Merging two datasets on approximate values - r

I need to merge (left join) two data sets x and y.
merge(x,y, by.x = "z", by.y = "zP", all.x = TRUE)
Every value of z is not there in zP but there must be nearest value in zP. So we need to use nearest value in zP for process of merging.
For example
z <- c(0.231, 0.045, 0.632, 0.217, 0.092, ...)
zP <- c(0.010,0.013, 0.017, 0.021, ...)
How can we do it in R ?

Based on the information you provided it sounds like you want to keep all the observations in x, and then for each observation in x you want to find the observation in y that minimizes the distance between columns z and zP. If that is what you are looking for then something like this might work
> library(data.table)
> x <- data.table(z = c(0.231, 0.045, 0.632, 0.217, 0.092), k = c("A","A","B","B","B"))
> y <- data.table(zP = c(0.010, 0.813, 0.017, 0.421), m = c(1,2,3,4))
> x
z k
1: 0.231 A
2: 0.045 A
3: 0.632 B
4: 0.217 B
5: 0.092 B
> y
zP m
1: 0.010 1
2: 0.813 2
3: 0.017 3
4: 0.421 4
> find.min.zP <- function(x){
+ y[which.min(abs(x - zP)), zP]
+ }
> x[, zP := find.min.zP(z), by = z]
> x
z k zP
1: 0.231 A 0.421
2: 0.045 A 0.017
3: 0.632 B 0.813
4: 0.217 B 0.017
5: 0.092 B 0.017
> merge(x, y, by="zP", all.x = T, all.y = F)
zP z k m
1: 0.017 0.045 A 3
2: 0.017 0.217 B 3
3: 0.017 0.092 B 3
4: 0.421 0.231 A 4
5: 0.813 0.632 B 2
This is the solution that popped into my head given that I use data.table quite a bit. Please note that using data.table here may or may not be the most elegant way and it may not even be the fastest way (although if x and y are large some solution involving data.table probably will be the fastest). Also note that this is likely an example of using data.table "badly" as I didn't make any effort to optimize for speed. If speed is important I would highly recommend reading the helpful documentation on the github wiki. Hope that helps.
Edit:
As I suspected, data.table provides a much better way, which Arun pointed out in the comments.
> setkey(x, z)
> setkey(y, zP)
> y[x, roll="nearest"]
zP m k
1: 0.045 3 A
2: 0.092 3 B
3: 0.217 3 B
4: 0.231 4 A
5: 0.632 2 B
The only difference is that the z column is now named zP and the original zP column is gone. If preserving that column is important you can always copy the zP column in y to a new column named z and join on that.
> y[, z := zP]
> setkey(x, z)
> setkey(y, z)
> y[x, roll='nearest']
zP m z k
1: 0.017 3 0.045 A
2: 0.017 3 0.092 B
3: 0.017 3 0.217 B
4: 0.421 4 0.231 A
5: 0.813 2 0.632 B
This is slightly less typing, but the real improvement is in compute times with large datasets.
> x <- data.table(z = runif(100000, 0, 100), k = sample(LETTERS, 100000, replace = T))
> y <- data.table(zP = runif(50000, 0, 100), m = sample(letters, 50000, replace = T))
> start <- proc.time()
> x[, zP := find.min.zP(z), by = z]
> slow <- merge(x, y, by="zP", all.x = T, all.y = F)
> proc.time() - start
user system elapsed
104.849 0.072 106.432
> x[, zP := NULL] # Drop the zP column we added to x doing the merge the slow way
> start <- proc.time()
> y[, z := zP]
> setkey(x, z)
> setkey(y, z)
> fast <- y[x, roll='nearest']
> proc.time() - start
user system elapsed
0.046 0.000 0.045
# Reorder the rows and columns so that we can compare the two data tables
> setkey(slow, z)
> setcolorder(slow, c("z", "zP", "k", "m"))
> setcolorder(fast, c("z", "zP", "k", "m"))
> all.equal(slow, fast)
TRUE
Notice, that the faster method is 2,365 times faster! I would expect the time gains to be even more dramatic for a data set with more than 100,000 observations (which is relatively small these days). This is why reading the data.table documentation is worth while if you are working with large data sets. You can often achieve very large speed ups by using the built in methods, but you won't know that they're there unless you look.

Related

Compare two rows of a data.table and show only columns with differences [duplicate]

This question already has answers here:
Remove constant columns with or without NAs
(8 answers)
How to remove data frame column with a single value
(3 answers)
Closed 3 years ago.
I am given a large data.table, which has columns of different types: e.g. numeric or character. E.g.
