menggabungkan dataframe berdasarkan beberapa kolom dan ambang batas

Saya memiliki dua data.frames dengan beberapa kolom yang umum (di sini: date, city, ctry, dan ( other_) number).

Sekarang saya ingin menggabungkan mereka pada kolom di atas tetapi mentolerir beberapa tingkat perbedaan:

threshold.numbers <- 3
threshold.date <- 5  # in days

Jika perbedaan antara dateentri adalah > threshold.date(dalam hari) atau > threshold.numbers , saya tidak ingin garis digabungkan. Demikian pula, jika entri dalam cityadalah substring dari dfentri orang lain di citykolom, saya ingin garis digabungkan. [Jika ada yang punya ide yang lebih baik untuk menguji kesamaan nama kota yang sebenarnya, saya akan senang mendengarnya.] (Dan menyimpan dfentri pertama dari date, citydan countrytetapi kedua ( other_) numberkolom dan semua kolom lainnya di df.

Perhatikan contoh berikut:

df1 <- data.frame(date = c("2003-08-29", "1999-06-12", "2000-08-29", "1999-02-24", "2001-04-17",
                           "1999-06-30", "1999-03-16", "1999-07-16", "2001-08-29", "2002-07-30"),
                  city = c("Berlin", "Paris", "London", "Rome", "Bern",
                           "Copenhagen", "Warsaw", "Moscow", "Tunis", "Vienna"),
                  ctry = c("Germany", "France", "UK", "Italy", "Switzerland",
                           "Denmark", "Poland", "Russia", "Tunisia", "Austria"),
                  number = c(10, 20, 30, 40, 50, 60, 70, 80, 90, 100),
                  col = c("apple", "banana", "pear", "banana", "lemon", "cucumber", "apple", "peach", "cherry", "cherry"))


df2 <- data.frame(date = c("2003-08-29", "1999-06-12", "2000-08-29", "1999-02-24", "2001-04-17", # all identical to df1
                           "1999-06-29", "1999-03-14", "1999-07-17", # all 1-2 days different
                           "2000-01-29", "2002-07-01"), # all very different (> 2 weeks)
                  city = c("Berlin", "East-Paris", "near London", "Rome", # same or slight differences
                           "Zurich", # completely different
                           "Copenhagen", "Warsaw", "Moscow", "Tunis", "Vienna"), # same
                  ctry = c("Germany", "France", "UK", "Italy", "Switzerland", # all the same 
                           "Denmark", "Poland", "Russia", "Tunisia", "Austria"),
                  other_number = c(13, 17, 3100, 45, 51, 61, 780, 85, 90, 101), # slightly different to very different
                  other_col = c("yellow", "green", "blue", "red", "purple", "orange", "blue", "red", "black", "beige"))

Sekarang, saya ingin menggabungkan data.framesdan menerima di dfmana garis digabungkan jika kondisi di atas terpenuhi.

(Kolom pertama hanya untuk kenyamanan Anda: di belakang digit pertama, yang menunjukkan case asli, ini menunjukkan apakah garis tempat digabungkan ( .) atau apakah garis berasal df1( 1) atau df2( 2).

          date        city        ctry number other_col other_number    other_col2          #comment
 1.  2003-08-29      Berlin     Germany     10     apple              13        yellow      # matched on date, city, number
 2.  1999-06-12       Paris      France     20    banana              17         green      # matched on date, city similar, number - other_number == threshold.numbers
 31  2000-08-29      London          UK     30      pear            <NA>          <NA>      # not matched: number - other_number > threshold.numbers
 32  2000-08-29 near London         UK    <NA>      <NA>            3100          blue      #
 41  1999-02-24        Rome       Italy     40    banana            <NA>          <NA>      # not matched: number - other_number > threshold.numbers
 42  1999-02-24        Rome       Italy   <NA>      <NA>              45           red      #
 51  2001-04-17        Bern Switzerland     50     lemon            <NA>          <NA>      # not matched: cities different (dates okay, numbers okay)
 52  2001-04-17      Zurich Switzerland   <NA>      <NA>              51        purple      #
 6.  1999-06-30  Copenhagen     Denmark     60  cucumber              61        orange      # matched: date difference < threshold.date (cities okay, dates okay)
 71  1999-03-16      Warsaw      Poland     70     apple            <NA>          <NA>      # not matched: number - other_number > threshold.numbers (dates okay)
 72  1999-03-14      Warsaw      Poland   <NA>      <NA>             780          blue      # 
 81  1999-07-16      Moscow      Russia     80     peach            <NA>          <NA>      # not matched: number - other_number > threshold.numbers (dates okay)
 82  1999-07-17      Moscow      Russia   <NA>      <NA>              85           red      #
 91  2001-08-29       Tunis     Tunisia     90    cherry            <NA>          <NA>      # not matched: date difference < threshold.date (cities okay, dates okay)
 92  2000-01-29       Tunis     Tunisia   <NA>      <NA>              90         black      #
101  2002-07-30      Vienna     Austria    100    cherry            <NA>          <NA>      # not matched: date difference < threshold.date (cities okay, dates okay)
102  2002-07-01      Vienna     Austria   <NA>      <NA>             101         beige      #

