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Minus operation of data frames

I have 2 data frames df1 and df2.

df1

df2

df1 <- data.frame(c1=c("a","b","c","d"),c2=c(1,2,3,4) )
df2 <- data.frame(c1=c("c","d","e","f"),c2=c(3,4,5,6) )

> df1
c1 c2
1 a 1
2 b 2
3 c 3
4 d 4

> df2
c1 c2
1 c 3
2 d 4
3 e 5
4 f 6


I need to perform set operation of these 2 data frames. I used merge(df1,df2,all=TRUE) and merge(df1,df2,all=FALSE) method to get the union and intersection of these data frames and got the required output. What is the function to get the minus of these data frames,that is all the positions existing on one data frame but not the other? I need the following output.

merge(df1,df2,all=TRUE)

merge(df1,df2,all=FALSE)

c1 c2
1 a 1
2 b 2


setdiff

data.frame

setdiff

c,3

d,4

setdiff(df1, df2)

df2

a,1 and b,2

setdiff

setdiff(df1, df2)

setdiff(df2, df1)

7 Answers
7

I remember coming across this exact issue quite a few months back. Managed to sift through my Evernote one-liners.

Note: This is not my solution. Credit goes to whoever wrote it (whom I can't seem to find at the moment).

If you don't worry about rownames then you can do:

rownames

df1[!duplicated(rbind(df2, df1))[-seq_len(nrow(df2))], ]
# c1 c2
# 1 a 1
# 2 b 2


Edit: A data.table solution:

data.table

dt1 <- data.table(df1, key="c1")
dt2 <- data.table(df2)
dt1[!dt2]


or better one-liner (from v1.9.6+):

setDT(df1)[!df2, on="c1"]


This returns all rows in df1 where df2$c1 doesn't have a match with df1$c1.

df1

df2$c1

df1$c1

data.table

setdiff

%in%

I prefer sqldf package:

sqldf

require(sqldf)
sqldf("select * from df1 except select * from df2")

## c1 c2
## 1 a 1
## 2 b 2


You can create identifier columnas then subset:

e.g.

df1 <- data.frame(c1=c("a","b","c","d"),c2=c(1,2,3,4), indf1 = rep("Y",4) )
df2 <- data.frame(c1=c("c","d","e","f"),c2=c(3,4,5,6),indf2 = rep("Y",4) )
merge(df1,df2)
# c1 c2 indf1 indf2
#1 c 3 Y Y
#2 d 4 Y Y

bigdf <- merge(df1,df2,all=TRUE)
# c1 c2 indf1 indf2
#1 a 1 Y <NA>
#2 b 2 Y <NA>
#3 c 3 Y Y
#4 d 4 Y Y
#5 e 5 <NA> Y
#6 f 6 <NA> Y


Then subset how you wish:

bigdf[is.na(bigdf$indf1) ,]
# c1 c2 indf1 indf2
#5 e 5 <NA> Y
#6 f 6 <NA> Y

bigdf[is.na(bigdf$indf2) ,] #<- output you requested those not in df2
# c1 c2 indf1 indf2
#1 a 1 Y <NA>
#2 b 2 Y <NA>


all.x=TRUE

indf2

If you're not planning on using any of the actual data in d2, then you don't need merge at all:

d2

merge

df1[!(df1$c1 %in% df2$c1), ]


You can check the values in both columns and subset like this (just adding another solution):

na.omit( df1[ sapply( 1:ncol(df1) , function(x) ! df1[,x] %in% df2[,x] ) , ] )
# c1 c2
#1 a 1
#2 b 2


One issue with https://stackoverflow.com/a/16144262/2055486 is it assumes neither data frame already has duplicated rows. The following function removes that limitation and also works with arbitrary user defined columns in x or y.

The implementation uses a similar idea to the implementation of duplicated.data.frame in concatenating the columns together with a separator. duplicated.data.frame uses "r", which can cause collisions if the entries have embedded "r" characters. This uses the ASCII record separator "30" which will have a much lower chance of appearing in input data.

duplicated.data.frame

duplicated.data.frame

"r"

"r"

"30"

setdiff.data.frame <- function(x, y,
by = intersect(names(x), names(y)),
by.x = by, by.y = by)
stopifnot(
is.data.frame(x),
is.data.frame(y),
length(by.x) == length(by.y))

!do.call(paste, c(x[by.x], sep = "30")) %in% do.call(paste, c(y[by.y], sep = "30"))


# Example usage
# remove all 4 or 6 cylinder 4 gear cars or 8 cylinder 3 gear rows
to_remove <- data.frame(cyl = c(4, 6, 8), gear = c(4, 4, 3))
mtcars[setdiff.data.frame(mtcars, to_remove), ]
#> mpg cyl disp hp drat wt qsec vs am gear carb
#> Hornet 4 Drive 21.4 6 258.0 110 3.08 3.215 19.44 1 0 3 1
#> Valiant 18.1 6 225.0 105 2.76 3.460 20.22 1 0 3 1
#> Toyota Corona 21.5 4 120.1 97 3.70 2.465 20.01 1 0 3 1
#> Porsche 914-2 26.0 4 120.3 91 4.43 2.140 16.70 0 1 5 2
#> Lotus Europa 30.4 4 95.1 113 3.77 1.513 16.90 1 1 5 2
#> Ford Pantera L 15.8 8 351.0 264 4.22 3.170 14.50 0 1 5 4
#> Ferrari Dino 19.7 6 145.0 175 3.62 2.770 15.50 0 1 5 6
#> Maserati Bora 15.0 8 301.0 335 3.54 3.570 14.60 0 1 5 8

# with differing column names
to_remove2 <- data.frame(a = c(4, 6, 8), b = c(4, 4, 3))
mtcars[setdiff.data.frame(mtcars, to_remove2, by.x = c("cyl", "gear"), by.y = c("a", "b")), ]
#> mpg cyl disp hp drat wt qsec vs am gear carb
#> Hornet 4 Drive 21.4 6 258.0 110 3.08 3.215 19.44 1 0 3 1
#> Valiant 18.1 6 225.0 105 2.76 3.460 20.22 1 0 3 1
#> Toyota Corona 21.5 4 120.1 97 3.70 2.465 20.01 1 0 3 1
#> Porsche 914-2 26.0 4 120.3 91 4.43 2.140 16.70 0 1 5 2
#> Lotus Europa 30.4 4 95.1 113 3.77 1.513 16.90 1 1 5 2
#> Ford Pantera L 15.8 8 351.0 264 4.22 3.170 14.50 0 1 5 4
#> Ferrari Dino 19.7 6 145.0 175 3.62 2.770 15.50 0 1 5 6
#> Maserati Bora 15.0 8 301.0 335 3.54 3.570 14.60 0 1 5 8


I think the simplest solution is with dplyr (tidyverse).

require(tidyverse)
anti_join(df1, df2)


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