data.table(name=c("A","A"),val1=c(1,2),val2=c(3,3),cat=c("u","v"))
name val1 val2 cat
1: A 1 3 u
2: A 2 3 v
As a results, I would like a data.table just with the columns, where the entries are different between the two rows:
data.table(val1=c(1,2),cat=c("u","v"))
val1 cat
1: 1 u
2: 2 v
With base R you could do:
library(data.table)
dt <- data.table(name=c("A","A"),val1=c(1,2),val2=c(3,3),cat=c("u","v"))
Filter(function(x) length(unique(x)) > 1, dt)
#> val1 cat
#> 1: 1 u
#> 2: 2 v
You can check whether there is only one value in the column and return only the ones with more than one value:
mydt <- data.table(name=c("A", "A"), val1=c(1, 2), val2=c(3, 3), cat=c("u", "v"))
mydt_red <- mydt[, lapply(.SD, function(x) if(length(unique(x))!=1) x else NULL)]
mydt_red
# val1 cat
#1: 1 u
#2: 2 v
EDIT
As mentionned by #kath, a more efficient way to get your result is to use min and max functions and to combine them with Filter:
mydt_red2 <- Filter(function(x) min(x)!=max(x), mydt)
Some basic benchmarking
# Data (inspired by https://stackoverflow.com/a/35746513/680068)
nrow=10000
ncol=10000
mydt <- data.frame(matrix(sample(1:(ncol*nrow),ncol*nrow,replace = FALSE), ncol = ncol))
setDT(mydt)
system.time(mydt_redUni <- mydt[, lapply(.SD, function(x) if(length(unique(x))>1) x else NULL)])
#utilisateur système écoulé
# 2.31 0.52 2.83
system.time(mydt_redFilt <- Filter(function(x) length(unique(x)) > 1, mydt))
#utilisateur système écoulé
# 1.65 0.22 1.87
system.time(mydt_redSort <- mydt[, lapply(.SD, function(x) {xs <- sort(x); if(xs[1]!=tail(xs, 1)) x else NULL})])
#utilisateur système écoulé
# 3.87 0.00 3.87
system.time(mydt_redMinMax <- mydt[, lapply(.SD, function(x) if(min(x)!=max(x)) x else NULL)])
#utilisateur système écoulé
# 0.67 0.00 0.67
system.time(mydt_redFiltminmax <- Filter(function(x) min(x)!=max(x), mydt))
#utilisateur système écoulé
# 0.13 0.01 0.14
system.time(mydt_redSotos <- Filter(function(i)var(as.numeric(as.factor(i))) != 0, mydt))
#utilisateur système écoulé
# 100.76 0.05 100.84
Here is a fun idea for the math junkies out there. If the values are the same, the variance will be 0. So under that assumption, we can do, (credit to #Joris Chau for the Filter method)
Filter(function(i)var(as.numeric(as.factor(i))) != 0, dt)
# val1 cat
#1: 1 u
#2: 2 v

Summarize distribution of factors in R data frame

Say I have a data.frame like this:
X1 X2 X3
1 A B A
2 A C B
3 B A B
4 A A C
I would like to count the occurrences of A, B, C, etc. in each column, and return the result as
A_count B_count C_count
X1 3 1 0
X2 2 1 1
X3 1 2 1
I'm sure this question has a thousand duplicates, but I can't seem to find an answer that works for me :(
By running
apply(mydata, 2, table)
I get something like
$X1
B A
1 3
$X2
A C B
2 1 1
But it's not exactly what I want and if I try to build it back into a data frame, it doesn't work because I don't get the same number of columns for every row (like $X1 above where there are no C's).
What am I missing?
Many thanks!
You can refactor to include the factor levels common to each column, then tabulate. I would also recommend using lapply() instead of apply(), as apply() is for matrices.
df <- read.table(text = "X1 X2 X3
1 A B A
2 A C B
3 B A B
4 A A C", h=T)
do.call(
rbind,
lapply(df, function(x) table(factor(x, levels=levels(unlist(df)))))
)
# A B C
# X1 3 1 0
# X2 2 1 1
# X3 1 2 1
Suppose your data frame is x, I would simply do:
do.call(rbind, tapply(unlist(x, use.names = FALSE),
rep(1:ncol(x), each = nrow(x)),
table))
# A B C
#1 3 1 0
#2 2 1 1
#3 1 2 1
Benchmarking
# a function to generate toy data
# `k` factor levels
# `n` row
# `p` columns
datsim <- function(n, p, k) {
as.data.frame(replicate(p, sample(LETTERS[1:k], n, TRUE), simplify = FALSE),
col.names = paste0("X",1:p), stringsAsFactors = TRUE)
}
# try `n = 100`, `p = 500` and `k = 3`
x <- datsim(100, 500, 3)
## DirtySockSniffer's answer
system.time(do.call(rbind, lapply(x, function(u) table(factor(u, levels=levels(unlist(x)))))))
# user system elapsed
# 21.240 0.068 21.365
## my answer
system.time(do.call(rbind, tapply(unlist(x, use.names = FALSE), rep(1:ncol(x), each = nrow(x)), table)))
# user system elapsed
# 0.108 0.000 0.111
Dirty's answer can be improved, by:
## improved DirtySockSniffer's answer
system.time({clevels <- levels(unlist(x, use.names = FALSE));
do.call(rbind, lapply(x, function(u) table(factor(u, levels=clevels))))})
# user system elapsed
# 0.108 0.000 0.108
Also consider user20650's answer:
## Let's try a large `n`, `p`, `k`
x <- datsim(200, 5000, 5)
system.time(t(table(stack(lapply(x, as.character)))))
# user system elapsed
# 0.592 0.052 0.646
While my answer does:
system.time(do.call(rbind, tapply(unlist(x, use.names = FALSE), rep(1:ncol(x), each = nrow(x)), table)))
# user system elapsed
# 1.844 0.056 1.904
Improved Dirty's answer does:
system.time({clevels <- levels(unlist(x, use.names = FALSE));
do.call(rbind, lapply(x, function(u) table(factor(u, levels=clevels))))})
# user system elapsed
# 1.240 0.012 1.263

Speedy/elegant way to unite many pairs of columns

Is there an elegant/fastR way to combine all pairs of columns in a data.frame?