Saya mencoba berbagai implementasi penggabungan tetapi tidak dapat menerapkan ambang.

Permintaan Maaf EDIT untuk formulasi tidak jelas - Saya ingin mempertahankan semua baris dan menerima indikator apakah baris cocok, tidak cocok dan dari df1 atau tidak cocok dan dari df2.

pseudo-code adalah:

  if there is a case where abs("date_df2" - "date_df1") <= threshold.date:
    if "ctry_df2" == "ctry_df1":
      if "city_df2" ~ "city_df1":
        if abs("number_df2" - "number_df1") <= threshold.numbers:
          merge and go to next row in df2
  else:
    add row to df1```

Inilah solusi yang menggunakan safejoin paket saya , membungkus dalam hal ini paket fuzzyjoin .

Kita dapat menggunakan byargumen untuk menentukan kondisi yang kompleks, menggunakan fungsi X()untuk mendapatkan nilai dari df1, dan Y()untuk mendapatkan nilai dari df2.

Jika tabel asli Anda besar, ini mungkin lambat atau tidak mungkin karena ia menghasilkan produk cartesian, tetapi di sini ia berfungsi dengan baik.

Yang kami inginkan adalah bergabung penuh (pertahankan semua baris, dan bergabunglah dengan apa yang bisa digabungkan), dan kami ingin mempertahankan nilai pertama ketika mereka bergabung, dan mengambil yang berikutnya dengan bijak, ini berarti kami ingin menangani konflik kolom dinamai identik dengan penggabungan, jadi kami menggunakan argumen conflict = dplyr::coalesce

# remotes::install_github("moodymudskipper/safejoin")


# with provides inputs date is a factor, this will cause issues, so we need to
# convert either to date or character, character will do for now.
df1$date <- as.character(df1$date)
df2$date <- as.character(df2$date)

# we want our joining columns named the same to make them conflicted and use our
# conflict agument on conflicted paires
names(df2)[1:4] <- names(df1)[1:4]

library(safejoin)
safe_full_join(
  df1, df2,  
  by = ~ {
    # must convert every type because fuzzy join uses a matrix so coerces all inputs to character
    # see explanation at the bottom
    city1 <- X("city")
    city2 <- Y("city")
    date1 <- as.Date(X("date"), origin = "1970-01-01")
    date2 <- as.Date(Y("date"), origin = "1970-01-01")
    number1 <- as.numeric(X("number"))
    number2 <- as.numeric(Y("number"))
    # join if one city name contains the other
    (mapply(grepl, city1, city2) | mapply(grepl, city2, city1)) &
    # and dates are close enough (need to work in seconds because difftime is dangerous)
      abs(difftime(date1, date2, "sec")) <= threshold.date*3600*24 &
    # and numbers are close enough
      abs(number1 - number2) <= threshold.numbers
    },
  conflict = dplyr::coalesce)

keluaran:

