Reduce computing time for reshape

2019-04-28 18:49发布

问题:

I have the following dataset, which I would like to reshape from wide to long format:

Name     Code  CURRENCY   01/01/1980   02/01/1980   03/01/1980   04/01/1980
Abengoa  4256  USD        1.53         1.54         1.51         1.52      
Adidas   6783  USD        0.23         0.54         0.61         0.62      

The data consists of stock prices for different firms on each day from 1980 to 2013. Therefore, I have 8,612 columns in my wide data (and a abou 3,000 rows). Now, I am using the following command to reshape the data into long format:

library(reshape)
data <- read.csv("data.csv")
data1 <- melt(data,id=c("Name","Code", "CURRENCY"),variable_name="Date")

However, for .csv files that are about 50MB big, it already takes about two hours. The computing time shouldn't be driven by weak hardware, since I am running this on a 2.7 GHz Intel Core i7 with 16GB of RAM. Is there any other more efficient way to do this?

Many thanks!

回答1:

While the testing is going on, I'll post my comment as an answer for you to consider. Try using stack as in:

data1 <- data.frame(data[1:3], stack(data[-c(1, 2, 3)]))

In many cases, stack works really efficiently with these types of operations, and adding back in the first few columns also works quickly because of how vectors are recycled in R.

For that matter, this might also be worth considering:

data.frame(data[1:3],
           vals = as.vector(as.matrix(data[-c(1, 2, 3)])),
           date = rep(names(data)[-c(1, 2, 3)], each = nrow(data)))

I'm cautious to benchmark on such a small sample of data though, because I suspect the results won't be quite comparable to benchmarking on your actual dataset.


Update: Results of some more benchmarks

Using @RicardoSaporta's benchmarking procedure, I have benchmarked data.table against what I've called "Manual" data.frame creation. You can see the results of the benchmarks here, on datasets ranging from 1000 rows to 3000 rows, in 500 row increments, and all with 8003 columns (8000 data columns, plus the three initial columns).

The results can be seen here: http://rpubs.com/mrdwab/reduce-computing-time

Ricardo's correct--there seems to be something about 3000 rows that makes a huge difference with the base R approaches (and it would be interesting if anyone has any explanation about what that might be). But this "Manual" approach is actually even faster than stack, if performance really is the primary concern.

Here are the results for just the last three runs:

data <- makeSomeData(2000, 8000)
dtt <- data.table(data)
suppressWarnings(benchmark(DT = eval(DT), Manual = eval(Manual), replications = 1, 
    columns = c("relative", "test", "elapsed", "user.self", "sys.self", "replications"), 
    order = "relative"))
##   relative   test elapsed user.self sys.self replications
## 2    1.000 Manual   0.908     0.696    0.108            1
## 1    3.963     DT   3.598     3.564    0.012            1

rm(data, dateCols, nvc, dtt)

data <- makeSomeData(2500, 8000)
dtt <- data.table(data)
suppressWarnings(benchmark(DT = eval(DT), Manual = eval(Manual), replications = 1, 
    columns = c("relative", "test", "elapsed", "user.self", "sys.self", "replications"), 
    order = "relative"))
##   relative   test elapsed user.self sys.self replications
## 2    1.000 Manual   2.841     1.044    0.296            1
## 1    1.694     DT   4.813     4.661    0.080            1

rm(data, dateCols, nvc, dtt)

data <- makeSomeData(3000, 8000)
dtt <- data.table(data)
suppressWarnings(benchmark(DT = eval(DT), Manual = eval(Manual), replications = 1, 
    columns = c("relative", "test", "elapsed", "user.self", "sys.self", "replications"), 
    order = "relative"))
##   relative   test elapsed user.self sys.self replications
## 1     1.00     DT   7.223     5.769    0.112            1
## 2    29.27 Manual 211.416     1.560    0.952            1

Ouch! data.table really turns the tables on that last run!



回答2:

Benchmarks Summary:

Using Stack (as suggested by @AnandaMahto) is definitely
the way to go for smaller data sets (N < 3,000).
As the data sets gets larger, data.table begins to outperform stack


Here is an option using data.table

dtt <- data.table(data)

# non value columns, ie, the columns to keep post reshape
nvc <- c("Name","Code", "CURRENCY")

# name of columns being transformed 
dateCols <- setdiff(names(data), nvc)

# use rbind list to combine subsets
dtt2 <- rbindlist(lapply(dateCols, function(d) {
    dtt[, Date := d]
    cols <- c(nvc, "Date", d)
    setnames(dtt[, cols, with=FALSE], cols, c(nvc, "Date", "value"))
}))

## Results: 

dtt2
#       Name Code CURRENCY         Date value
# 1: Abengoa 4256      USD X_01_01_1980  1.53
# 2:  Adidas 6783      USD X_01_01_1980  0.23
# 3: Abengoa 4256      USD X_02_01_1980  1.54
# 4:  Adidas 6783      USD X_02_01_1980  0.54
# 5: ... <cropped>

Updated Benchmarks with larger sample data

As per the suggestion from @AnandaMahto, below are benchmarks using a large (larger) sample data. Please feel free to improve any of the methods used below and/or add new methods.