For example, using mapply() and paste() we can turn this data.frame:
mydf <- data.frame(a.1 = letters, a.2 = 26:1, b.1 = letters, b.2 = 1:26)
head(mydf)
a.1 a.2 b.1 b.2
1 a 26 a 1
2 b 25 b 2
3 c 24 c 3
4 d 23 d 4
5 e 22 e 5
6 f 21 f 6
into this data.frame:
mydf2 <- mapply(function(x, y) {
paste(x, y, sep = ".")},
mydf[ ,seq(1, ncol(mydf), by = 2)],
mydf[ ,seq(2, ncol(mydf), by = 2)])
head(mydf2)
a.1 b.1
[1,] "a.26" "a.1"
[2,] "b.25" "b.2"
[3,] "c.24" "c.3"
[4,] "d.23" "d.4"
[5,] "e.22" "e.5"
[6,] "f.21" "f.6"
However, this feels clumsy and is a bit slow when applied to big datasets. Any suggestions, perhaps using a Hadley package?
EDIT:
The ideal solution would easily scale to large numbers of columns, such that the names of the columns would not need to be included in the function call. Thanks!
It's amusing to note that the OP's solution appears to be the fastest one:
f1 <- function(mydf) {
mapply(function(x, y) {
paste(x, y, sep = ".")},
mydf[ ,seq(1, ncol(mydf), by = 2)],
mydf[ ,seq(2, ncol(mydf), by = 2)])
}
f.thelatemail <- function(mydf) {
mapply(paste,mydf[c(TRUE,FALSE)],mydf[c(FALSE,TRUE)],sep=".")
}
require(dplyr)
f.on_the_shores_of_linux_sea <- function(mydf) {
transmute(mydf,x1=paste0( a.1,'.', a.2),x2=paste0( b.1,'.', b.2))
}
f.jazurro <- function(mydf) {
odd <- seq(1, ncol(mydf), 2);
lapply(odd, function(x) paste(mydf[,x], mydf[,x+1], sep = ".")) %>%
do.call(cbind,.)
}
library(data.table)
f.akrun <- function(mydf) {
res <- as.data.table(matrix(, ncol=ncol(mydf)/2, nrow=nrow(mydf)))
indx <- seq(1, ncol(mydf), 2)
setDT(mydf)
for(j in seq_along(indx)){
set(res, i=NULL, j=j, value= paste(mydf[[indx[j]]],
mydf[[indx[j]+1]], sep='.'))
}
res
}
mydf <- data.frame(a.1 = letters, a.2 = 26:1, b.1 = letters, b.2 = 1:26)
mydf <- mydf[rep(1:nrow(mydf),5000),]
library(rbenchmark)
benchmark(f1(mydf),f.thelatemail(mydf),f.on_the_shores_of_linux_sea(mydf),f.jazurro(mydf),f.akrun(mydf))
Results:
# test replications elapsed relative user.self sys.self user.child sys.child
# 5 f.akrun(mydf) 100 14.000 75.269 13.673 0.296 0 0
# 4 f.jazurro(mydf) 100 0.388 2.086 0.314 0.071 0 0
# 3 f.on_the_shores_of_linux_sea(mydf) 100 15.585 83.790 15.293 0.280 0 0
# 2 f.thelatemail(mydf) 100 26.416 142.022 25.736 0.639 0 0
# 1 f1(mydf) 100 0.186 1.000 0.169 0.017 0 0
[Updated Benchmark]
I've added one solution from #thelatemail, which I missed in the original answer, and one solution from #akrun:
f.thelatemail2 <- function(mydf) {
data.frame(Map(paste,mydf[c(TRUE,FALSE)],mydf[c(FALSE,TRUE)],sep="."))
}
f.akrun2 <- function(mydf) {
setDT(mydf)
indx <- as.integer(seq(1, ncol(mydf), 2))
mydf2 <- copy(mydf)
for(j in indx){
set(mydf2, i=NULL, j=j, value= paste(mydf2[[j]],
mydf2[[j+1]], sep="."))
}
mydf2[,indx, with=FALSE]
}
Benchmark:
library(rbenchmark)
benchmark(f1(mydf),f.thelatemail(mydf), f.thelatemail2(mydf), f.on_the_shores_of_linux_sea(mydf),f.jazurro(mydf),f.akrun(mydf),f.akrun2(mydf))
# test replications elapsed relative user.self sys.self user.child sys.child
# 6 f.akrun(mydf) 100 13.247 69.356 12.897 0.340 0 0
# 7 f.akrun2(mydf) 100 12.746 66.733 12.405 0.339 0 0
# 5 f.jazurro(mydf) 100 0.327 1.712 0.254 0.073 0 0
# 4 f.on_the_shores_of_linux_sea(mydf) 100 16.347 85.586 15.838 0.445 0 0
# 2 f.thelatemail(mydf) 100 26.307 137.733 25.536 0.708 0 0
# 3 f.thelatemail2(mydf) 100 15.938 83.445 15.136 0.750 0 0
# 1 f1(mydf) 100 0.191 1.000 0.156 0.036 0 0
I'm not sure this is the best approach. See if the below code gives any speed improvement
require(dplyr)
transmute(mydf,x1=paste0( a.1,'.', a.2),x2=paste0( b.1,'.', b.2))
Answer updated based on comment :-)
An option using set from data.table. It should be fast for large datasets as it modifies by reference and the overhead of [.data.table is avoided. Assuming that the columns are ordered for each pair of columns.
library(data.table)
res <- as.data.table(matrix(, ncol=ncol(mydf)/2, nrow=nrow(mydf)))
indx <- seq(1, ncol(mydf), 2)
setDT(mydf)
for(j in seq_along(indx)){
set(res, i=NULL, j=j, value= paste(mydf[[indx[j]]],
mydf[[indx[j]+1]], sep='.'))
