Вопрос
How can I achieve a cross join in R ? I know that "merge" can do inner join, outer join. But I do not know how to achieve a cross join in R.
Thanks
Решение
Is it just all=TRUE
?
x<-data.frame(id1=c("a","b","c"),vals1=1:3)
y<-data.frame(id2=c("d","e","f"),vals2=4:6)
merge(x,y,all=TRUE)
From documentation of merge
:
If by or both by.x and by.y are of length 0 (a length zero vector or NULL), the result, r, is the Cartesian product of x and y, i.e., dim(r) = c(nrow(x)*nrow(y), ncol(x) + ncol(y)).
Другие советы
If speed is an issue, I suggest checking out the excellent data.table
package. In the example at the end it's ~90x faster than merge
.
You didn't provide example data. If you just want to get all combinations of two (or more individual) columns, you can use CJ
(cross join):
library(data.table)
CJ(x=1:2,y=letters[1:3])
# x y
#1: 1 a
#2: 1 b
#3: 1 c
#4: 2 a
#5: 2 b
#6: 2 c
If you want to do a cross join on two tables, I haven't found a way to use CJ(). But you can still use data.table
:
x2<-data.table(id1=letters[1:3],vals1=1:3)
y2<-data.table(id2=letters[4:7],vals2=4:7)
res<-setkey(x2[,c(k=1,.SD)],k)[y2[,c(k=1,.SD)],allow.cartesian=TRUE][,k:=NULL]
res
# id1 vals1 id2 vals2
# 1: a 1 d 4
# 2: b 2 d 4
# 3: c 3 d 4
# 4: a 1 e 5
# 5: b 2 e 5
# 6: c 3 e 5
# 7: a 1 f 6
# 8: b 2 f 6
# 9: c 3 f 6
#10: a 1 g 7
#11: b 2 g 7
#12: c 3 g 7
Explanation of the res
line:
- Basically you add a dummy column (k in this example) to one table and set it as the key (
setkey(tablename,keycolumns)
), add the dummy column to the other table, and then join them. - The data.table structure uses column positions and not names in the join, so you have to put the dummy column at the beginning. The
c(k=1,.SD)
part is one way that I have found to add columns at the beginning (the default is to add them to the end). - A standard data.table join has a format of
X[Y]
. The X in this case issetkey(x2[,c(k=1,.SD)],k)
, and the Y isy2[,c(k=1,.SD)]
. allow.cartesian=TRUE
tellsdata.table
to ignore the duplicate key values, and perform a cartesian join (prior versions didn't require this)- The
[,k:=NULL]
at the end just removes the dummy key from the result.
You can also turn this into a function, so it's cleaner to use:
# Version 1; easier to write:
CJ.table.1 <- function(X,Y)
setkey(X[,c(k=1,.SD)],k)[Y[,c(k=1,.SD)],allow.cartesian=TRUE][,k:=NULL]
CJ.table.1(x2,y2)
# id1 vals1 id2 vals2
# 1: a 1 d 4
# 2: b 2 d 4
# 3: c 3 d 4
# 4: a 1 e 5
# 5: b 2 e 5
# 6: c 3 e 5
# 7: a 1 f 6
# 8: b 2 f 6
# 9: c 3 f 6
#10: a 1 g 7
#11: b 2 g 7
#12: c 3 g 7
# Version 2; faster but messier:
CJ.table.2 <- function(X,Y) {
eval(parse(text=paste0("setkey(X[,c(k=1,.SD)],k)[Y[,c(k=1,.SD)],list(",paste0(unique(c(names(X),names(Y))),collapse=","),")][,k:=NULL]")))
}
Here are some speed benchmarks:
# Create a bigger (but still very small) example:
n<-1e3
x3<-data.table(id1=1L:n,vals1=sample(letters,n,replace=T))
y3<-data.table(id2=1L:n,vals2=sample(LETTERS,n,replace=T))
library(microbenchmark)
microbenchmark(merge=merge.data.frame(x3,y3,all=TRUE),
CJ.table.1=CJ.table.1(x3,y3),
CJ.table.2=CJ.table.2(x3,y3),
times=3, unit="s")
#Unit: seconds
# expr min lq median uq max neval
# merge 4.03710225 4.23233688 4.42757152 5.57854711 6.72952271 3
# CJ.table.1 0.06227603 0.06264222 0.06300842 0.06701880 0.07102917 3
# CJ.table.2 0.04740142 0.04812997 0.04885853 0.05433146 0.05980440 3
Note that these data.table
methods are much faster than the merge
method suggested by @danas.zuokas. The two tables with 1,000 rows in this example result in a cross-joined table with 1 million rows. So even if your original tables are small, the result can get big quickly and speed becomes important.
