Efficiently building a large (200 MM line) dataframe - r

I am attempting to build a large (~200 MM line) dataframe in R. Each entry in the dataframe will consist of approximately 10 digits (e.g. 1234.12345). The code is designed to walk through a list, subtract an item in position [i] from every item after [i], but not the items before [i] (If I was putting the output into a matrix it would be a triangular matrix). The code is simple and works fine on smaller lists, but I am wondering if there is a faster or more efficient way to do this? I assume the first part of the answer is going to entail "don't use a nested for loop," but I am not sure what the alternatives are.
The idea is that this will be an "edge list" for a social network analysis graph. Once I have 'outlist' I will reduce the number of edges based on some criteria(<,>,==,) so the final list (and graph) won't be quite so ponderous.
#Fake data of same approximate dimensions as real data
dlist<-sample(1:20,20, replace=FALSE)
#purge the output list before running the loop
rm(outlist)
outlist<-data.frame()
for(i in 1:(length(dlist)-1)){
for(j in (i+1):length(dlist)){
outlist<-rbind(outlist, c(dlist[i],dlist[j], dlist[j]-dlist[i]))
}
}

IIUC your final dataset will be ~200 million rows by 3 columns, all of type numeric, which takes a total space of:
200e6 (rows) * 3 (cols) * 8 (bytes) / (1024 ^ 3)
# ~ 4.5GB
That's quite a big data, where it's essential to avoid copies wherever possible.
Here's a method that uses data.table package's unexported (internal) vecseq function (written in C and is fast + memory efficient) and makes use of it's assignment by reference operator :=, to avoid copies.
fn1 <- function(x) {
require(data.table) ## 1.9.2
lx = length(x)
vx = as.integer(lx * (lx-1)/2)
# R v3.1.0 doesn't copy on doing list(.) - so should be even more faster there
ans = setDT(list(v1 = rep.int(head(x,-1L), (lx-1L):1L),
v2=x[data.table:::vecseq(2:lx, (lx-1L):1, vx)]))
ans[, v3 := v2-v1]
}
Benchmarking:
I'll benchmark with functions from other answers on your data dimensions. Note that my benchmark is on R v3.0.2, but fn1() should give better performance (both speed and memory) on R v3.1.0 because list(.) doesn't result in copy anymore.
fn2 <- function(x) {
diffmat <- outer(x, x, "-")
ss <- which(upper.tri(diffmat), arr.ind = TRUE)
data.frame(v1 = x[ss[,1]], v2 = x[ss[,2]], v3 = diffmat[ss])
}
fn3 <- function(x) {
idx <- combn(seq_along(x), 2)
out2 <- data.frame(v1=x[idx[1, ]], v2=x[idx[2, ]])
out2$v3 <- out2$v2-out2$v1
out2
}
set.seed(45L)
x = runif(20e3L)
system.time(ans1 <- fn1(x)) ## 18 seconds + ~8GB (peak) memory usage
system.time(ans2 <- fn2(x)) ## 158 seconds + ~19GB (peak) memory usage
system.time(ans3 <- fn3(x)) ## 809 seconds + ~12GB (peak) memory usage
Note that fn2() due to use of outer requires quite a lot of memory (peak memory usage was >=19GB) and is slower than fn1(). fn3() is just very very slow (due to combn, and unnecessary copy).

Another way to create that data is
#Sample Data
N <- 20
set.seed(15) #for reproducibility
dlist <- sample(1:N,N, replace=FALSE)
we could do
idx <- combn(1:N,2)
out2 <- data.frame(i=dlist[idx[1, ]], j=dlist[idx[2, ]])
out2$dist <- out2$j-out2$i
This uses combn to create all paris of indices in the data.set rather than doing loops. This allows us to build the data.frame all at once rather than adding a row at a time.
We compare that to
out1 <- data.frame()
for(i in 1:(length(dlist)-1)){
for(j in (i+1):length(dlist)){
out1<-rbind(out1, c(dlist[i],dlist[j], dlist[j]-dlist[i]))
}
}
we see that
all(out1==out2)
# [1] TRUE
Plus, if we compare with microbenchmark we see that
microbenchmark(loops(), combdata())
# Unit: microseconds
# expr min lq median uq max neval
# loops() 30888.403 32230.107 33764.7170 34821.2850 82891.166 100
# combdata() 684.316 800.384 873.5015 940.9215 4285.627 100
The method that doesn't use loops is much faster.