}
head(res)
# V1 V2
#1: a.26 a.1
#2: b.25 b.2
#3: c.24 c.3
#4: d.23 d.4
#5: e.22 e.5
#6: f.21 f.6
Instead of creating a new result dataset, we can also update the same or a copy of the original dataset. There will be some warnings about type conversion, but I guess this would be a bit faster (not benchmarked)
setDT(mydf)
mydf2 <- copy(mydf)
for(j in indx){
set(mydf2, i=NULL, j=j, value= paste(mydf2[[j]],
mydf2[[j+1]], sep="."))
}
mydf2[,indx, with=FALSE]
Benchmarks
I tried the benchmarks on a slightly bigger data with many columns.
data
set.seed(24)
d1 <- as.data.frame(matrix(sample(letters,500*10000, replace=TRUE),
ncol=500), stringsAsFactors=FALSE)
set.seed(4242)
d2 <- as.data.frame(matrix(sample(1:200,500*10000,
replace=TRUE), ncol=500))
d3 <- cbind(d1,d2)
mydf <- d3[,order(c(1:ncol(d1), 1:ncol(d2)))]
mydf1 <- copy(mydf)
Compared f1, f.jazurro (fastest) (from #Marat Talipov's post) with f.akrun2
microbenchmark(f1(mydf), f.jazurro(mydf), f.akrun2(mydf1),
unit='relative', times=20L)
#Unit: relative
# expr min lq mean median uq max neval
# f1(mydf) 3.420448 2.3217708 2.714495 2.653178 2.819952 2.736376 20
#f.jazurro(mydf) 1.000000 1.0000000 1.000000 1.000000 1.000000 1.000000 20
#f.akrun2(mydf1) 1.204488 0.8015648 1.031248 1.042262 1.097136 1.066671 20
#cld
#b
#a
#a
In this, f.jazurro is slighly better than f.akrun2. I think if I increase the group size, nrows etc, it would be an interesting comparison
For what its worth seven years later, here is a trick using the glue package and eval() + parse(). I don't know how it compares to other answers, but it works pretty darn well for me.
mydf <- data.frame(a.1 = letters, a.2 = 26:1, b.1 = letters, b.2 = 1:26)
mydf2 <- mydf
vars <- c('a', 'b')
eval(parse(text = glue::glue('mydf2 <- mydf2 |> unite({vars}, c(`{vars}.1`, `{vars}.2`), na.rm = T, sep = ".")')))
mydf2

Complicated reshaping

I want to reshape my dataframe from long to wide format and I loose some data that I'd like to keep.
For the following example:
df <- data.frame(Par1 = unlist(strsplit("AABBCCC","")),
Par2 = unlist(strsplit("DDEEFFF","")),
ParD = unlist(strsplit("foo,bar,baz,qux,bla,xyz,meh",",")),
Type = unlist(strsplit("pre,post,pre,post,pre,post,post",",")),
Val = c(10,20,30,40,50,60,70))
# Par1 Par2 ParD Type Val
# 1 A D foo pre 10
# 2 A D bar post 20
# 3 B E baz pre 30
# 4 B E qux post 40
# 5 C F bla pre 50
# 6 C F xyz post 60
# 7 C F meh post 70
dfw <- dcast(df,
formula = Par1 + Par2 ~ Type,
value.var = "Val",
fun.aggregate = mean)
# Par1 Par2 post pre
# 1 A D 20 10
# 2 B E 40 30
# 3 C F 65 50
this is almost what I need but I would like to have
some field keeping data from ParD field (for example, as single merged string),
number of observations used for aggregation.
i.e. I would like the resulting data.frame to be as follows:
# Par1 Par2 post pre Num.pre Num.post ParD
# 1 A D 20 10 1 1 foo_bar
# 2 B E 40 30 1 1 baz_qux
# 3 C F 65 50 1 2 bla_xyz_meh
I would be grateful for any ideas. For example, I tried to solve the second task by writing in dcast: fun.aggregate=function(x) c(Val=mean(x),Num=length(x)) - but this causes an error.
Late to the party, but here's another alternative using data.table:
require(data.table)
dt <- data.table(df, key=c("Par1", "Par2"))
dt[, list(pre=mean(Val[Type == "pre"]),
post=mean(Val[Type == "post"]),
pre.num=length(Val[Type == "pre"]),
post.num=length(Val[Type == "post"]),
ParD = paste(ParD, collapse="_")),
by=list(Par1, Par2)]
# Par1 Par2 pre post pre.num post.num ParD
# 1: A D 10 20 1 1 foo_bar
# 2: B E 30 40 1 1 baz_qux
# 3: C F 50 65 1 2 bla_xyz_meh
[from Matthew] +1 Some minor improvements to save repeating the same ==, and to demonstrate local variables inside j.
