r - Dplyr:清洁双管姓氏

Question

我有一个 data.frame 名称，如下所示，其中有一些姓氏的样本，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"))

我想把那些双管齐下的姓氏全部拉出来，稍微清理一下:

1. 找出带有连字符 (-) 或两个姓氏(例如 Lopez Garcia)的任何一个。
2. 我需要用 Lopez-加西亚 M。而 Smith Dawson E 应该是 Smith-Dawson E.

输出看起来像:

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，这里的挑战与其说是解析 character 字符串，不如说是将逻辑定义为

• 在代表性标签 ("Lopez-Garcia M "); 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映射 表。

文字映射

文字映射涉及在其真正代表的名称旁边键入每个已知变体。

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

一旦您有了映射，一个简单的dplyr::*_join()应该可以解决问题

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，您可以只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?)?)?$

有了这个映射后，您将需要一个 fuzzyjoin::regex_*_join()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 ，我可能不会推荐 stringdist在这种情况下的方法。每个名字不仅在拼写上不同，而且在结构上也不同。两个不同人的两个结构相似的条目完全有可能

<表类="s-表"><头>变体真名<正文>加西亚A安迪加西亚加西亚麦克洛佩兹-加西亚中号洛佩兹-加西亚中号洛佩兹-加西亚中号

与相同名称上的两个不同结构变体相比，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