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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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