dt[, list(pre=mean(Val[.pre <- Type=="pre"]), # save .pre
post=mean(Val[.post <- Type=="post"]), # save .post
pre.num=sum(.pre), # reuse .pre
post.num=sum(.post), # reuse .post
ParD = paste(ParD, collapse="_")),
by=list(Par1, Par2)]
# Par1 Par2 pre post pre.num post.num ParD
# 1: A D 10 20 1 1 foo_bar
# 2: B E 30 40 1 1 baz_qux
# 3: C F 50 65 1 2 bla_xyz_meh
dt[, { .pre <- Type=="pre" # or save .pre and .post up front
.post <- Type=="post"
list(pre=mean(Val[.pre]),
post=mean(Val[.post]),
pre.num=sum(.pre),
post.num=sum(.post),
ParD = paste(ParD, collapse="_")) }
, by=list(Par1, Par2)]
# Par1 Par2 pre post pre.num post.num ParD
# 1: A D 10 20 1 1 foo_bar
# 2: B E 30 40 1 1 baz_qux
# 3: C F 50 65 1 2 bla_xyz_meh
And if a list column is ok rather than a paste, then this should be faster :
dt[, { .pre <- Type=="pre"
.post <- Type=="post"
list(pre=mean(Val[.pre]),
post=mean(Val[.post]),
pre.num=sum(.pre),
post.num=sum(.post),
ParD = list(ParD)) } # list() faster than paste()
, by=list(Par1, Par2)]
# Par1 Par2 pre post pre.num post.num ParD
# 1: A D 10 20 1 1 foo,bar
# 2: B E 30 40 1 1 baz,qux
# 3: C F 50 65 1 2 bla,xyz,meh
Solution in 2 steps using ddply ( i am not happy with , but I get the result)
dat <- ddply(df,.(Par1,Par2),function(x){
data.frame(ParD=paste(paste(x$ParD),collapse='_'),
Num.pre =length(x$Type[x$Type =='pre']),
Num.post = length(x$Type[x$Type =='post']))
})
merge(dfw,dat)
Par1 Par2 post pre ParD Num.pre Num.post
1 A D 2.0 1 foo_bar 1 1
2 B E 4.0 3 baz_qux 1 1
3 C F 6.5 5 bla_xyz_meh 1 2
You could do a merge of two dcasts and an aggregate, here all wrapped into one large expression mostly to avoid having intermediate objects hanging around afterwards:
Reduce(merge, list(
dcast(df, formula = Par1+Par2~Type, value.var="Val",
fun.aggregate=mean),
setNames(dcast(df, formula = Par1+Par2~Type, value.var="Val",
fun.aggregate=length), c("Par1", "Par2", "Num.post",
"Num.pre")),
aggregate(df["ParD"], df[c("Par1", "Par2")], paste, collapse="_")
))
I'll post but agstudy's puts me to shame:
step1 <- with(df, split(df, list(Par1, Par2)))
step2 <- step1[sapply(step1, nrow) > 0]
step3 <- lapply(step2, function(x) {
piece1 <- tapply(x$Val, x$Type, mean)
piece2 <- tapply(x$Type, x$Type, length)
names(piece2) <- paste0("Num.", names(piece2))
out <- x[1, 1:2]
out[, 3:6] <- c(piece1, piece2)
names(out)[3:6] <- names(c(piece1, piece2))
out$ParD <- paste(unique(x$ParD), collapse="_")
out
})
data.frame(do.call(rbind, step3), row.names=NULL)
Yielding:
Par1 Par2 post pre Num.post Num.pre ParD
1 A D 2.0 1 1 1 foo_bar
2 B E 4.0 3 1 1 baz_qux
3 C F 6.5 5 2 1 bla_xyz_meh
What a great opprotunity to benchmark!
Below are some runs of the plyr method (as suggested by #agstudy) compared with the data.table method (as suggested by #Arun)
using different sample sizes (N = 900, 2700, 10800)
Summary:
The data.table method outperforms the plyr method by a factor of 7.5
#-------------------#
# M E T H O D S #
#-------------------#
# additional methods below, in the updates
# Method 1 -- suggested by #agstudy
plyrMethod <- quote({
dfw<-dcast(df,
formula = Par1+Par2~Type,
value.var="Val",
fun.aggregate=mean)
dat <- ddply(df,.(Par1,Par2),function(x){
data.frame(ParD=paste(paste(x$ParD),collapse='_'),
Num.pre =length(x$Type[x$Type =='pre']),
Num.post = length(x$Type[x$Type =='post']))
})
merge(dfw,dat)
})
# Method 2 -- suggested by #Arun
dtMethod <- quote(
dt[, list(pre=mean(Val[Type == "pre"]),
post=mean(Val[Type == "post"]),
Num.pre=length(Val[Type == "pre"]),
Num.post=length(Val[Type == "post"]),
ParD = paste(ParD, collapse="_")),
by=list(Par1, Par2)]
)
# Method 3 -- suggested by #regetz
reduceMethod <- quote(
Reduce(merge, list(
dcast(df, formula = Par1+Par2~Type, value.var="Val",
fun.aggregate=mean),
setNames(dcast(df, formula = Par1+Par2~Type, value.var="Val",
fun.aggregate=length), c("Par1", "Par2", "Num.post",
"Num.pre")),
aggregate(df["ParD"], df[c("Par1", "Par2")], paste, collapse="_")
))
)
# Method 4 -- suggested by #Ramnath
castddplyMethod <- quote(
reshape::cast(Par1 + Par2 + ParD ~ Type,
data = ddply(df, .(Par1, Par2), transform,
ParD = paste(ParD, collapse = "_")),
fun = c(mean, length)
)
)
# SAMPLE DATA #
#-------------#
library(data.table)
library(plyr)
library(reshape2)
library(rbenchmark)
# for Par1, ParD
LLL <- apply(expand.grid(LETTERS, LETTERS, LETTERS, stringsAsFactors=FALSE), 1, paste0, collapse="")
lll <- apply(expand.grid(letters, letters, letters, stringsAsFactors=FALSE), 1, paste0, collapse="")
# max size is 17568 with current sample data setup, ie: floor(length(LLL) / 18) * 18
size <- 17568
size <- 10800
size <- 900
set.seed(1)
