in U.S. with state and place but no county
no_county <- (
# country is U.S.
!is.na(t_offices$country)
& t_offices$country == "United States"
# with state
& !is.na(t_offices$administrative_area_level_1)
# blank county
& is.na(t_offices$administrative_area_level_2)
# with place
& !is.na(t_offices$locality))
# update blank counties
t_offices$administrative_area_level_2[no_county] <-
# unambiguous counties
places_statename$county_name_1[is.na(places_statename$county_name_2)
# locality matches place
& match(t_offices$locality[no_county], places_statename$place_nameshort)
# administrative_area_level_1 matches state
& match(t_offices$administrative_area_level_1[no_county],places_statename$state_name)]
答案 0 :(得分：0)
这是我的长期解决方案。可能更短，更优雅。
# split multiple counties into columns
library(splitstackshape)
t_places <- cSplit(t_places, "county_name", sep = ", ", drop = F, type.convert = F)
# subset original places with single county
places_singlecounty <- t_places[is.na(places_statename$county_name_2), c(1,8,9)]
# subset truncated places with single county
library(data.table)
setDT(places_singlecounty)
places_singlecounty <- merge(places_singlecounty,
places_singlecounty[, .N, by = c("state_ab", "place_nameshort")][N == 1, 1:2])
# merge state names into single-county truncated places
places_statename <- merge(places_singlecounty, r_states[,2:3], by = "state_ab")
# define condition to select t_offices records in U.S. with state and place but no county
no_county <- (
# country is U.S.
!is.na(t_offices$country)
& t_offices$country == "United States"
# with state
& !is.na(t_offices$administrative_area_level_1)
# NA county
& is.na(t_offices$administrative_area_level_2)
# with place
& !is.na(t_offices$locality))
# update t_offices NA counties based on single-county truncated places
setDT(t_offices)
t_offices[no_county, administrative_area_level_2 :=
places_statename[.(.SD), county_name_1,
on = c(state_name = "administrative_area_level_1",
place_nameshort = "locality")]]