Benchmarks

 Resh <- quote(reshape::melt(data,id=c("Name","Code", "CURRENCY"),variable_name="Date"))
 Resh2 <- quote(reshape2::melt(data,id=c("Name","Code", "CURRENCY"),variable_name="Date"))
 DT <- quote({    nvc <- c("Name","Code", "CURRENCY"); dateCols <- setdiff(names(data), nvc); rbindlist(lapply(dateCols, function(d) { dtt[, Date := d]; cols <- c(nvc, "Date", d); setnames(dtt[, cols, with=FALSE], cols, c(nvc, "Date", "value"))}))})
 Stack <- quote(data.frame(data[1:3], stack(data[-c(1, 2, 3)])))


 # SAMPLE SIZE: ROWS = 900; COLS = 380 + 3; 
 dtt <- data.table(data);  
 benchmark(Resh=eval(Resh),Resh2=eval(Resh2),DT=eval(DT), Stack=eval(Stack), replications=5, columns=c("relative", "test", "elapsed", "user.self", "sys.self", "replications"), order="relative")
 # relative  test elapsed user.self sys.self replications
 #    1.000 Stack   0.813     0.623    0.192            5
 #    2.530    DT   2.057     2.035    0.026            5
 #   40.470  Resh  32.902    18.410   14.602            5
 #   40.578 Resh2  32.990    18.419   14.728            5

 # SAMPLE SIZE: ROWS = 3,500; COLS = 380 + 3; 
 dtt <- data.table(data);  
 benchmark(DT=eval(DT), Stack=eval(Stack), replications=5, columns=c("relative", "test", "elapsed", "user.self", "sys.self", "replications"), order="relative")
 #  relative  test elapsed user.self sys.self replications
 #      1.00    DT   2.407     2.336    0.076            5
 #      1.08 Stack   2.600     1.626    0.983            5

 # SAMPLE SIZE: ROWS = 27,000; COLS = 380 + 3; 
 dtt <- data.table(data);  
 benchmark(DT=eval(DT), Stack=eval(Stack), replications=5, columns=c("relative", "test", "elapsed", "user.self", "sys.self", "replications"), order="relative")
 # relative  test elapsed user.self sys.self replications
 #    1.000    DT  10.450     7.418    3.058            5
 #    2.232 Stack  23.329    14.180    9.266            5

Sample Data Creation

  # rm(list=ls(all=TRUE))
  set.seed(1)
  LLLL <- apply(expand.grid(LETTERS, LETTERS[10:15], LETTERS[1:20], LETTERS[1:5], stringsAsFactors=FALSE), 1, paste0, collapse="")

  size <- 900
  dateSamples <- 380
  startDate <- as.Date("1980-01-01")

  Name <- apply(matrix(LLLL[1:(2*size)], ncol=2), 1, paste0, collapse="")
  Code <- sample(1e3:max(1e4-1, size+1e3), length(Name))
  CURRENCY <- sample(c("USD", "EUR", "YEN"), length(Name), TRUE)

  Dates <- seq(startDate, length.out=dateSamples, by="mon")
  Values <- sample(c(1:1e2, 1:5e2), size=size*dateSamples, TRUE) / 1e2

  # Calling the sample dataframe `data` to keep consistency, but I dont like this practice
  data <- data.frame(Name, Code, CURRENCY,       
                     matrix(Values, ncol=length(Dates), dimnames=list(c(), as.character(Dates)))
                    ) 

  data[1:6, 1:8]
  #        Name Code CURRENCY X1980.01.01 X1980.02.01 X1980.03.01 X1980.04.01 X1980.05.01
  # 1  AJAAQNFA 3389      YEN        0.37        0.33        3.58        4.33        1.06
  # 2  BJAARNFA 4348      YEN        1.14        2.69        2.57        0.27        3.02
  # 3  CJAASNFA 6154      USD        2.47        3.72        3.32        0.36        4.85
  # 4  DJAATNFA 9171      USD        2.22        2.48        0.71        0.79        2.85
  # 5  EJAAUNFA 2814      USD        2.63        2.17        1.66        0.55        3.12
  # 6  FJAAVNFA 9081      USD        1.92        1.47        3.51        3.23        3.68


回答3:

From the question :

data <- read.csv("data.csv")

and

... for .csv files that are about 50MB big, it already takes about two hours ...

So although stack/melt/reshape comes into play, I'm guessing (since this is your fist ever S.O. question) that the biggest factor here is read.csv. Assuming you're including that in your timing as well as melt (it isn't clear).

Default arguments to read.csv are well known to be slow. A few quick searches should reveal hint and tips (e.g. stringsAsFactors, colClasses) such as :

  • http://cran.r-project.org/doc/manuals/R-data.html
  • Quickly reading very large tables as dataframes in R

But I'd suggest the new fread function in data.table 1.8.7. Install from the R-Forge repo (as it's not yet on CRAN). To get a feel for fread, without installing, its manual page in raw text form is here : https://r-forge.r-project.org/scm/viewvc.php/pkg/man/fread.Rd?view=markup&root=datatable

The examples section there, as it happens, has a 50MB example shown to be read in 3 seconds instead of up to 60. And benchmarks are starting to appear in other answers which is great to see.

Then the stack/reshape/melt answers are next order, if I guessed correctly.