#>          date        city        ctry number      col other_col
#> 1  2003-08-29      Berlin     Germany     10    apple    yellow
#> 2  1999-06-12       Paris      France     20   banana     green
#> 3  1999-06-30  Copenhagen     Denmark     60 cucumber    orange
#> 4  2000-08-29      London          UK     30     pear      <NA>
#> 5  1999-02-24        Rome       Italy     40   banana      <NA>
#> 6  2001-04-17        Bern Switzerland     50    lemon      <NA>
#> 7  1999-03-16      Warsaw      Poland     70    apple      <NA>
#> 8  1999-07-16      Moscow      Russia     80    peach      <NA>
#> 9  2001-08-29       Tunis     Tunisia     90   cherry      <NA>
#> 10 2002-07-30      Vienna     Austria    100   cherry      <NA>
#> 11 2000-08-29 near London          UK   3100     <NA>      blue
#> 12 1999-02-24        Rome       Italy     45     <NA>       red
#> 13 2001-04-17      Zurich Switzerland     51     <NA>    purple
#> 14 1999-03-14      Warsaw      Poland    780     <NA>      blue
#> 15 1999-07-17      Moscow      Russia     85     <NA>       red
#> 16 2000-01-29       Tunis     Tunisia     90     <NA>     black
#> 17 2002-07-01      Vienna     Austria    101     <NA>     beige

^{Dibuat pada 2019-11-13 oleh paket reprex (v0.3.0)}

Sayangnya fuzzyjoin memaksa semua kolom dalam matriks ketika melakukan multi join, dan safejoin membungkus fuzzyjoin sehingga kita harus mengonversi variabel ke tipe yang sesuai di dalam argumen, ini menjelaskan baris pertama dalam byargumen.

Lebih lanjut tentang safejoin : https://github.com/moodymudskipper/safejoin

Saya pertama-tama mengubah nama kota menjadi vektor karakter, karena (jika saya mengerti dengan benar) Anda ingin memasukkan nama kota yang terdapat dalam df2.

df1$city<-as.character(df1$city)
df2$city<-as.character(df2$city)

Kemudian gabungkan mereka menurut negara:

df = merge(df1, df2, by = ("ctry"))

> df
          ctry     date.x     city.x number      col     date.y      city.y other_number other_col
1      Austria 2002-07-30     Vienna    100   cherry 2002-07-01      Vienna          101     beige
2      Denmark 1999-06-30 Copenhagen     60 cucumber 1999-06-29  Copenhagen           61    orange
3       France 1999-06-12      Paris     20   banana 1999-06-12  East-Paris           17     green
4      Germany 2003-08-29     Berlin     10    apple 2003-08-29      Berlin           13    yellow
5        Italy 1999-02-24       Rome     40   banana 1999-02-24        Rome           45       red
6       Poland 1999-03-16     Warsaw     70    apple 1999-03-14      Warsaw          780      blue
7       Russia 1999-07-16     Moscow     80    peach 1999-07-17      Moscow           85       red
8  Switzerland 2001-04-17       Bern     50    lemon 2001-04-17      Zurich           51    purple
9      Tunisia 2001-08-29      Tunis     90   cherry 2000-01-29       Tunis           90     black
10          UK 2000-08-29     London     30     pear 2000-08-29 near London         3100      blue

Perpustakaan stringrakan memungkinkan Anda untuk melihat apakah city.x ada di dalam city.y di sini (lihat kolom terakhir):

library(stringr)
df$city_keep<-str_detect(df$city.y,df$city.x) # this returns logical vector if city.x is contained in city.y (works one way)
> df
          ctry     date.x     city.x number      col     date.y      city.y other_number other_col city_keep
1      Austria 2002-07-30     Vienna    100   cherry 2002-07-01      Vienna          101     beige      TRUE
2      Denmark 1999-06-30 Copenhagen     60 cucumber 1999-06-29  Copenhagen           61    orange      TRUE
3       France 1999-06-12      Paris     20   banana 1999-06-12  East-Paris           17     green      TRUE
4      Germany 2003-08-29     Berlin     10    apple 2003-08-29      Berlin           13    yellow      TRUE
5        Italy 1999-02-24       Rome     40   banana 1999-02-24        Rome           45       red      TRUE
6       Poland 1999-03-16     Warsaw     70    apple 1999-03-14      Warsaw          780      blue      TRUE
7       Russia 1999-07-16     Moscow     80    peach 1999-07-17      Moscow           85       red      TRUE
8  Switzerland 2001-04-17       Bern     50    lemon 2001-04-17      Zurich           51    purple     FALSE
9      Tunisia 2001-08-29      Tunis     90   cherry 2000-01-29       Tunis           90     black      TRUE
10          UK 2000-08-29     London     30     pear 2000-08-29 near London         3100      blue      TRUE

