（通过使用AGREP R）创建由名称的模糊匹配的唯一ID(Create a unique ID by

使用R，我想通过一年的城市构成的数据集上的人的名字相匹配。由于一些拼写错误，精确匹配是不可能的，所以我试图用AGREP（）来模糊匹配的名字。

数据集的样品块的结构如下：

df <- data.frame(matrix( c("1200013","1200013","1200013","1200013","1200013","1200013","1200013","1200013",                             "1996","1996","1996","1996","2000","2000","2004","2004","AGUSTINHO FORTUNATO FILHO","ANTONIO PEREIRA NETO","FERNANDO JOSE DA COSTA","PAULO CEZAR FERREIRA DE ARAUJO","PAULO CESAR FERREIRA DE ARAUJO","SEBASTIAO BOCALOM RODRIGUES","JOAO DE ALMEIDA","PAULO CESAR FERREIRA DE ARAUJO"), ncol=3,dimnames=list(seq(1:8),c("citycode","year","candidate")) ))

整洁的版本：

  citycode year                      candidate
1  1200013 1996      AGUSTINHO FORTUNATO FILHO
2  1200013 1996           ANTONIO PEREIRA NETO
3  1200013 1996         FERNANDO JOSE DA COSTA
4  1200013 1996 PAULO CEZAR FERREIRA DE ARAUJO
5  1200013 2000 PAULO CESAR FERREIRA DE ARAUJO
6  1200013 2000    SEBASTIAO BOCALOM RODRIGUES
7  1200013 2004                JOAO DE ALMEIDA
8  1200013 2004 PAULO CESAR FERREIRA DE ARAUJO

我想在每个城市分别检查，是否有出现在数年的候选人。例如，在示例中，

PAULO塞萨尔·费雷拉·德阿劳霍
保罗·塞萨尔费雷拉·德阿劳霍

出现两次（带拼写错误）。整个数据集中各候选应被分配一个唯一的数字考生ID。该数据集是相当大（5500个城市，约10万项），所以有点高效的编码将是有益的。任何建议，如何实现这一点？

编辑：这是我尝试（从迄今的评论帮助）是实现手头的任务非常慢（低效率）。任何建议，改进呢？

f <- function(x) {matches <- lapply(levels(x), agrep, x=levels(x),fixed=TRUE, value=FALSE)
                  levels(x) <- levels(x)[unlist(lapply(matches, function(x) x[1]))]
                  x
                }

temp <- tapply(df$candidate, df$citycode, f, simplify=TRUE)
df$candidatenew <- unlist(temp)
df$spellerror <- ifelse(as.character(df$candidate)==as.character(df$candidatenew), 0, 1)

编辑2：现在，在良好的运行速度。问题是比较许多因素在每一步（感谢指出了这一点，蓝魔导师）。减少的比较仅在一组（即市）考生运行80,000线在5秒内命令 - 我能忍受的速度。

df$candidate <- as.character(df$candidate)

f <- function(x) {x <- as.factor(x)
                  matches <- lapply(levels(x), agrep, x=levels(x),fixed=TRUE, value=FALSE)
                  levels(x) <- levels(x)[unlist(lapply(matches, function(x) x[1]))]
                  as.character(x)
                }

temp <- tapply(df$candidate, df$citycode, f, simplify=TRUE)
df$candidatenew <- unlist(temp)
df$spellerror <- ifelse(as.character(df$candidate)==as.character(df$candidatenew), 0, 1)

Answer 1:

这里是我的投篮吧。它可能不是很有效，但我认为它会完成这项工作。我认为df$candidates是阶级因素。

#fuzzy matches candidate names to other candidate names
#compares each pair of names only once
##by looking at names that have a greater index
matches <- unlist(lapply(1:(length(levels(df[["candidate"]]))-1),
    function(x) {max(x,x + agrep(
        pattern=levels(df[["candidate"]])[x], 
        x=levels(df[["candidate"]])[-seq_len(x)]
    ))}
))
#assigns new levels (omits the last level because that doesn't change)
levels(df[["candidate"]])[-length(levels(df[["candidate"]]))] <- 
    levels(df[["candidate"]])[matches]