Question

这与question @DavidArenburg关于条件键控连接的问题非常类似，另外还有一个我无法理解的错误。

基本上，除了条件连接之外，我还想定义一个标志，说明匹配过程的哪一步发生了匹配;我的问题是我只能获得标志来定义所有值，而不是匹配的值。

以下是我希望的最小工作示例：

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，每年首先尝试匹配上一年的名字/姓氏;如果失败，那么尝试匹配名字/首页名称。我想定义flag1表示完全匹配，flag2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)    
}

我希望在我定义TRUE时在j参数中包含id值，只有匹配的name s（例如，Joe在第一步））将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的行'flag值？理想输出如下：

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

Answer 1

我认为这里的旗帜很乱;更好地简单地识别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
)]

给出了

     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

数据的原始排序会丢失，但如果您需要，则很容易恢复。

Answer 2

关键（没有双关语意）我认为合并正在为错过的ID返回NA，所以我应该在每一步添加flag到unmatched，例如，在步骤1：

unmatched <- dt[.(1:(yr - 1L))
                ][!id %in% existing_ids,
                  .SD[.N], by = id][ , flag1 := TRUE]
dt[year == yr, c("id", "flag1") := 
     unmatched[.SD, .(id, flag1), on = "name,surname"]]

最后，这会产生：

> 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    NA    NA
 6: Carol   Klein      Clymer  3    2    NA  TRUE
 7:   Joe   Smith              1    2  TRUE FALSE
 8:   Ann  Cotter              4    3  TRUE FALSE
 9:  Beth   Brown              5    3    NA    NA
10:   Joe   Smith              1    4  TRUE FALSE
11:   Joe   Smith              1    4  TRUE FALSE

剩下的一个问题是，应该F的某些标记已重置为NA;能够设置nomatch=F会很高兴，但我并不太担心这种副作用 - 对我而言，关键是知道每个标志何时为T。

条件键控连接/更新_and_更新匹配的标志列

2 个答案: