2016-11-22 171 views
1

我很難過。如何根據行是否匹配將數據複製到ID變量中的另一行。我正在處理數千個歷史地址,並不是所有地址都完美匹配。但任何差異通常都在地址的末尾,因此使用該值的前4或5個字符應該可以處理它。我想用合適的道路代碼填寫新航。我一直在嘗試dplyr解決方案,並沒有獲得任何地方。任何想法將不勝感激。根據列值複製行

ID<-c(50,50,50,71,71,71) 
ID_Y<-c(505,506,507,715,716,717) 
address<-c("325 Park St N","325 Park St","325 Park","616 Holly","616 Holly Dr","510 Walnut Dr") 
tract<-c(110,NA,NA,223,NA,989) 

AD567<-data.frame(ID,ID_Y,address,tract) 
AD567 
    ID ID_Y  address tract 
1 50 505 325 Park St N 110 
2 50 506 325 Park St NA 
3 50 507  325 Park NA 
4 71 715  616 Holly 223 
5 71 716 616 Holly Dr NA 
6 71 717 510 Walnut Dr 989 

試圖讓這裏:

ID ID_Y  address tract 
1 50 505 325 Park St N 110 
2 50 506 325 Park St 110 
3 50 507  325 Park 110 
4 71 715  616 Holly 223 
5 71 716 616 Holly Dr 223 
6 71 717 510 Walnut Dr 989 
+0

嘗試'na.locf(df)'。它在動物園圖書館。 –

+0

爲什麼你將混合類型存儲在矩陣中?爲什麼不是數據框?無論哪種方式,似乎你需要某種類型的模糊聯接,谷歌它。 –

+0

我會和@David一起去。根據地址對數據幀進行排序並使用'na.locf()'。試試看看。 –

回答

1

這是一個沒有任何附加的庫

# introduce an additional column which serves as heuristic key 
AD567$prefix = substr(AD567$address, 1, 8) 

# extract all records which have a tract code 
TRACT = AD567[! is.na(AD567$tract),c("prefix", "tract")] 
# check if the record is unique per prefix 
aggregate(tract ~ prefix, TRACT, length) 
# ... one may use only those records further on which are unique ... 

# merge both data frames to inject the tract code; make sure nothing 
# is lost from AD567 
AD567 = merge(AD567, TRACT, by="prefix", suffixes = c("", ".ref"), all.x = TRUE) 

# copy over tract code 
AD567$tract = AD567$tract.ref 

# remove utility columns 
AD567 = AD567[, ! colnames(AD567) %in% c("prefix", "tract.ref")] 

請記住,這是一個非常貧窮的啓發式的解決方案。不精確或模糊的數據匹配本身就是一門科學。