Maka Anda bisa mendapatkan perbedaan hari antara tanggal:

df$dayDiff<-abs(as.POSIXlt(df$date.x)$yday - as.POSIXlt(df$date.y)$yday)

dan perbedaan angka:

df$numDiff<-abs(df$number - df$other_number)

Inilah tampilan kerangka data yang dihasilkan:

> df
          ctry     date.x     city.x number      col     date.y      city.y other_number other_col city_keep dayDiff numDiff
1      Austria 2002-07-30     Vienna    100   cherry 2002-07-01      Vienna          101     beige      TRUE      29       1
2      Denmark 1999-06-30 Copenhagen     60 cucumber 1999-06-29  Copenhagen           61    orange      TRUE       1       1
3       France 1999-06-12      Paris     20   banana 1999-06-12  East-Paris           17     green      TRUE       0       3
4      Germany 2003-08-29     Berlin     10    apple 2003-08-29      Berlin           13    yellow      TRUE       0       3
5        Italy 1999-02-24       Rome     40   banana 1999-02-24        Rome           45       red      TRUE       0       5
6       Poland 1999-03-16     Warsaw     70    apple 1999-03-14      Warsaw          780      blue      TRUE       2     710
7       Russia 1999-07-16     Moscow     80    peach 1999-07-17      Moscow           85       red      TRUE       1       5
8  Switzerland 2001-04-17       Bern     50    lemon 2001-04-17      Zurich           51    purple     FALSE       0       1
9      Tunisia 2001-08-29      Tunis     90   cherry 2000-01-29       Tunis           90     black      TRUE     212       0
10          UK 2000-08-29     London     30     pear 2000-08-29 near London         3100      blue      TRUE       0    3070

Tapi kami ingin meletakkan hal-hal di mana city.x tidak ditemukan di dalam city.y, di mana perbedaan hari lebih besar dari 5 atau perbedaan angka lebih besar dari 3:

df<-df[df$dayDiff<=5 & df$numDiff<=3 & df$city_keep==TRUE,]

> df
     ctry     date.x     city.x number      col     date.y     city.y other_number other_col city_keep dayDiff numDiff
2 Denmark 1999-06-30 Copenhagen     60 cucumber 1999-06-29 Copenhagen           61    orange      TRUE       1       1
3  France 1999-06-12      Paris     20   banana 1999-06-12 East-Paris           17     green      TRUE       0       3
4 Germany 2003-08-29     Berlin     10    apple 2003-08-29     Berlin           13    yellow      TRUE       0       3

Yang tersisa adalah tiga baris yang Anda miliki di atas (yang berisi titik-titik di kolom 1).

Sekarang kita bisa menjatuhkan tiga kolom yang kita buat, dan tanggal dan kota dari df2:

> df<-subset(df, select=-c(city.y, date.y, city_keep, dayDiff, numDiff))
> df
     ctry     date.x     city.x number      col other_number other_col
2 Denmark 1999-06-30 Copenhagen     60 cucumber           61    orange
3  France 1999-06-12      Paris     20   banana           17     green
4 Germany 2003-08-29     Berlin     10    apple           13    yellow

Langkah 1: Menggabungkan data berdasarkan "kota" dan "ctry":

df = merge(df1, df2, by = c("city", "ctry"))

Langkah 2: Hapus baris jika perbedaan antara entri tanggal> threshold.date (dalam hari):

date_diff = abs(as.numeric(difftime(strptime(df$date.x, format = "%Y-%m-%d"),
                                    strptime(df$date.y, format = "%Y-%m-%d"), units="days")))
index_remove = date_diff > threshold.date
df = df[-index_remove,]

Langkah 3: Hapus baris jika perbedaan antara angka adalah> threshhold.number:

number_diff = abs(df$number - df$other_number) 
index_remove = number_diff > threshold.numbers
df = df[-index_remove,]

Data harus digabung sebelum menerapkan kondisi, jika baris tidak cocok.