You can always start with a triangular matrix and then make your dataframe directly from that:
vec <- 1:10
diffmat <- outer(vec,vec,"-")
ss <- which(upper.tri(diffmat),arr.ind = TRUE)
data.frame(one = vec[ss[,1]],
two = vec[ss[,2]],
diff = diffmat[ss])

You need to preallocate out list, this will significantly increase the speed of your code. By preallocating I mean creating an output structure that already has the desired size, but filled with for example NA's.

Related

Looping a function through a list of dataframes is very slow

I have a list, which contains 4438 dataframes with different sizes. I am not sure how to make a reproducible example, but the way I obtained the list is using the expand.grid function to have a dataframe with all the possible combination of elements:
citation <- citation %>%
map_depth(., 1, expand.grid)
List before applying expand.grid
List after applying expand.grid
What I am going to achieve is for each dataframe, counting the number of unique values per row, and finding the minimum number of unique values in the dataframe.
First, I write the function below
fun1 <- function(res){
min(apply(res,1,function(x) length(unique(x))))
}
Then, apply the function to each dataframe:
library(furrr)
plan(multisession, workers = 4)
min_set <- c()
min_set <- citation %>% future_map_dbl(fun1)
However, the calculation is super slow, almost 30 mins to complete. I would like to find another way to accelerate the performance. Looking forward to hear the solution from you guys. Thank you in advance
To speed up the current approach of enumerating the combinations, use rowTabulate from the Rfast package (or rowTabulates from the matrixStats package).
However, it will be much faster to get the desired results with the setcover function in the adagio package, which solves the set cover problem directly (i.e., without the use of expand.grid) via integer linear programming with lp from the lpSolve package.
library(Rfast) # for the rowTabulate function
library(adagio) # for the setcover function
# reproducible example data
set.seed(1141593349)
citation1 <- list(
lapply(c(5,2,8,12,6,38), function(size) sample(50, size)),
lapply(c(5,2,8,12,7), function(size) sample(50, size))
)
# get all combinations of the indices of the unique values for each list in citation1
citation2 <- lapply(citation1, function(x) expand.grid(lapply(x, match, table = unique(unlist(x)))))
# original solution
fun1 <- function(res) min(apply(res, 1, function(x) length(unique(x))))
# faster version of the original solution
fun2 <- function(res) min(rowsums(rowTabulate(as.matrix(res)) > 0L))
# linear programming solution (uses citation1 rather than citation2)
fun3 <- function(res) {
v <- unlist(res)
m <- matrix(0L, max(v), length(res))
m[cbind(v, rep.int(seq_along(res), lengths(res)))] <- 1L
setcover(m)$objective
}
microbenchmark::microbenchmark(fun1 = sapply(citation2, fun1),
fun2 = as.integer(sapply(citation2, fun2)),
fun3 = as.integer(sapply(citation1, fun3)),
times = 10,
check = "identical")
#> Unit: milliseconds
#> expr min lq mean median uq max
#> fun1 1110.4976 1162.003601 1217.049501 1204.608151 1281.121601 1331.057001
#> fun2 101.5173 113.123501 142.265371 145.964502 165.788700 187.196301
#> fun3 1.4038 1.461101 1.734781 1.850701 1.870801 1.888702

