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我正在尝试使用 recordLinkage 包将两个数据集链接在一起，其中一个数据集倾向于给出多个姓氏/中间名，另一个只给出一个姓氏。目前正在使用的字符串比较函数是 Jaro-Winkler 函数，但是返回的分数取决于字符串如何偶然匹配，而不是较短字符串的内容是否包含在较长字符串中的任何位置。这导致创建了许多质量较差的链接。错误权重的可重现示例如下：

library(RecordLinkage)
data1 <- as.data.frame(list("lname" = c("lolli gaggen nazeem", "lolli gaggen nazeem", "lolli gaggen nazeem"),
                           "bday" = c("1908-08-08", "1979-12-12", "1560-06-06") ) )

data2 <- as.data.frame(list("lname" = c("lolli", "gaggen", "nazeem"),
                           "bday" = c("1908-08-08", "1979-12-12", "1560-06-06") ) )

blocking_variable <- c("bday")
pass <- compare.linkage(data1, data2, blockfld = blocking_variable, strcmp = T)
pass_weights <- epiWeights(pass)
getPairs(pass_weights, single.rows = TRUE)

  id1              lname.1     bday.1 id2 lname.2     bday.2    Weight
1   1 lolli gaggen nazheem 1908-08-08   1   lolli 1908-08-08 0.9162463
2   2 lolli gaggen nazheem 1979-12-12   2  gaggen 1979-12-12 0.8697165
3   3 lolli gaggen nazheem 1560-06-06   3 nazheem 1560-06-06 0.6995502

我希望 id 的 2 和 3 获得与 id #1 大致相同的权重，但目前它们要低得多，因为它们的姓在两个数据集中的位置并不完全相同（尽管内容是一致的）。有没有办法可以修改此处使用的字符串比较函数/数据结构，以便考虑不同的顺序？

补充说明：

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2 回答 2

1

您是否考虑过以下方法？

我知道你会知道记录链接和名称很困难。理想情况下，您希望阻止其他可用信息（性别、唯一标识符、dob、位置信息等），然后对名称进行字符串比较。

您提到了具有数百万条记录的大型数据集。只需看看 data.table伟大的 Matt Dowle 的包装 ( https://stackoverflow.com/users/403310/matt-dowle )。

RecordLinkage 包比较慢。您可以轻松改进以下代码，以考虑使用 soundex、双变音、nysii 等的字符串散列技术。

# install.packages("data.table")
library(RecordLinkage)
library(data.table)

data1 <- as.data.frame(list("lname" = c("lolli gaggen nazeeem", "lolli gaggen nazeem", "lollly gaggen nazeem", "matt dowle", "john-smith"),
                           "bday" = c("1908-08-08", "1979-12-12", "1560-06-06", "1979-12-12", "1560-06-06") ) )

data2 <- as.data.frame(list("lname" = c("lolli", "gaggen", "nazeem", "m dowl", "johnny smith"),
                           "bday" = c("1908-08-08", "1979-12-12", "1560-06-06", "1979-12-12", "1560-06-06") ) )


# Coerce to data.tables
setDT(data1)
setDT(data2)

# Define a regex split (we will split all words based on space or hyphen)
split <- " |-"

# Apply a blocking strategy based on bday. Ideally your dataset would allow for additional blocking strategies(?).
block_pairs <- merge(data1, data2, by = "bday", all = T,
            sort = TRUE, suffixes = c(".x", ".y"))

# Store the split up components of each comparison variable.
split1 <- strsplit(block_pairs[["lname.x"]], split)
split2 <- strsplit(block_pairs[["lname.y"]], split)

# Perform jarowinkler comparisons on each combination of components of each string
fc <- jarowinkler(block_pairs[["lname.x"]], block_pairs[["lname.y"]])
pc <- mapply(function(x, y) max(outer(x, y, jarowinkler)), split1, split2)

# Store the max of the full and partial comparisons
block_pairs[, ("winkler.lname") := mapply(function(x,y) max(x,y), fc, pc)]


# Sort by the jarowinkler score
block_pairs <- block_pairs[order(winkler.lname)]

# Inspect
block_pairs

# 0.96 is an appropriate threshold in this instance
block_pairs <- block_pairs[winkler.lname >= 0.96]
于 2018-12-13T22:22:17.647 回答
0

如评论中所述，我对 Khayenes 的回答进行了补充：

library(gtools)

...

# Store the split up components of each comparison variable.
split1 <- strsplit(block_pairs[["lname.x"]], split)
split2 <- strsplit(block_pairs[["lname.y"]], split)

# Recombine tokens into all possible orderings:
make_combinations <- function(x) {
      # Use permutations from the gtools package
      split_names <- permutations(length(x),length(x),x)
      apply(X=split_names, MARGIN=1, FUN=paste0, collapse=' ')
}

split1 <- lapply(X=split1, FUN=`make_combinations`)
split2 <- lapply(X=split2, FUN=`make_combinations`)

# Perform jarowinkler comparisons on each string combination and append it to the table
block_pairs[ ,("winkler.lname") := mapply(function(x, y) max(outer(x, y, jarowinkler)), split1, split2)]

# Sort by the jarowinkler score
block_pairs <- block_pairs[order(winkler.lname)]

# 0.85 is an appropriate threshold in this instance
block_pairs <- block_pairs[winkler.lname >= 0.85]


      bday           lname.x             lname.y    winkler.lname
1: 1908-08-08  lolli gaggen nazeem         lolli     0.8526316
2: 1560-06-06  lolli gaggen nazeem        nazeem     0.8631579
3: 1979-12-12  lolli gaggen nazeem        gaggen     0.8631579
4: 1979-12-12           matt dowle        m dowl     0.9200000
5: 1560-06-06           john-smith  johnny smith     0.9666667
于 2019-02-11T01:22:56.140 回答