regex - r stringdist 或 levenshtein.distance 替换字符串

Question

我有一个大型数据集，其中包含约一百万个观测值，并以定义的观测值类型作为键控。在数据集中，有大约 900,000 个观察类型错误的观察，50 种可接受的观察类型有大约 850 个(不正确的)变化。

keys <- c("DAY", "EVENING","SUNSET", "DUSK","NIGHT", "MIDNIGHT", "TWILIGHT", "DAWN","SUNRISE", "MORNING")

entries <- c("Day", "day", "SUNSET/DUSK", "DAYS", "dayy", "EVEN", "Evening", "early dusk", "late day", "nite", "red dawn", "Evening Sunset", "mid-night", "midnight", "midnite","DAY", "EVENING","SUNSET", "DUSK","NIGHT", "MIDNIGHT", "TWILIGHT", "DAWN","SUNRISE", "MORNING")

使用 gsub 类似于用手铲挖地下室，在我自己的例子中，是一把断柄的铲子，因为我对 r 和复杂的正则表达式还很陌生。简单的回退(对我来说)是为每个接受的观察类型编写一个 gsub 语句，但这似乎不必要地费力，因为它需要 50 个语句。

我想使用 levenshtein.distance或 stringdist用最短距离字符串替换有问题的条目。正在运行 z <- for (i in length(y)) { z[i] = levenshtein.distance(y[i], x)}不起作用，因为它试图将 (length(x)) 结果传递给每个 y[i]。

如何返回最小距离的结果？我见过function(x) x[2]返回系列中的第二个结果，但如何获得最低的结果？

Answer 1

你可以试试:

library(stringdist)
m <- stringdistmatrix(entries, keys, method = "lv")
a <- keys[apply(m, 1, which.min)]

如果您想尝试不同的算法，请查看 ?'stringdist-metrics'

---

或者按照@RHertel 在评论中提到的那样:

b <- keys[apply(adist(entries, keys), 1, which.min)]

来自 adist() 文档:

Compute the approximate string distance between character vectors. The distance is a generalized Levenshtein (edit) distance, giving the minimal possibly weighted number of insertions, deletions and substitutions needed to transform one string into another.

这两种方法产生相同的结果:

> identical(a, b)
#[1] TRUE