Dplyr：干净的双管姓氏

Question

我有一个如下名称的 data.frame，其中包含一些姓氏的样本，后跟一个首字母（例如 Smith S 或 Lopez-Garcia M）：

df<-data.frame(names=c("Adu-Amankwah E",
"Smith Dawson E",
"Lopez-Garcia M",
"Lopez Garcia MA",
"Garcia MAC",
"Lopez Garcia MA",
"Garcia MAC"))

我想把那些双管姓的名字都抽出来，做一点清理：

1. 拔出带有连字符 (-) 或两个姓氏（例如 Lopez Garcia）的任何名称。
2. 我需要用 Lopez- 替换以下任何一项：Lopez Garcia MA、Lopez-Garcia MA 或 Garcia MAC加西亚 M。 Smith Dawson E 应该是 Smith-Dawson E.

输出如下：

df<-data.frame(names=c("Adu-Amankwah E",
"Smith-Dawson E",
"Lopez-Garcia M",
"Lopez-Garcia M",
"Lopez-Garcia M",
"Lopez-Garcia M",
"Lopez-Garcia M"))

Answer 1

正如我在my comments 中提到的，这里的挑战不是解析 character 字符串，而是定义逻辑 到

• 在代表标签（"Lopez-Garcia M"）下关联同名变体（例如"Garcia MAC"、"Lopez Garcia MA"）； 仍然
• 避免将不同名称（如"Andy Garcia"）的相似变体（如"Garcia A"）混为一谈。

因此，您最好的方法可能是为名称的已知变体定义一个mapping 表。

文字映射

文字映射涉及在其真正代表的名称旁边输入每个已知的变体。

mapping_lit <- data.frame(
  True_Name = c("Adu-Amankwah E", "Smith-Dawson E", "Lopez-Garcia M", "Lopez-Garcia M",  "Lopez-Garcia M"),
  Variant   = c("Adu-Amankwah E", "Smith Dawson E", "Lopez-Garcia M", "Lopez Garcia MA", "Garcia MAC")
)

mapping_lit
#>        True_Name         Variant
#> 1 Adu-Amankwah E  Adu-Amankwah E
#> 2 Smith-Dawson E  Smith Dawson E
#> 3 Lopez-Garcia M  Lopez-Garcia M
#> 4 Lopez-Garcia M Lopez Garcia MA
#> 5 Lopez-Garcia M      Garcia MAC

拥有mapping 后，一个简单的dplyr::*_join() 就可以解决问题

library(dplyr)

# The LEFT JOIN preserves any names without matches, so you can handle them as you wish.
left_join(
  df,
  mapping_lit,
  by = c("names" = "Variant")
)

结果如下：

            names      True_Name
1  Adu-Amankwah E Adu-Amankwah E
2  Smith Dawson E Smith-Dawson E
3  Lopez-Garcia M Lopez-Garcia M
4 Lopez Garcia MA Lopez-Garcia M
5      Garcia MAC Lopez-Garcia M
6 Lopez Garcia MA Lopez-Garcia M
7      Garcia MAC Lopez-Garcia M

正则表达式映射

如果您对regular expressions 足够熟练，您可以定义一个正则表达式来表示每个True_Name 上的所有变体：

mapping_rgx <- data.frame(
  True_Name = c("Adu-Amankwah E",             "Smith-Dawson E",             "Lopez-Garcia M"),
  Pattern   = c("^(Adu[- ]?)?Amankwah( E)?$", "^(Smith[- ]?)?Dawson( E)?$", "^(Lopez[- ]?)?Garcia( M(AC?)?)?$")
)

mapping_rgx
#>        True_Name                          Pattern
#> 1 Adu-Amankwah E       ^(Adu[- ]?)?Amankwah( E)?$
#> 2 Smith-Dawson E       ^(Smith[- ]?)?Dawson( E)?$
#> 3 Lopez-Garcia M ^(Lopez[- ]?)?Garcia( M(AC?)?)?$

一旦你有了这个映射，你就需要一个fuzzyjoin::regex_*_join() 来匹配变体

library(fuzzyjoin)

# The LEFT JOIN preserves any names without matches, so you can handle them as you wish.
regex_left_join(
  df,
  mapping_rgx,
  by = c("names" = "Pattern"),
  # Account for typos in capitalization.
  ignore_case = TRUE
)

结果如下：

            names      True_Name                          Pattern
1  Adu-Amankwah E Adu-Amankwah E         (Adu[- ]?)?Amankwah( E)?
2  Smith Dawson E Smith-Dawson E         (Smith[- ]?)?Dawson( E)?
3  Lopez-Garcia M Lopez-Garcia M ^(Lopez[- ]?)?Garcia( M(AC?)?)?$
4 Lopez Garcia MA Lopez-Garcia M ^(Lopez[- ]?)?Garcia( M(AC?)?)?$
5      Garcia MAC Lopez-Garcia M ^(Lopez[- ]?)?Garcia( M(AC?)?)?$
6 Lopez Garcia MA Lopez-Garcia M ^(Lopez[- ]?)?Garcia( M(AC?)?)?$
7      Garcia MAC Lopez-Garcia M ^(Lopez[- ]?)?Garcia( M(AC?)?)?$

警告

我也是commented，在这种情况下我可能不推荐stringdist 方法。每个名称不仅拼写不同，结构也不同。两个不同人的两个结构相似的条目完全有可能

Variant	True_Name
Garcia A	Andy Garcia
Garcia MAC	Lopez-Garcia M
Lopez-Garcia M	Lopez-Garcia M

与相同名称的两个不同结构的变体相比，字符串距离更短：

# Run the full gamut of methods for 'stringdist::stringdist()'.
methods <- c(
  "osa", "lv", "dl", "hamming", "lcs", "qgram",
  "cosine", "jaccard", "jw", "soundex"
)


# Display string distances for variants of the same and of different names:
rbind(
  # Compare different names.
  sapply(X = methods, FUN = function(x) {stringdist::stringdist(
    a = "Garcia MAC", b = "Garcia A",
    method = x
  )}),
  # Compare variations on the same name.
  sapply(X = methods, FUN = function(x) {stringdist::stringdist(
    a = "Garcia MAC", b = "Lopez-Garcia M",
    method = x
  )})
)

#>      osa lv dl hamming lcs qgram     cosine   jaccard         jw soundex
#> [1,]   2  2  2     Inf   2     2 0.08712907 0.2222222 0.06666667       1
#> [2,]   8  8  8     Inf   8     8 0.27831216 0.5333333 0.20952381       1