Opsi menggunakan data.table(penjelasan sebaris):

library(data.table)
setDT(df1)
setDT(df2)

#dupe columns and create ranges for non-equi joins
df1[, c("n", "ln", "un", "d", "ld", "ud") := .(
    number, number - threshold.numbers, number + threshold.numbers,
    date, date - threshold.date, date + threshold.date)]
df2[, c("n", "ln", "un", "d", "ld", "ud") := .(
    other_number, other_number - threshold.numbers, other_number + threshold.numbers,
    date, date - threshold.date, date + threshold.date)]

#perform non-equi join using ctry, num, dates in both ways
res <- rbindlist(list(
    df1[df2, on=.(ctry, n>=ln, n<=un, d>=ld, d<=ud),
        .(date1=x.date, date2=i.date, city1=x.city, city2=i.city, ctry1=x.ctry, ctry2=i.ctry, number, col, other_number, other_col)],
    df2[df1, on=.(ctry, n>=ln, n<=un, d>=ld, d<=ud),
        .(date1=i.date, date2=x.date, city1=i.city, city2=x.city, ctry1=i.ctry, ctry2=x.ctry, number, col, other_number, other_col)]),
    use.names=TRUE, fill=TRUE)

#determine if cities are substrings of one and another
res[, city_match := {
    i <- mapply(grepl, city1, city2) | mapply(grepl, city2, city1)
    replace(i, is.na(i), TRUE)
}]

#just like SQL coalesce (there is a version in dev in rdatatable github)
coalesce <- function(...) Reduce(function(x, y) fifelse(!is.na(y), y, x), list(...))

#for rows that are matching or no matches to be found
ans1 <- unique(res[(city_match), .(date=coalesce(date1, date2),
    city=coalesce(city1, city2),
    ctry=coalesce(ctry1, ctry2),
    number, col, other_number, other_col)])

#for rows that are close in terms of dates and numbers but are diff cities
ans2 <- res[(!city_match), .(date=c(.BY$date1, .BY$date2),
        city=c(.BY$city1, .BY$city2),
        ctry=c(.BY$ctry1, .BY$ctry2),
        number=c(.BY$number, NA),
        col=c(.BY$col, NA),
        other_number=c(NA, .BY$other_number),
        other_col=c(NA, .BY$other_col)),
    names(res)][, seq_along(names(res)) := NULL]

#final desired output
setorder(rbindlist(list(ans1, ans2)), date, city, number, na.last=TRUE)[]

keluaran:

          date        city        ctry number      col other_number other_col
 1: 1999-02-24        Rome       Italy     40   banana           NA      <NA>
 2: 1999-02-24        Rome       Italy     NA     <NA>           45       red
 3: 1999-03-14      Warsaw      Poland     NA     <NA>          780      blue
 4: 1999-03-16      Warsaw      Poland     70    apple           NA      <NA>
 5: 1999-06-12  East-Paris      France     20   banana           17     green
 6: 1999-06-29  Copenhagen     Denmark     60 cucumber           61    orange
 7: 1999-07-16      Moscow      Russia     80    peach           NA      <NA>
 8: 1999-07-17      Moscow      Russia     NA     <NA>           85       red
 9: 2000-01-29       Tunis     Tunisia     NA     <NA>           90     black
10: 2000-08-29      London          UK     30     pear           NA      <NA>
11: 2000-08-29 near London          UK     NA     <NA>         3100      blue
12: 2001-04-17        Bern Switzerland     50    lemon           NA      <NA>
13: 2001-04-17      Zurich Switzerland     NA     <NA>           51    purple
14: 2001-08-29       Tunis     Tunisia     90   cherry           NA      <NA>
15: 2002-07-01      Vienna     Austria     NA     <NA>          101     beige
16: 2002-07-30      Vienna     Austria    100   cherry           NA      <NA>
17: 2003-08-29      Berlin     Germany     10    apple           13    yellow

Anda dapat menguji citypertandingan dengan grepldan ctrydengan sederhana ==. Bagi mereka yang cocok hingga di sini Anda dapat menghitung perbedaan tanggal dengan mengonversi datemenggunakan as.Datedan membandingkannya dengan a difftime. The numberPerbedaan ini dilakukan dengan cara yang sama.