df<-data.frame(Par1=rep(LLL[1:(size/2)], times=rep(c(2,2,3), size)[1:(size/2)])[1:(size)]
, Par2=rep(lll[1:(size/2)], times=rep(c(2,2,3), size)[1:(size/2)])[1:(size)]
, ParD=sample(unlist(lapply(c("f", "b"), paste0, lll)), size, FALSE)
, Type=rep(c("pre","post"), size/2)
, Val =sample(seq(10,100,10), size, TRUE)
)
dt <- data.table(df, key=c("Par1", "Par2"))
# Confirming Same Results #
#-------------------------#
# Evaluate
DF1 <- eval(plyrMethod)
DF2 <- eval(dtMethod)
# Convert to DF and sort columns and sort ParD levels, for use in identical
colOrder <- sort(names(DF1))
DF1 <- DF1[, colOrder]
DF2 <- as.data.frame(DF2)[, colOrder]
DF2$ParD <- factor(DF2$ParD, levels=levels(DF1$ParD))
identical((DF1), (DF2))
# [1] TRUE
#-------------------------#
RESULTS
#--------------------#
# BENCHMARK #
#--------------------#
benchmark(plyr=eval(plyrMethod), dt=eval(dtMethod), reduce=eval(reduceMethod), castddply=eval(castddplyMethod),
replications=5, columns=c("relative", "test", "elapsed", "user.self", "sys.self", "replications"),
order="relative")
# SAMPLE SIZE = 900
relative test elapsed user.self sys.self replications
1.000 reduce 0.392 0.375 0.018 5
1.003 dt 0.393 0.377 0.016 5
7.064 plyr 2.769 2.721 0.047 5
8.003 castddply 3.137 3.030 0.106 5
# SAMPLE SIZE = 2,700
relative test elapsed user.self sys.self replications
1.000 dt 1.371 1.327 0.090 5
2.205 reduce 3.023 2.927 0.102 5
7.291 plyr 9.996 9.644 0.377 5
# SAMPLE SIZE = 10,800
relative test elapsed user.self sys.self replications
1.000 dt 8.678 7.168 1.507 5
2.769 reduce 24.029 23.231 0.786 5
6.946 plyr 60.277 52.298 7.947 5
13.796 castddply 119.719 113.333 10.816 5
# SAMPLE SIZE = 17,568
relative test elapsed user.self sys.self replications
1.000 dt 27.421 13.042 14.470 5
4.030 reduce 110.498 75.853 34.922 5
5.414 plyr 148.452 105.776 43.156 5
Update : Added results for baseMethod1
# Used only sample size of 90, as it was taking long
relative test elapsed user.self sys.self replications
1.000 dt 0.044 0.043 0.001 5
7.773 plyr 0.342 0.339 0.003 5
65.614 base1 2.887 2.866 0.028 5
Where
baseMethod1 <- quote({
step1 <- with(df, split(df, list(Par1, Par2)))
step2 <- step1[sapply(step1, nrow) > 0]
step3 <- lapply(step2, function(x) {
piece1 <- tapply(x$Val, x$Type, mean)
piece2 <- tapply(x$Type, x$Type, length)
names(piece2) <- paste0("Num.", names(piece2))
out <- x[1, 1:2]
out[, 3:6] <- c(piece1, piece2)
names(out)[3:6] <- names(c(piece1, piece2))
out$ParD <- paste(unique(x$ParD), collapse="_")
out
})
data.frame(do.call(rbind, step3), row.names=NULL)
})
Update 2: Added keying the DT as part of the metric
Adding the indexing step to the benchmark for fairness as per #MatthewDowle s comment.
However, presumably, if data.table is used, it will be in place of the data.frame and
hence the indexing will occur once and not simply for this procedure
dtMethod.withkey <- quote({
dt <- data.table(df, key=c("Par1", "Par2"))
dt[, list(pre=mean(Val[Type == "pre"]),
post=mean(Val[Type == "post"]),
Num.pre=length(Val[Type == "pre"]),
Num.post=length(Val[Type == "post"]),
ParD = paste(ParD, collapse="_")),
by=list(Par1, Par2)]
})
# SAMPLE SIZE = 10,800
relative test elapsed user.self sys.self replications
1.000 dt 9.155 7.055 2.137 5
1.043 dt.withkey 9.553 7.245 2.353 5
3.567 reduce 32.659 31.196 1.586 5
6.703 plyr 61.364 54.080 7.600 5
Update 3: Benchmarking #MD's edits to #Arun's original answer
dtMethod.MD1 <- quote(
dt[, list(pre=mean(Val[.pre <- Type=="pre"]), # save .pre
post=mean(Val[.post <- Type=="post"]), # save .post
pre.num=sum(.pre), # reuse .pre
post.num=sum(.post), # reuse .post
ParD = paste(ParD, collapse="_")),
by=list(Par1, Par2)]
)
dtMethod.MD2 <- quote(
dt[, { .pre <- Type=="pre" # or save .pre and .post up front
.post <- Type=="post"
list(pre=mean(Val[.pre]),
post=mean(Val[.post]),
pre.num=sum(.pre),
post.num=sum(.post),
ParD = paste(ParD, collapse="_")) }
, by=list(Par1, Par2)]
)
dtMethod.MD3 <- quote(
dt[, { .pre <- Type=="pre"
.post <- Type=="post"
list(pre=mean(Val[.pre]),
post=mean(Val[.post]),
pre.num=sum(.pre),
post.num=sum(.post),
ParD = list(ParD)) } # list() faster than paste()
, by=list(Par1, Par2)]
)
benchmark(dt.M1=eval(dtMethod.MD1), dt.M2=eval(dtMethod.MD2), dt.M3=eval(dtMethod.MD3), dt=eval(dtMethod),
replications=5, columns=c("relative", "test", "elapsed", "user.self", "sys.self", "replications"),
order="relative")
#--------------------#
Comparing the different data.table methods amongst themselves
# SAMPLE SIZE = 900
relative test elapsed user.self sys.self replications
1.000 dt.M3 0.198 0.197 0.001 5 <~~~ "list()" Method
1.242 dt.M1 0.246 0.243 0.004 5
1.253 dt.M2 0.248 0.242 0.007 5
1.884 dt 0.373 0.367 0.007 5