Assigning value to dataframe in R - for loop speed

I have the following code:
n <- 1e6
no_clm <- rpois(n,30)
hold <- data.frame("x" = double(n))
c = 1
for (i in no_clm){
ctl <- sum(rgamma(i,30000)-2000)
hold[c,1] <- ctl
#hold <- rbind(hold,df)
c = c +1
}
Unfortunately the speed of this code is quite slow. I've narrowed down the speed to hold[c,1] <- ctl. If I remove this then the code runs near instantly.
How can I make this efficient? I need to store the results to some sort of dataframe or list in a fast fashion. In reality the actual code is more complex than this but the slowing point is the assigning.
Note that the above is just an example, in reality I have multiple calculations on the rgamma samples and each of these calculations are then stored in a large dataframe.
Try this
hold=data.frame(sapply(no_clm,function(x){
return(sum(rgamma(x,30000)-2000))
}))
It looks like you can just use one call to rgamma, as you are iterating over the number of observations parameter.
So if you do one call and the split the vector to the lengths required (no_clm) you can then just iterate over that list and sum
n <- 1e6
no_clm <- rpois(n, 30)
hold <- data.frame("x" = double(n))
# total observations to use for rgamma
total_clm <- sum(no_clm)
# get values
gammas <- rgamma(total_clm, 30000) - 2000
# split into list of lengths dictated by no_clm
hold$x <- sapply(split(gammas, cumsum(sequence(no_clm) == 1)), sum)
This took 5.919892 seconds
Move into sapply() loop instead of a for loop and then realise 2000 * no_clm can be moved outside the loop (to minimise number of function calls).
n <- 1e6
no_clm <- rpois(n, 30)
hold <- data.frame(x = sapply(no_clm, function(i) sum(rgamma(i, 30000))) - 2000 * no_clm)
You may observe a speed pickup using data.table:
dt = data.table(no_clm)
dt[, hold := sapply(no_clm, function(x) sum(rgamma(x, 30000)-2000))]

Optimising sapply() or for(), paste(), to efficiently transform sparse triplet matrix to a libsvm format

I have a piece of R code I want to optimise for speed working with larger datasets. It currently depends on sapply cycling through a vector of numbers (which correspond to rows of a sparse matrix). The reproducible example below gets at the nub of the problem; it is the three line function expensive() that chews up the time, and its obvious why (lots of matching big vectors to eachother, and two nested paste statements for each cycle of the loop). Before I give up and start struggling with doing this bit of the work in C++, is there something I'm missing? Is there a way to vectorize the sapply call that will make it an order of magnitude or three faster?
library(microbenchmark)
# create an example object like a simple_triple_matrix
# number of rows and columns in sparse matrix:
n <- 2000 # real number is about 300,000
ncols <- 1000 # real number is about 80,000
# number of non-zero values, about 10 per row:
nonzerovalues <- n * 10
stm <- data.frame(
i = sample(1:n, nonzerovalues, replace = TRUE),
j = sample(1:ncols, nonzerovalues, replace = TRUE),
v = sample(rpois(nonzerovalues, 5), replace = TRUE)
)
# It seems to save about 3% of time to have i, j and v as objects in their own right
i <- stm$i
j <- stm$j
v <- stm$v
expensive <- function(){
sapply(1:n, function(k){
# microbenchmarking suggests quicker to have which() rather than a vector of TRUE and FALSE:
whichi <- which(i == k)
paste(paste(j[whichi], v[whichi], sep = ":"), collapse = " ")
})
}
microbenchmark(expensive())
The output of expensive is a character vector, of n elements, that looks like this:
[1] "344:5 309:3 880:7 539:6 338:1 898:5 40:1"
[2] "307:3 945:2 949:1 130:4 779:5 173:4 974:7 566:8 337:5 630:6 567:5 750:5 426:5 672:3 248:6 300:7"
[3] "407:5 649:8 507:5 629:5 37:3 601:5 992:3 377:8"
For what its worth, the motivation is to efficiently write data from a sparse matrix format - either from slam or Matrix, but starting with slam - into libsvm format (which is the format above, but with each row beginning with a number representing a target variable for a support vector machine - omitted in this example as it's not part of the speed problem). Trying to improve on the answers to this question. I forked one of the repositories referred to from there and adapted its approach to work with sparse matrices with these functions. The tests show that it works fine; but it doesn't scale up.
Use package data.table. Its by combined with the fast sorting saves you from finding the indices of equal i values.
res1 <- expensive()
library(data.table)
cheaper <- function() {
setDT(stm)
res <- stm[, .(i, jv = paste(j, v, sep = ":"))
][, .(res = paste(jv, collapse = " ")), keyby = i][["res"]]
setDF(stm) #clean-up which might not be necessary
res
}
res2 <- cheaper()
all.equal(res1, res2)
#[1] TRUE
microbenchmark(expensive(),
cheaper())
#Unit: milliseconds
# expr min lq mean median uq max neval cld
# expensive() 127.63343 135.33921 152.98288 136.13957 138.87969 222.36417 100 b
# cheaper() 15.31835 15.66584 16.16267 15.98363 16.33637 18.35359 100 a