Lastly, recent versions of data.table
require you to add the allow.cartesian=TRUE
(as in CJ.table.1) or specify the names of the columns that should be returned (CJ.table.2). The second method (CJ.table.2) seems to be faster, but requires some more complicated code if you want to automatically specify all the column names. And it may not work with duplicate column names. (Feel free to suggest a simpler version of CJ.table.2)
This was asked years ago, but you can use tidyr::crossing()
to do a cross-join. Definitely the simplest solution of the bunch.
library(tidyr)
league <- c("MLB", "NHL", "NFL", "NBA")
season <- c("2018", "2017")
tidyr::crossing(league, season)
#> # A tibble: 8 x 2
#> league season
#> <chr> <chr>
#> 1 MLB 2017
#> 2 MLB 2018
#> 3 NBA 2017
#> 4 NBA 2018
#> 5 NFL 2017
#> 6 NFL 2018
#> 7 NHL 2017
#> 8 NHL 2018
Created on 2018-12-08 by the reprex package (v0.2.0).
If you want to do it via data.table, this is one way:
cjdt <- function(a,b){
cj = CJ(1:nrow(a),1:nrow(b))
cbind(a[cj[[1]],],b[cj[[2]],])
}
A = data.table(ida = 1:10)
B = data.table(idb = 1:10)
cjdt(A,B)
Having said the above, if you are doing many little joins, and you don't need a data.table
object and the overhead of producing it, a significant speed increase can be achieved by writing a c++
code block using Rcpp
and the like:
// [[Rcpp::export]]
NumericMatrix crossJoin(NumericVector a, NumericVector b){
int szA = a.size(),
szB = b.size();
int i,j,r;
NumericMatrix ret(szA*szB,2);
for(i = 0, r = 0; i < szA; i++){
for(j = 0; j < szB; j++, r++){
ret(r,0) = a(i);
ret(r,1) = b(j);
}
}
return ret;
}
To compare, firstly for a large join:
C++
n = 1
a = runif(10000)
b = runif(10000)
system.time({for(i in 1:n){
crossJoin(a,b)
}})
user system elapsed 1.033 0.424 1.462
data.table
system.time({for(i in 1:n){
CJ(a,b)
}})
user system elapsed 0.602 0.569 2.452
Now for lots of little joins:
C++
n = 1e5
a = runif(10)
b = runif(10)
system.time({for(i in 1:n){
crossJoin(a,b)
}})
user system elapsed 0.660 0.077 0.739
data.table
system.time({for(i in 1:n){
CJ(a,b)
}})
user system elapsed 26.164 0.056 26.271
Usig sqldf
:
x <- data.frame(id1 = c("a", "b", "c"), vals1 = 1:3)
y <- data.frame(id2 = c("d", "e", "f"), vals2 = 4:6)
library(sqldf)
sqldf("SELECT * FROM x
CROSS JOIN y")
Output:
id1 vals1 id2 vals2
1 a 1 d 4
2 a 1 e 5
3 a 1 f 6
4 b 2 d 4
5 b 2 e 5
6 b 2 f 6
7 c 3 d 4
8 c 3 e 5
9 c 3 f 6
Just for the record, with the base package, we can use the by= NULL
instead of all=TRUE
:
merge(x, y, by= NULL)
By using the merge function and its optional parameters:
Inner join: merge(df1, df2) will work for these examples because R automatically joins the frames by common variable names, but you would most likely want to specify merge(df1, df2, by = "CustomerId") to make sure that you were matching on only the fields you desired. You can also use the by.x and by.y parameters if the matching variables have different names in the different data frames.
