这与 question 非常相似@DavidArenburg 询问了条件键连接,还有一个我似乎无法解决的问题。
基本上,除了条件连接之外,我想定义一个标志,说明匹配发生在匹配过程的哪一步;我的问题是我只能获取为所有值定义的标志,而不是匹配值。
我希望这是一个最小的工作示例:
DT = data.table(
name = c("Joe", "Joe", "Jim", "Carol", "Joe",
"Carol", "Ann", "Ann", "Beth", "Joe", "Joe"),
surname = c("Smith", "Smith", "Jones",
"Clymer", "Smith", "Klein", "Cotter",
"Cotter", "Brown", "Smith", "Smith"),
maiden_name = c("", "", "", "", "", "Clymer",
"", "", "", "", ""),
id = c(1, 1:3, rep(NA, 7)),
year = rep(1:4, c(4, 3, 2, 2)),
flag1 = NA, flag2 = NA, key = "year"
)
DT
# name surname maiden_name id year flag1 flag2
# 1: Joe Smith 1 1 FALSE FALSE
# 2: Joe Smith 1 1 FALSE FALSE
# 3: Jim Jones 2 1 FALSE FALSE
# 4: Carol Clymer 3 1 FALSE FALSE
# 5: Joe Smith NA 2 FALSE FALSE
# 6: Carol Klein Clymer NA 2 FALSE FALSE
# 7: Ann Cotter NA 2 FALSE FALSE
# 8: Ann Cotter NA 3 FALSE FALSE
# 9: Beth Brown NA 3 FALSE FALSE
# 10: Joe Smith NA 4 FALSE FALSE
# 11: Joe Smith NA 4 FALSE FALSE
我的方法是,对于每一年,首先尝试匹配前一年的名字/姓氏;如果失败,则尝试匹配名字/娘家姓。我想定义 flag1
来表示完全匹配,flag2
来表示婚姻。
for (yr in 2:4) {
#which ids have we hit so far?
existing_ids = DT[.(yr), unique(id)]
#find people in prior years appearing to
# correspond to those people
unmatched =
DT[.(1:(yr - 1))][!id %in% existing_ids, .SD[.N], by = id]
setkey(unmatched, name, surname)
#merge a la Arun, define flag1
setkey(DT, name, surname)
DT[year == yr, c("id", "flag1") := unmatched[.SD, .(id, TRUE)]]
setkey(DT, year)
#repeat, this time keying on name/maiden_name
existing_ids = DT[.(yr), unique(id)]
unmatched =
DT[.(1:(yr - 1))][!id %in% existing_ids, .SD[.N],by=id]
setkey(unmatched, name, surname)
#now define flag2 = TRUE
setkey(DT, name, maiden_name)
DT[year==yr & is.na(id), c("id", "flag2") := unmatched[.SD, .(id, TRUE)]]
setkey(DT, year)
#this is messy, but I'm trying to increment id
# for "new" individuals
setkey(DT, name, surname, maiden_name)
DT[year == yr & is.na(id),
id := unique(
DT[year == yr & is.na(id)],
by = c("name", "surname", "maiden_name")
)[ , count := .I][.SD, count] + DT[ , max(id, na.rm = TRUE)]
]
#re-sort by year at the end
setkey(DT, year)
}
我希望通过在定义 id
时在 j
参数中包含 TRUE
值,只有匹配的 name
s(例如,第一步的 Joe)会将它们的 flag
更新为 TRUE
,但事实并非如此——它们都已更新:
DT[]
# name surname maiden_name id year flag1 flag2
# 1: Carol Clymer 3 1 FALSE FALSE
# 2: Jim Jones 2 1 FALSE FALSE
# 3: Joe Smith 1 1 FALSE FALSE
# 4: Joe Smith 1 1 FALSE FALSE
# 5: Ann Cotter 4 2 TRUE TRUE
# 6: Carol Klein Clymer 3 2 TRUE TRUE
# 7: Joe Smith 1 2 TRUE FALSE
# 8: Ann Cotter 4 3 TRUE FALSE
# 9: Beth Brown 5 3 TRUE TRUE
# 10: Joe Smith 1 4 TRUE FALSE
# 11: Joe Smith 1 4 TRUE FALSE
有没有办法只更新匹配行的 flag
值?理想的输出如下:
DT[]
# name surname maiden_name id year flag1 flag2
# 1: Carol Clymer 3 1 FALSE FALSE
# 2: Jim Jones 2 1 FALSE FALSE
# 3: Joe Smith 1 1 FALSE FALSE
# 4: Joe Smith 1 1 FALSE FALSE
# 5: Ann Cotter 4 2 FALSE FALSE
# 6: Carol Klein Clymer 3 2 FALSE TRUE
# 7: Joe Smith 1 2 TRUE FALSE
# 8: Ann Cotter 4 3 TRUE FALSE
# 9: Beth Brown 5 3 FALSE FALSE
# 10: Joe Smith 1 4 TRUE FALSE
# 11: Joe Smith 1 4 TRUE FALSE
最佳答案
我觉得这里的旗帜很乱;最好简单地识别 id
的来源:
dt[,c("flag1","flag2"):=NULL]
# create name -> id table
namemap <- unique(dt[,.(maiden_name,id,year),keyby=.(name,surname)],by=NULL)
# tag original ids
namemap[!is.na(id),src:="original"]
# carried over from earlier years
namemap[, has_oid := any(!is.na(id)), by=key(namemap)]
namemap[(has_oid),`:=`(
id = id[!is.na(id)],
src = ifelse(is.na(id), "history", src)
),by=.(name,surname)]
# carry over for surname changes on marriage
namemap[maiden_name!="",`:=`(
id = namemap[.BY]$id,
src = "maiden"
),by=.(name,maiden_name)]
# create new ids where none exists
namemap[is.na(id),`:=`(
id = .GRP+max(dt$id,na.rm=TRUE),
src = "new"
),by=.(name,surname)]
# copy back to the original table
setkey(dt,name,surname,year)
setkey(namemap,name,surname,year)
dt[namemap,`:=`(
id = i.id,
src = src
)]
给出
name surname maiden_name id year src
1: Ann Cotter 4 2 new
2: Ann Cotter 4 3 new
3: Beth Brown 5 3 new
4: Carol Clymer 3 1 original
5: Carol Klein Clymer 3 2 maiden
6: Jim Jones 2 1 original
7: Joe Smith 1 1 original
8: Joe Smith 1 1 original
9: Joe Smith 1 2 history
10: Joe Smith 1 4 history
11: Joe Smith 1 4 history
数据的原始排序丢失了,但如果你想要它很容易恢复。
关于r - 条件键连接/更新_and_更新匹配的标志列,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/31329939/