Constructing an R data.table by selecting each row from an array of tables

Assume I have a list of length D containing data.table objects. Each data.table has the same columns (X, Y) and same number of rows N. I'd like to construct another table with N rows, with the individual rows taken from the tables specified by an index vector also of length N. Restated, each row in the final table is taken from one and only one of the tables in the array, with the index of the source table specified by an existing vector.
N = 100 # rows in each table (actual ~1000000 rows)
D = 4 # number of tables in array (actual ~100 tables)
tableArray = vector("list", D)
for (d in 1:D) {
tableArray[[d]] = data.table(X=rnorm(N), Y=d) # actual ~100 columns
}
tableIndexVector = sample.int(D, N, replace=TRUE) # length N of random 1:D
finalTable = copy(tableArray[[1]]) # just for length and column names
for (n in 1:N) {
finalTable[n] = tableArray[[tableIndexVector[n]]][n]
}
This seems to work the way I want, but the array within array notation is hard to understand, and I presume the performance of the for loop isn't going to be very good. It seems like there should be some elegant way of doing this, but I haven't stumbled across it yet. Is there another way of doing this that is efficient and less arcane?
(In case you are wondering, each table in the array represents simulated counterfactual observations for a subject under a particular regime of treatment, and I want to sample from these with different probabilities to test the behavior of different regression approaches with different ratios of regimes observed.)
for loops work just fine with data.table but we can improve the performance of your specific loop significantly (I believe) using the following approaches.
Approach # 1
Use set instead, as it avoids the [.data.table overhead
Don't loop over 1:N because you can simplify your loop to run only on unique values of tableIndexVector and assign all the corresponding values at once. This should decrease the run time by at least x10K (as N is of size 1MM and D is only of size 100, while unique(tableIndexVector) <= D)
So you basically could convert your loop to the following
for (i in unique(tableIndexVector)) {
indx <- which(tableIndexVector == i)
set(finalTable, i = indx, j = 1:2, value = tableArray[[i]][indx])
}
Approach # 2
Another approach is to use rbindlist and combine all the tables into one big data.table while adding the new idcol parameter in order to identify the different tables within the big table. You will need the devel version for that. This will avoid the loop as requested, but the result will be ordered by the tables appearance
temp <- rbindlist(tableArray, idcol = "indx")
indx <- temp[, .I[which(tableIndexVector == indx)], by = indx]$V1
finalTable <- temp[indx]
Here's a benchmark on bigger data set
N = 100000
D = 10
tableArray = vector("list", D)
set.seed(123)
for (d in 1:D) {
tableArray[[d]] = data.table(X=rnorm(N), Y=d)
}
set.seed(123)
tableIndexVector = sample.int(D, N, replace=TRUE)
finalTable = copy(tableArray[[1]])
finalTable2 = copy(tableArray[[1]])
## Your approach
system.time(for (n in 1:N) {
finalTable[n] = tableArray[[tableIndexVector[n]]][n]
})
# user system elapsed
# 154.79 33.14 191.57
## My approach # 1
system.time(for (i in unique(tableIndexVector)) {
indx <- which(tableIndexVector == i)
set(finalTable2, i = indx, j = 1:2, value = tableArray[[i]][indx])
})
# user system elapsed
# 0.01 0.00 0.02
## My approach # 2
system.time({
temp <- rbindlist(tableArray, idcol = "indx")
indx <- temp[, .I[which(tableIndexVector == indx)], by = indx]$V1
finalTable3 <- temp[indx]
})
# user system elapsed
# 0.11 0.00 0.11
identical(finalTable, finalTable2)
## [1] TRUE
identical(setorder(finalTable, X), setorder(finalTable3[, indx := NULL], X))
## [1] TRUE
So to conclusion
My first approach is by far the fastest and elapses x15K times faster
than your original one. It is also returns identical result
My second approach is still x1.5K times faster than your original approach but avoids the loop (which you don't like for some reason). Though the result is order by the tables appearance, so the order isn't identical to your result.

