# Comparing two data frames with different number of rows

January 24, 2013
Categorized as: R, R-Bloggers, comparison

I posted a question over on StackOverflow on an efficient way of comparing two data frames with the same column structure, but with different rows. What I would like to end up with is an n x m logical matrix where n and m are the number of rows in the first and second data frames, respectively; and the value at the ith row and jth column indicates whether all the values from row i from data frame one is equal to row j from data frame two. To provide some context, this will be used in a propensity score matching algorithm to identify candidate matches that match exactly on any number of covariates. In addition to the approaches I had, joran provided an approach using the Vectorize function (thanks again as I learned another nice function). I decided to put three approaches to a race…

To understand what I need, I’ll start with a small example with two data frames, one with 4 rows, the other with 3, and each has two variables, one logical and the other numeric. As an aside, I only need this to work for integers, factors, characters, and logical types therefore avoiding issues of comparing numerics.

> df1 <- data.frame(row.names=1:4, var1=c(TRUE, TRUE, FALSE, FALSE), var2=c(1,2,3,4))
> df2 <- data.frame(row.names=5:7, var1=c(FALSE, TRUE, FALSE), var2=c(5,2,3))
> df1
var1 var2
1  TRUE    1
2  TRUE    2
3 FALSE    3
4 FALSE    4
> df2
var1 var2
5 FALSE    5
6  TRUE    2
7 FALSE    3

First, let’s consider the case when there is only one variable:

> system.time({
+ 	df3 <- sapply(df2$var1, FUN=function(x) { x == df1$var1 })
+ 	dimnames(df3) <- list(row.names(df1), row.names(df2))
+ })
user  system elapsed
0       0       0
> df3
5     6     7
1 FALSE  TRUE FALSE
2 FALSE  TRUE FALSE
3  TRUE FALSE  TRUE
4  TRUE FALSE  TRUE

This is pretty straight forward. Now I want the same type of result, but to compare more than one column (in the final implementation I need to handle any number of columns so not necessarily limited to one or two).

The first approach uses nested apply functions.

> system.time({
+ 	m1 <- t(as.matrix(df1))
+ 	m2 <- as.matrix(df2)
+ 	df4 <- apply(m2, 1, FUN=function(x) { apply(m1, 2, FUN=function(y) { all(x == y) } ) })
+ })
user  system elapsed
0.001   0.000   0.001
> df4
5     6     7
1 FALSE FALSE FALSE
2 FALSE  TRUE FALSE
3 FALSE FALSE  TRUE
4 FALSE FALSE FALSE

Secondly, using the Vectorize and outer functions.

> system.time({
+ 	foo <- Vectorize(function(x,y) { all(df1[x,] == df2[y,]) })
+ 	df5 <- outer(1:nrow(df1), 1:nrow(df2), FUN=foo)
+ })
user  system elapsed
0.005   0.000   0.006
> df5
[,1]  [,2]  [,3]
[1,] FALSE FALSE FALSE
[2,] FALSE  TRUE FALSE
[3,] FALSE FALSE  TRUE
[4,] FALSE FALSE FALSE

Lastly, we’ll create a new character vector by pasting the other variables together.

> system.time({
+ 	df1$var3 <- apply(df1, 1, paste, collapse='.') + df2$var3 <- apply(df2, 1, paste, collapse='.')
+ 	df6 <- sapply(df2$var3, FUN=function(x) { x == df1$var3 })
+ 	dimnames(df6) <- list(row.names(df1), row.names(df2))
+ })
user  system elapsed
0.000   0.000   0.001
> df6
5     6     7
1 FALSE FALSE FALSE
2 FALSE  TRUE FALSE
3 FALSE FALSE  TRUE
4 FALSE FALSE FALSE

We can already see with this small example that the Vectorize approach is the slowest. However, let’s try a larger example. First we’ll create two data frames, one with 1,000 rows and the second with 1,500. The resulting matrix will be 1,000 x 1,500.

set.seed(2112)
df1 <- data.frame(row.names=1:1000,
var1=sample(c(TRUE,FALSE), 1000, replace=TRUE),
var2=sample(1:10, 1000, replace=TRUE) )
df2 <- data.frame(row.names=1001:2500,
var1=sample(c(TRUE,FALSE), 1500, replace=TRUE),
var2=sample(1:10, 1500, replace=TRUE))

Nested apply functions approach:

> system.time({
+ 	m1 <- t(as.matrix(df1))
+ 	m2 <- as.matrix(df2)
+ 	df4 <- apply(m2, 1, FUN=function(x) { apply(m1, 2, FUN=function(y) { all(x == y) } ) })
+ })
user  system elapsed
10.807   0.043  11.096 

Vectorize approach:

> system.time({
+ 	foo <- Vectorize(function(x,y) { all(df1[x,] == df2[y,]) })
+ 	df5 <- outer(1:nrow(df1), 1:nrow(df2), FUN=foo)
+ })
user  system elapsed
390.904   0.808 392.134 

Combined columns approach:

> system.time({
+ 	df1$var3 <- apply(df1, 1, paste, collapse='.') + df2$var3 <- apply(df2, 1, paste, collapse='.')
+ 	df6 <- sapply(df2$var3, FUN=function(x) { x == df1$var3 })
+ 	dimnames(df6) <- list(row.names(df1), row.names(df2))
+ })
user  system elapsed
0.421   0.000   0.422 

The combined column approach is by far the fasted way, and it makes good since. It is a bit surprising (at least to me), how much worse the Vectorize and outer functions are. Moreover, I am a bit concerned about potential issues with the paste method and doing comparisons on those results. Please feel free to leave comments below if there are other approaches.