通过模糊匹配名称创建唯一ID(通过使用R的agrep)

Question

使用R,我尝试在按年份和城市构建的数据集中匹配人名.由于一些拼写错误,无法进行精确匹配,因此我尝试使用agrep()来模糊匹配名称.

数据集的样本块结构如下:

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

我想分别检查每个城市,是否有候选人出现在几年.例如,在示例中,

PAULO CEZAR FERREIRA DE ARAUJO

PAULO CESAR FERREIRA DE ARAUJO

出现两次(拼写错误).应为整个数据集中的每个候选者分配唯一的数字候选ID.数据集相当大(5500个城市,大约100K条目),因此稍微有效的编码会有所帮助.有关如何实现这一点的任何建议？

编辑:这是我在实现手头任务时非常缓慢(效率低下)的尝试(在迄今为止的评论的帮助下).有关改进的建议吗？

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:现在以良好的速度运行.问题在于每一步都与许多因素进行比较(感谢你指出这一点,Blue Magister).将比较减少到只有一组中的候选者(即一个城市),在5秒内运行命令,持续80,000行 - 这是我可以忍受的速度.

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

这是我的尝试。它可能不是很有效，但我认为它会完成工作。我认为这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]