# SAMPLE SIZE = 17,568
relative test elapsed user.self sys.self replications
1.000 dt.M3 33.492 24.487 9.122 5 <~~~ "list()" Method
1.086 dt.M1 36.388 11.442 25.086 5
1.086 dt.M2 36.388 10.845 25.660 5
1.126 dt 37.701 13.256 24.535 5
Comparing MD3 ("list" method) with MD1 (best of DT non-list methods)
Using a clean session (ie, removing string cache)
_Note: Ran the following twice, fresh session each time, with practically identical results
Then re-ran in the *same* session, with reps=5. Results very different._
benchmark(dt.M1=eval(dtMethod.MD1), dt.M3=eval(dtMethod.MD3), replications=1, columns=c("relative", "test", "elapsed", "user.self", "sys.self", "replications"), order="relative")
# SAMPLE SIZE=17,568; CLEAN SESSION
relative test elapsed user.self sys.self replications
1.000 dt.M1 8.885 4.260 4.617 1
1.633 dt.M3 14.506 12.821 1.677 1
# SAMPLE SIZE=17,568; *SAME* SESSION
relative test elapsed user.self sys.self replications
1.000 dt.M1 33.443 10.200 23.226 5
1.048 dt.M3 35.060 26.127 8.915 5
#--------------------#
New benchmarks against previous methods
_Note: Not using the "list method" as results are not the same as other methods_
# SAMPLE SIZE = 900
relative test elapsed user.self sys.self replications
1.000 dt.M1 0.254 0.247 0.008 5
1.705 reduce 0.433 0.425 0.010 5
11.280 plyr 2.865 2.842 0.031 5
# SAMPLE SIZE = 17,568
relative test elapsed user.self sys.self replications
1.000 dt.M1 24.826 10.427 14.458 5
4.348 reduce 107.935 70.107 38.314 5
5.942 plyr 147.508 106.958 41.083 5
One Step solution combining reshape::cast with plyr::ddply
cast(Par1 + Par2 + ParD ~ Type, data = ddply(df, .(Par1, Par2), transform,
ParD = paste(ParD, collapse = "_")), fun = c(mean, length))
NOTE that the dcast function in reshape2 does not allow multiple aggregate functions to be passed, while the cast function in reshape does.
I believe this base R solution is comparable with #Arun's data table solution. (Which isn't to say I would prefer it; that code is much simpler!)
baseMethod2 <- quote({
is <- unname(split(1:nrow(df), with(df, paste(Par1, Par2, sep="\b"))))
i1 <- sapply(is, `[`, 1)
out <- with(df, data.frame(Par1=Par1[i1], Par2=Par2[i1]))
js <- lapply(is, function(i) split(i, df$Type[i]))
out$post <- sapply(js, function(j) mean(df$Val[j$post]))
out$pre <- sapply(js, function(j) mean(df$Val[j$pre]))
out$Num.pre <- sapply(js, function(j) length(j$pre))
out$Num.post <- sapply(js, function(j) length(j$post))
out$ParD <- sapply(is, function(x) paste(df$ParD[x], collapse="_"))
out
})
Using #RicardoSaporta's timing code with 900, 2700, and 10,800, respectively:
> relative test elapsed user.self sys.self replications
3 1.000 baseMethod2 0.230 0.229 0 5
1 1.130 dt 0.260 0.257 0 5
2 8.752 plyr 2.013 2.006 0 5
> relative test elapsed user.self sys.self replications
3 1.000 baseMethod2 0.877 0.872 0 5
1 1.068 dt 0.937 0.934 0 5
2 8.060 plyr 7.069 7.043 0 5
> relative test elapsed user.self sys.self replications
1 1.000 dt 6.232 6.178 0.031 5
3 1.085 baseMethod2 6.763 6.683 0.054 5
2 7.263 plyr 45.261 44.983 0.104 5
Trying to wrap different aggregation expressions into a self-contained function (expressions should yield atomic values)...
multi.by <- function(X, INDEX,...) {
expressions <- substitute(...())
duplicates <- duplicated(INDEX)
res <- do.call(rbind,sapply(split(X,cumsum(!duplicates),drop=T), function(part)
sapply(expressions,eval,part,simplify=F),simplify=F))
if (is.data.frame(INDEX)) res <- cbind(INDEX[!duplicates,],res)
else rownames(res) <- INDEX[!duplicates]
res
}
multi.by(df,df[,1:2],
pre=mean(Val[Type=="pre"]),
post=mean(Val[Type=="post"]),
Num.pre=sum(Type=="pre"),
Num.post=sum(Type=="post"),
ParD=paste(ParD, collapse="_"))

Efficiently replicate matrices in R

I have a matrix and look for an efficient way to replicate it n times (where n is the number of observations in the dataset). For example, if I have a matrix A
A <- matrix(1:15, nrow=3)
then I want an output of the form
rbind(A, A, A, ...) #n times.
Obviously, there are many ways to construct such a large matrix, for example using a for loop or apply or similar functions. However, the call to the "matrix-replication-function" takes place in the very core of my optimization algorithm where it is called tens of thousands of times during one run of my program. Therefore, loops, apply-type of functions and anything similar to that are not efficient enough. (Such a solution would basically mean that a loop over n is performed tens of thousands of times, which is obviously inefficient.) I already tried to use the ordinary rep function, but haven't found a way to arrange the output of rep in a matrix of the desired format.
