通过为大于n个字符的单词部分匹配两列来设置子行

时间:2016-08-17 19:59:47

标签: python regex r pandas stringr

我愿意在python pandas中这样做,但在R中我有以下df:

    result<-structure(list(traffic_Count_Street = c("San Angelo", "W Commerce St", 
"W Commerce St", "S Gevers St", "Austin Hwy", "W Evergreen St"
), unit_Street = c("San Pedro Ave", "W Commerce", "W Commerce", 
"S New Braunfels", "Austin Highway", "W Cypress")), .Names = c("traffic_Count_Street", 
"unit_Street"), row.names = c(1L, 17L, 18L, 34L, 260L, 273L), class = "data.frame")

1             San Angelo   San Pedro Ave
17         W Commerce St      W Commerce
18         W Commerce St      W Commerce
34           S Gevers St S New Braunfels
260           Austin Hwy  Austin Highway
273       W Evergreen St       W Cypress

对于每一行,如果其中一个字符(大于3个字符)与另一个字符匹配,我希望将第1列与第2列部分匹配。

我会删除:

1             San Angelo   San Pedro Ave
34           S Gevers St   S New Braunfels
273       W Evergreen St   W Cypress

并保持:

17         W Commerce St      W Commerce
18         W Commerce St      W Commerce
260           Austin Hwy  Austin Highway

我尝试以下列方式使用stringR,但它不起作用:

result$unit_Street[str_detect(result$traffic_Count_Street, "\\w{3}")]

1 个答案:

答案 0 :(得分:1)

创建具有阈值调整的距离过滤器。然后你可以调整,直到你得到你想要的结果。在这种情况下,Levenshtein距离为5表现良好:

distanceFilter <- function(df, thresh=5) {
  ind <- apply(df, 1, function(x) adist(x[1], x[2]) < thresh )
  df[ind,]
}

distanceFilter(result, 5)
#     traffic_Count_Street    unit_Street
# 17         W Commerce St     W Commerce
# 18         W Commerce St     W Commerce
# 260           Austin Hwy Austin Highway

要了解详情,请参阅the wiki pageR doc help page