How to append rows to an R data frame

I have looked around StackOverflow, but I cannot find a solution specific to my problem, which involves appending rows to an R data frame.
I am initializing an empty 2-column data frame, as follows.
df = data.frame(x = numeric(), y = character())
Then, my goal is to iterate through a list of values and, in each iteration, append a value to the end of the list. I started with the following code.
for (i in 1:10) {
df$x = rbind(df$x, i)
df$y = rbind(df$y, toString(i))
}
I also attempted the functions c, append, and merge without success. Please let me know if you have any suggestions.
Update from comment:
I don't presume to know how R was meant to be used, but I wanted to ignore the additional line of code that would be required to update the indices on every iteration and I cannot easily preallocate the size of the data frame because I don't know how many rows it will ultimately take. Remember that the above is merely a toy example meant to be reproducible. Either way, thanks for your suggestion!
Update
Not knowing what you are trying to do, I'll share one more suggestion: Preallocate vectors of the type you want for each column, insert values into those vectors, and then, at the end, create your data.frame.
Continuing with Julian's f3 (a preallocated data.frame) as the fastest option so far, defined as:
# pre-allocate space
f3 <- function(n){
df <- data.frame(x = numeric(n), y = character(n), stringsAsFactors = FALSE)
for(i in 1:n){
df$x[i] <- i
df$y[i] <- toString(i)
}
df
}
Here's a similar approach, but one where the data.frame is created as the last step.
# Use preallocated vectors
f4 <- function(n) {
x <- numeric(n)
y <- character(n)
for (i in 1:n) {
x[i] <- i
y[i] <- i
}
data.frame(x, y, stringsAsFactors=FALSE)
}
microbenchmark from the "microbenchmark" package will give us more comprehensive insight than system.time:
library(microbenchmark)
microbenchmark(f1(1000), f3(1000), f4(1000), times = 5)
# Unit: milliseconds
# expr min lq median uq max neval
# f1(1000) 1024.539618 1029.693877 1045.972666 1055.25931 1112.769176 5
# f3(1000) 149.417636 150.529011 150.827393 151.02230 160.637845 5
# f4(1000) 7.872647 7.892395 7.901151 7.95077 8.049581 5
f1() (the approach below) is incredibly inefficient because of how often it calls data.frame and because growing objects that way is generally slow in R. f3() is much improved due to preallocation, but the data.frame structure itself might be part of the bottleneck here. f4() tries to bypass that bottleneck without compromising the approach you want to take.
Original answer
This is really not a good idea, but if you wanted to do it this way, I guess you can try:
for (i in 1:10) {
df <- rbind(df, data.frame(x = i, y = toString(i)))
}
Note that in your code, there is one other problem:
You should use stringsAsFactors if you want the characters to not get converted to factors. Use: df = data.frame(x = numeric(), y = character(), stringsAsFactors = FALSE)
Let's benchmark the three solutions proposed:
# use rbind
f1 <- function(n){
df <- data.frame(x = numeric(), y = character())
for(i in 1:n){
df <- rbind(df, data.frame(x = i, y = toString(i)))
}
df
}
# use list
f2 <- function(n){
df <- data.frame(x = numeric(), y = character(), stringsAsFactors = FALSE)
for(i in 1:n){
df[i,] <- list(i, toString(i))
}
df
}
# pre-allocate space
f3 <- function(n){
df <- data.frame(x = numeric(1000), y = character(1000), stringsAsFactors = FALSE)
for(i in 1:n){
df$x[i] <- i
df$y[i] <- toString(i)
}
df
}
system.time(f1(1000))
# user system elapsed
# 1.33 0.00 1.32
system.time(f2(1000))
# user system elapsed
# 0.19 0.00 0.19
system.time(f3(1000))