i1 <- seq_len(nrow(df1)) #Store all rows 
i2 <- seq_len(nrow(df2))
res <- do.call(rbind, sapply(seq_len(nrow(df1)), function(i) { #Loop over all rows in df1
  t1 <- which(df1$ctry[i] == df2$ctry) #Match ctry
  t2 <- grepl(df1$city[i], df2$city[t1]) | sapply(df2$city[t1], grepl, df1$city[i]) #Match city
  t1 <- t1[t2 & abs(as.Date(df1$date[i]) - as.Date(df2$date[t1[t2]])) <=
    as.difftime(threshold.date, units = "days") & #Test for date difference
    abs(df1$number[i] - df2$other_number[t1[t2]]) <= threshold.numbers] #Test for number difference
  if(length(t1) > 0) { #Match found
    i1 <<- i1[i1!=i] #Remove row as it was found
    i2 <<- i2[i2!=t1]
    cbind(df1[i,], df2[t1,c("other_number","other_col")], match=".") 
  }
}))
rbind(res
    , cbind(df1[i1,], other_number=NA, other_col=NA, match="1")
    , cbind(df2[i2,1:3], number=NA, col=NA, other_number=df2[i2,4]
            , other_col=df2[i2,5], match="2"))
#          date        city        ctry number      col other_number other_col match
#1   2003-08-29      Berlin     Germany     10    apple           13    yellow     .
#2   1999-06-12       Paris      France     20   banana           17     green     .
#6   1999-06-30  Copenhagen     Denmark     60 cucumber           61    orange     .
#3   2000-08-29      London          UK     30     pear           NA      <NA>     1
#4   1999-02-24        Rome       Italy     40   banana           NA      <NA>     1
#5   2001-04-17        Bern Switzerland     50    lemon           NA      <NA>     1
#7   1999-03-16      Warsaw      Poland     70    apple           NA      <NA>     1
#8   1999-07-16      Moscow      Russia     80    peach           NA      <NA>     1
#9   2001-08-29       Tunis     Tunisia     90   cherry           NA      <NA>     1
#10  2002-07-30      Vienna     Austria    100   cherry           NA      <NA>     1
#31  2000-08-29 near London          UK     NA     <NA>         3100      blue     2
#41  1999-02-24        Rome       Italy     NA     <NA>           45       red     2
#51  2001-04-17      Zurich Switzerland     NA     <NA>           51    purple     2
#71  1999-03-14      Warsaw      Poland     NA     <NA>          780      blue     2
#81  1999-07-17      Moscow      Russia     NA     <NA>           85       red     2
#91  2000-01-29       Tunis     Tunisia     NA     <NA>           90     black     2
#101 2002-07-01      Vienna     Austria     NA     <NA>          101     beige     2

Berikut ini adalah pendekatan fleksibel yang memungkinkan Anda menentukan kumpulan kriteria penggabungan yang Anda pilih.

Persiapan kerja

Saya memastikan bahwa semua string df1dan df2string, bukan faktor (seperti dicatat dalam beberapa jawaban lainnya). Saya juga membungkus kurma as.Dateuntuk membuat mereka kencan nyata.

Tentukan kriteria gabungan

Buat daftar daftar. Setiap elemen dari daftar utama adalah satu kriteria; anggota kriteria adalah

• final.col.name: nama kolom yang kita inginkan di tabel akhir
• col.name.1: nama kolom di df1
• col.name.2: nama kolom di df2
• exact: boolean; haruskah kita melakukan pencocokan tepat pada kolom ini?
• threshold: threshold (jika kami tidak melakukan pencocokan tepat)
• match.function: fungsi yang mengembalikan apakah baris cocok atau tidak (untuk kasus khusus seperti menggunakan grepluntuk pencocokan string; perhatikan bahwa fungsi ini harus di -vektor)

merge.criteria = list(
  list(final.col.name = "date",
       col.name.1 = "date",
       col.name.2 = "date",
       exact = F,
       threshold = 5),
  list(final.col.name = "city",
       col.name.1 = "city",
       col.name.2 = "city",
       exact = F,
       match.function = function(x, y) {
         return(mapply(grepl, x, y) |
                  mapply(grepl, y, x))
       }),
  list(final.col.name = "ctry",
       col.name.1 = "ctry",
       col.name.2 = "ctry",
       exact = T),
  list(final.col.name = "number",
       col.name.1 = "number",
       col.name.2 = "other_number",
       exact = F,
       threshold = 3)
)

