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After posting “Removing Records by Duplicate Values” yesterday, I had an interesting communication thread with my friend Jeffrey Allard tonight regarding how to code this in R, a combination of order() and duplicated() or sqldf().

Afterward, I did a simple efficiency comparison between two methods as below. The comparison result is pretty self-explanatory. In terms of “user time”, dedup1() is at least 10 times more efficient than dedup2().

> library(sqldf)
> df1 <- read.table("../data/credit_count.txt", header = TRUE, sep = ",")
> cat(nrow(df1), ncol(df1), '\n')
13444 14
> # DEDUP WITH ORDER() AND DUPLICATED()
> dedup1 <- function(n){
+   for (i in 1:n){
+     df12 <- df1[order(df1$MAJORDRG, df1$INCOME), ]
+     df13 <- df12[!duplicated(df12\$MAJORDRG), ]
+   }
+ }
> # DEDUP WITH SQLDF()
> dedup2 <- function(n){
+   for (i in 1:n){
+     df22 <- sqldf("select * from df1 order by MAJORDRG, INCOME")
+     df23 <- sqldf("select a.* from df22 as a inner join (select MAJORDRG, min(rowid) as min_id from df22 group by MAJORDRG) as b on a.MAJORDRG = b.MAJORDRG and a.rowid = b.min_id")
+   }
+ }
> # RUN BOTH METHODS 100 TIMES AND COMPARE CPU TIMES
> system.time(dedup2(100))
user  system elapsed
22.581   1.684  26.965
> system.time(dedup1(100))
user  system elapsed
1.732   0.080   2.033


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