Outer join: merge(x = df1, y = df2, by = "CustomerId", all = TRUE)
Left outer: merge(x = df1, y = df2, by = "CustomerId", all.x = TRUE)
Right outer: merge(x = df1, y = df2, by = "CustomerId", all.y = TRUE)
Cross join: merge(x = df1, y = df2, by = NULL)
I don't know of a built-in way to do it with data.frame
's but it isn't hard to make.
@danas showed there is an easy built-in way, but I'll leave my answer here in case it is useful for other purposes.
cross.join <- function(a, b) {
idx <- expand.grid(seq(length=nrow(a)), seq(length=nrow(b)))
cbind(a[idx[,1],], b[idx[,2],])
}
and showing that it works with some built-in data sets:
> tmp <- cross.join(mtcars, iris)
> dim(mtcars)
[1] 32 11
> dim(iris)
[1] 150 5
> dim(tmp)
[1] 4800 16
> str(tmp)
'data.frame': 4800 obs. of 16 variables:
$ mpg : num 21 21 22.8 21.4 18.7 18.1 14.3 24.4 22.8 19.2 ...
$ cyl : num 6 6 4 6 8 6 8 4 4 6 ...
$ disp : num 160 160 108 258 360 ...
$ hp : num 110 110 93 110 175 105 245 62 95 123 ...
$ drat : num 3.9 3.9 3.85 3.08 3.15 2.76 3.21 3.69 3.92 3.92 ...
$ wt : num 2.62 2.88 2.32 3.21 3.44 ...
$ qsec : num 16.5 17 18.6 19.4 17 ...
$ vs : num 0 0 1 1 0 1 0 1 1 1 ...
$ am : num 1 1 1 0 0 0 0 0 0 0 ...
$ gear : num 4 4 4 3 3 3 3 4 4 4 ...
$ carb : num 4 4 1 1 2 1 4 2 2 4 ...
$ Sepal.Length: num 5.1 5.1 5.1 5.1 5.1 5.1 5.1 5.1 5.1 5.1 ...
$ Sepal.Width : num 3.5 3.5 3.5 3.5 3.5 3.5 3.5 3.5 3.5 3.5 ...
$ Petal.Length: num 1.4 1.4 1.4 1.4 1.4 1.4 1.4 1.4 1.4 1.4 ...
$ Petal.Width : num 0.2 0.2 0.2 0.2 0.2 0.2 0.2 0.2 0.2 0.2 ...
$ Species : Factor w/ 3 levels "setosa","versicolor",..: 1 1 1 1 1 1 1 1 1 1 ...
I'd love to know if there exists a convenient way to crossjoin two data.tables. I do this so often I ended up rolling my own function that others may find helpful
library(data.table)
cartesian_join <- function(i, j){
# Cartesian join of two data.tables
# If i has M rows and j has N rows, the result will have M*N rows
# Example: cartesian_join(as.data.table(iris), as.data.table(mtcars))
# Check inputs
if(!is.data.table(i)) stop("'i' must be a data.table")
if(!is.data.table(j)) stop("'j' must be a data.table")
if(nrow(i) == 0) stop("'i' has 0 rows. Not sure how to handle cartesian join")
if(nrow(j) == 0) stop("'j' has 0 rows. Not sure how to handle cartesian join")
# Do the join (use a join column name that's unlikely to clash with a pre-existing column name)
i[, MrJoinyJoin := 1L]
j[, MrJoinyJoin := 1L]
result <- j[i, on = "MrJoinyJoin", allow.cartesian = TRUE]
result[, MrJoinyJoin := NULL]
i[, MrJoinyJoin := NULL]
j[, MrJoinyJoin := NULL]
return(result[])
}
foo <- data.frame(Foo = c(1,2,3))
foo
Foo
1 1
2 2
3 3
bar <- data.frame(Bar = c("a", "b", "c"))
bar
Bar
1 a
2 b
3 c
cartesian_join(as.data.table(foo), as.data.table(bar))
Bar Foo
1: a 1
2: b 1
3: c 1
4: a 2
5: b 2
6: c 2
7: a 3
8: b 3
9: c 3
For data.table use
dt1[, as.list(dt2), by = names(dt1)]
Note that this only works if there are no duplicate rows.