The solution
do.call("rbind", replicate(n, A, simplify=F))
is also too inefficient because rbind is used too often in this case. (Then, about 30% of the total runtime of my program are spent performing the rbinds.)
Does anyone know a better solution?
Two more solutions:
The first is a modification of the example in the question
do.call("rbind", rep(list(A), n))
The second involves unrolling the matrix, replicating it, and reassembling it.
matrix(rep(t(A),n), ncol=ncol(A), byrow=TRUE)
Since efficiency is what was requested, benchmarking is necessary
library("rbenchmark")
A <- matrix(1:15, nrow=3)
n <- 10
benchmark(rbind(A, A, A, A, A, A, A, A, A, A),
do.call("rbind", replicate(n, A, simplify=FALSE)),
do.call("rbind", rep(list(A), n)),
apply(A, 2, rep, n),
matrix(rep(t(A),n), ncol=ncol(A), byrow=TRUE),
order="relative", replications=100000)
which gives:
test replications elapsed
1 rbind(A, A, A, A, A, A, A, A, A, A) 100000 0.91
3 do.call("rbind", rep(list(A), n)) 100000 1.42
5 matrix(rep(t(A), n), ncol = ncol(A), byrow = TRUE) 100000 2.20
2 do.call("rbind", replicate(n, A, simplify = FALSE)) 100000 3.03
4 apply(A, 2, rep, n) 100000 7.75
relative user.self sys.self user.child sys.child
1 1.000 0.91 0 NA NA
3 1.560 1.42 0 NA NA
5 2.418 2.19 0 NA NA
2 3.330 3.03 0 NA NA
4 8.516 7.73 0 NA NA
So the fastest is the raw rbind call, but that assumes n is fixed and known ahead of time. If n is not fixed, then the fastest is do.call("rbind", rep(list(A), n). These were for a 3x5 matrix and 10 replications. Different sized matrices might give different orderings.
EDIT:
For n=600, the results are in a different order (leaving out the explicit rbind version):
A <- matrix(1:15, nrow=3)
n <- 600
benchmark(do.call("rbind", replicate(n, A, simplify=FALSE)),
do.call("rbind", rep(list(A), n)),
apply(A, 2, rep, n),
matrix(rep(t(A),n), ncol=ncol(A), byrow=TRUE),
order="relative", replications=10000)
giving
test replications elapsed
4 matrix(rep(t(A), n), ncol = ncol(A), byrow = TRUE) 10000 1.74
3 apply(A, 2, rep, n) 10000 2.57
2 do.call("rbind", rep(list(A), n)) 10000 2.79
1 do.call("rbind", replicate(n, A, simplify = FALSE)) 10000 6.68
relative user.self sys.self user.child sys.child
4 1.000 1.75 0 NA NA
3 1.477 2.54 0 NA NA
2 1.603 2.79 0 NA NA
1 3.839 6.65 0 NA NA
If you include the explicit rbind version, it is slightly faster than the do.call("rbind", rep(list(A), n)) version, but not by much, and slower than either the apply or matrix versions. So a generalization to arbitrary n does not require a loss of speed in this case.
Probably this is more efficient:
apply(A, 2, rep, n)
There's also this way:
rep(1, n) %x% A
You can use indexing
A[rep(seq(nrow(A)), n), ]
I came here for the same reason as the original poster and ultimately updated #Brian Diggs comparison to include all of the other posted answers. Hopefully I did this correctly:
#install.packages("rbenchmark")
library("rbenchmark")
A <- matrix(1:15, nrow=3)
n <- 600
benchmark(do.call("rbind", replicate(n, A, simplify=FALSE)),
do.call("rbind", rep(list(A), n)),
apply(A, 2, rep, n),
matrix(rep(t(A),n), ncol=ncol(A), byrow=TRUE),
A[rep(seq(nrow(A)), n), ],
rep(1, n) %x% A,
apply(A, 2, rep, n),
matrix(rep(as.integer(t(A)),n),nrow=nrow(A)*n,byrow=TRUE),
order="relative", replications=10000)
# test replications elapsed relative user.self sys.self user.child sys.child
#5 A[rep(seq(nrow(A)), n), ] 10000 0.32 1.000 0.33 0.00 NA NA
#8 matrix(rep(as.integer(t(A)), n), nrow = nrow(A) * n, byrow = TRUE) 10000 0.36 1.125 0.35 0.02 NA NA
#4 matrix(rep(t(A), n), ncol = ncol(A), byrow = TRUE) 10000 0.38 1.188 0.37 0.00 NA NA
#3 apply(A, 2, rep, n) 10000 0.59 1.844 0.56 0.03 NA NA
#7 apply(A, 2, rep, n) 10000 0.61 1.906 0.58 0.03 NA NA
#6 rep(1, n) %x% A 10000 1.44 4.500 1.42 0.02 NA NA
#2 do.call("rbind", rep(list(A), n)) 10000 1.67 5.219 1.67 0.00 NA NA
#1 do.call("rbind", replicate(n, A, simplify = FALSE)) 10000 5.03 15.719 5.02 0.01 NA NA
what about transforming it into an array, replicate the content and create a new matrix with the updated number of rows?
A <- matrix(...)
n = 2 # just a test
a = as.integer(A)
multi.a = rep(a,n)
multi.A = matrix(multi.a,nrow=nrow(A)*n,byrow=T)

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