# user system elapsed
# 0.14 0.00 0.14
The best solution is to pre-allocate space (as intended in R). The next-best solution is to use list, and the worst solution (at least based on these timing results) appears to be rbind.
Suppose you simply don't know the size of the data.frame in advance. It can well be a few rows, or a few millions. You need to have some sort of container, that grows dynamically. Taking in consideration my experience and all related answers in SO I come with 4 distinct solutions:
rbindlist to the data.frame
Use data.table's fast set operation and couple it with manually doubling the table when needed.
Use RSQLite and append to the table held in memory.
data.frame's own ability to grow and use custom environment (which has reference semantics) to store the data.frame so it will not be copied on return.
Here is a test of all the methods for both small and large number of appended rows. Each method has 3 functions associated with it:
create(first_element) that returns the appropriate backing object with first_element put in.
append(object, element) that appends the element to the end of the table (represented by object).
access(object) gets the data.frame with all the inserted elements.
rbindlist to the data.frame
That is quite easy and straight-forward:
create.1<-function(elems)
{
return(as.data.table(elems))
}
append.1<-function(dt, elems)
{
return(rbindlist(list(dt, elems),use.names = TRUE))
}
access.1<-function(dt)
{
return(dt)
}
data.table::set + manually doubling the table when needed.
I will store the true length of the table in a rowcount attribute.
create.2<-function(elems)
{
return(as.data.table(elems))
}
append.2<-function(dt, elems)
{
n<-attr(dt, 'rowcount')
if (is.null(n))
n<-nrow(dt)
if (n==nrow(dt))
{
tmp<-elems[1]
tmp[[1]]<-rep(NA,n)
dt<-rbindlist(list(dt, tmp), fill=TRUE, use.names=TRUE)
setattr(dt,'rowcount', n)
}
pos<-as.integer(match(names(elems), colnames(dt)))
for (j in seq_along(pos))
{
set(dt, i=as.integer(n+1), pos[[j]], elems[[j]])
}
setattr(dt,'rowcount',n+1)
return(dt)
}
access.2<-function(elems)
{
n<-attr(elems, 'rowcount')
return(as.data.table(elems[1:n,]))
}
SQL should be optimized for fast record insertion, so I initially had high hopes for RSQLite solution
This is basically copy&paste of Karsten W. answer on similar thread.
create.3<-function(elems)
{
con <- RSQLite::dbConnect(RSQLite::SQLite(), ":memory:")
RSQLite::dbWriteTable(con, 't', as.data.frame(elems))
return(con)
}
append.3<-function(con, elems)
{
RSQLite::dbWriteTable(con, 't', as.data.frame(elems), append=TRUE)
return(con)
}
access.3<-function(con)
{
return(RSQLite::dbReadTable(con, "t", row.names=NULL))
}
data.frame's own row-appending + custom environment.
create.4<-function(elems)
{
env<-new.env()
env$dt<-as.data.frame(elems)
return(env)
}
append.4<-function(env, elems)
{
env$dt[nrow(env$dt)+1,]<-elems
return(env)
}
access.4<-function(env)
{
return(env$dt)
}
The test suite:
For convenience I will use one test function to cover them all with indirect calling. (I checked: using do.call instead of calling the functions directly doesn't makes the code run measurable longer).
test<-function(id, n=1000)
{
n<-n-1
el<-list(a=1,b=2,c=3,d=4)
o<-do.call(paste0('create.',id),list(el))
s<-paste0('append.',id)
for (i in 1:n)
{
o<-do.call(s,list(o,el))
}
return(do.call(paste0('access.', id), list(o)))
}
Let's see the performance for n=10 insertions.
I also added a 'placebo' functions (with suffix 0) that don't perform anything - just to measure the overhead of the test setup.
r<-microbenchmark(test(0,n=10), test(1,n=10),test(2,n=10),test(3,n=10), test(4,n=10))
autoplot(r)
For 1E5 rows (measurements done on Intel(R) Core(TM) i7-4710HQ CPU # 2.50GHz):
nr function time
4 data.frame 228.251