Berfungsi untuk menggabungkan

Fungsi ini membutuhkan tiga argumen: dua frame data yang ingin kami gabungkan, dan daftar kriteria kecocokan. Itu hasil sebagai berikut:

1. Iterasi melalui kriteria pertandingan dan tentukan pasangan baris mana yang memenuhi atau tidak memenuhi semua kriteria. (Terinspirasi oleh jawaban @ GKi, ia menggunakan indeks baris alih-alih melakukan gabungan luar penuh, yang mungkin kurang intensif memori untuk kumpulan data besar.)
2. Buat kerangka data kerangka hanya dengan baris yang kita inginkan (baris yang digabungkan dalam kasus yang cocok, baris yang tidak digabungkan untuk catatan yang tidak cocok).
3. Iterasi melalui kolom frame data asli dan gunakan untuk mengisi kolom yang diinginkan dalam bingkai data baru. (Lakukan ini terlebih dahulu untuk kolom yang muncul dalam kriteria kecocokan, lalu untuk kolom lainnya yang tersisa.)

library(dplyr)
merge.data.frames = function(df1, df2, merge.criteria) {
  # Create a data frame with all possible pairs of rows from df1 and rows from
  # df2.
  row.decisions = expand.grid(df1.row = 1:nrow(df1), df2.row = 1:nrow(df2))
  # Iterate over the criteria in merge.criteria.  For each criterion, flag row
  # pairs that don't meet the criterion.
  row.decisions$merge = T
  for(criterion in merge.criteria) {
    # If we're looking for an exact match, test for equality.
    if(criterion$exact) {
      row.decisions$merge = row.decisions$merge &
        df1[row.decisions$df1.row,criterion$col.name.1] == df2[row.decisions$df2.row,criterion$col.name.2]
    }
    # If we're doing a threshhold test, test for difference.
    else if(!is.null(criterion$threshold)) {
      row.decisions$merge = row.decisions$merge &
        abs(df1[row.decisions$df1.row,criterion$col.name.1] - df2[row.decisions$df2.row,criterion$col.name.2]) <= criterion$threshold
    }
    # If the user provided a function, use that.
    else if(!is.null(criterion$match.function)) {
      row.decisions$merge = row.decisions$merge &
        criterion$match.function(df1[row.decisions$df1.row,criterion$col.name.1],
                                 df2[row.decisions$df2.row,criterion$col.name.2])
    }
  }
  # Create the new dataframe.  Just row numbers of the source dfs to start.
  new.df = bind_rows(
    # Merged rows.
    row.decisions %>% filter(merge) %>% select(-merge),
    # Rows from df1 only.
    row.decisions %>% group_by(df1.row) %>% summarize(matches = sum(merge)) %>% filter(matches == 0) %>% select(df1.row),
    # Rows from df2 only.
    row.decisions %>% group_by(df2.row) %>% summarize(matches = sum(merge)) %>% filter(matches == 0) %>% select(df2.row)
  )
  # Iterate over the merge criteria and add columns that were used for matching
  # (from df1 if available; otherwise from df2).
  for(criterion in merge.criteria) {
    new.df[criterion$final.col.name] = coalesce(df1[new.df$df1.row,criterion$col.name.1],
                                                df2[new.df$df2.row,criterion$col.name.2])
  }
  # Now add all the columns from either data frame that weren't used for
  # matching.
  for(other.col in setdiff(colnames(df1),
                           sapply(merge.criteria, function(x) x$col.name.1))) {
    new.df[other.col] = df1[new.df$df1.row,other.col]
  }
  for(other.col in setdiff(colnames(df2),
                           sapply(merge.criteria, function(x) x$col.name.2))) {
    new.df[other.col] = df2[new.df$df2.row,other.col]
  }
  # Return the result.
  return(new.df)
}

Terapkan fungsinya, dan kita selesai

df = merge.data.frames(df1, df2, merge.criteria)