3 sqlite 133.716
2 data.table 3.059
1 rbindlist 169.998
0 placebo 0.202
It looks like the SQLite-based sulution, although regains some speed on large data, is nowhere near data.table + manual exponential growth. The difference is almost two orders of magnitude!
Summary
If you know that you will append rather small number of rows (n<=100), go ahead and use the simplest possible solution: just assign the rows to the data.frame using bracket notation and ignore the fact that the data.frame is not pre-populated.
For everything else use data.table::set and grow the data.table exponentially (e.g. using my code).
Update with purrr, tidyr & dplyr
As the question is already dated (6 years), the answers are missing a solution with newer packages tidyr and purrr. So for people working with these packages, I want to add a solution to the previous answers - all quite interesting, especially .
The biggest advantage of purrr and tidyr are better readability IMHO.
purrr replaces lapply with the more flexible map() family,
tidyr offers the super-intuitive method add_row - just does what it says :)
map_df(1:1000, function(x) { df %>% add_row(x = x, y = toString(x)) })
This solution is short and intuitive to read, and it's relatively fast:
system.time(
map_df(1:1000, function(x) { df %>% add_row(x = x, y = toString(x)) })
)
user system elapsed
0.756 0.006 0.766
It scales almost linearly, so for 1e5 rows, the performance is:
system.time(
map_df(1:100000, function(x) { df %>% add_row(x = x, y = toString(x)) })
)
user system elapsed
76.035 0.259 76.489
which would make it rank second right after data.table (if your ignore the placebo) in the benchmark by #Adam Ryczkowski:
nr function time
4 data.frame 228.251
3 sqlite 133.716
2 data.table 3.059
1 rbindlist 169.998
0 placebo 0.202
A more generic solution for might be the following.
extendDf <- function (df, n) {
withFactors <- sum(sapply (df, function(X) (is.factor(X)) )) > 0
nr <- nrow (df)
colNames <- names(df)
for (c in 1:length(colNames)) {
if (is.factor(df[,c])) {
col <- vector (mode='character', length = nr+n)
col[1:nr] <- as.character(df[,c])
col[(nr+1):(n+nr)]<- rep(col[1], n) # to avoid extra levels
col <- as.factor(col)
} else {
col <- vector (mode=mode(df[1,c]), length = nr+n)
class(col) <- class (df[1,c])
col[1:nr] <- df[,c]
}
if (c==1) {
newDf <- data.frame (col ,stringsAsFactors=withFactors)
} else {
newDf[,c] <- col
}
}
names(newDf) <- colNames
newDf
}
The function extendDf() extends a data frame with n rows.
As an example:
aDf <- data.frame (l=TRUE, i=1L, n=1, c='a', t=Sys.time(), stringsAsFactors = TRUE)
extendDf (aDf, 2)
# l i n c t
# 1 TRUE 1 1 a 2016-07-06 17:12:30
# 2 FALSE 0 0 a 1970-01-01 01:00:00
# 3 FALSE 0 0 a 1970-01-01 01:00:00
system.time (eDf <- extendDf (aDf, 100000))
# user system elapsed
# 0.009 0.002 0.010
system.time (eDf <- extendDf (eDf, 100000))
# user system elapsed
# 0.068 0.002 0.070
Lets take a vector 'point' which has numbers from 1 to 5
point = c(1,2,3,4,5)
if we want to append a number 6 anywhere inside the vector then below command may come handy
i) Vectors
new_var = append(point, 6 ,after = length(point))
ii) columns of a table
new_var = append(point, 6 ,after = length(mtcars$mpg))
The command append takes three arguments:
the vector/column to be modified.
value to be included in the modified vector.
a subscript, after which the values are to be appended.
simple...!!
Apologies in case of any...!
My solution is almost the same as the original answer but it doesn't worked for me.
So, I gave names for the columns and it works:
painel <- rbind(painel, data.frame("col1" = xtweets$created_at,
"col2" = xtweets$text))

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