What's the fastest way to merge/join data.frames in R?

前端 未结 5 1453
走了就别回头了
走了就别回头了 2020-11-27 08:58

For example (not sure if most representative example though):

N <- 1e6
d1 <- data.frame(x=sample(N,N), y1=rnorm(N))
d2 <- data.frame(x=sample(N,N),          


        
相关标签:
5条回答
  • 2020-11-27 09:42

    For simple task (unique values on both sides of join) I use match:

    system.time({
        d <- d1
        d$y2 <- d2$y2[match(d1$x,d2$x)]
    })
    

    It's far more faster than merge (on my machine 0.13s to 3.37s).

    My timings:

    • merge: 3.32s
    • plyr: 0.84s
    • match: 0.12s
    0 讨论(0)
  • 2020-11-27 09:44

    The match approach works when there is a unique key in the second data frame for each key value in the first. If there are duplicates in the second data frame then the match and merge approaches are not the same. Match is, of course, faster since it is not doing as much. In particular it never looks for duplicate keys. (continued after code)

    DF1 = data.frame(a = c(1, 1, 2, 2), b = 1:4)
    DF2 = data.frame(b = c(1, 2, 3, 3, 4), c = letters[1:5])
    merge(DF1, DF2)
        b a c
      1 1 1 a
      2 2 1 b
      3 3 2 c
      4 3 2 d
      5 4 2 e
    DF1$c = DF2$c[match(DF1$b, DF2$b)]
    DF1$c
    [1] a b c e
    Levels: a b c d e
    
    > DF1
      a b c
    1 1 1 a
    2 1 2 b
    3 2 3 c
    4 2 4 e
    

    In the sqldf code that was posted in the question, it might appear that indexes were used on the two tables but, in fact, they are placed on tables which were overwritten before the sql select ever runs and that, in part, accounts for why its so slow. The idea of sqldf is that the data frames in your R session constitute the data base, not the tables in sqlite. Thus each time the code refers to an unqualified table name it will look in your R workspace for it -- not in sqlite's main database. Thus the select statement that was shown reads d1 and d2 from the workspace into sqlite's main database clobbering the ones that were there with the indexes. As a result it does a join with no indexes. If you wanted to make use of the versions of d1 and d2 that were in sqlite's main database you would have to refer to them as main.d1 and main.d2 and not as d1 and d2. Also, if you are trying to make it run as fast as possible then note that a simple join can't make use of indexes on both tables so you can save the time of creating one of the indexes. In the code below we illustrate these points.

    Its worthwhile to notice that the precise computation can make a huge difference on which package is fastest. For example, we do a merge and an aggregate below. We see that the results are nearly reversed for the two. In the first example from fastest to slowest we get: data.table, plyr, merge and sqldf whereas in the second example sqldf, aggregate, data.table and plyr -- nearly the reverse of the first one. In the first example sqldf is 3x slower than data.table and in the second its 200x faster than plyr and 100 times faster than data.table. Below we show the input code, the output timings for the merge and the output timings for the aggregate. Its also worthwhile noting that sqldf is based on a database and therefore can handle objects larger than R can handle (if you use the dbname argument of sqldf) while the other approaches are limited to processing in main memory. Also we have illustrated sqldf with sqlite but it also supports the H2 and PostgreSQL databases as well.

    library(plyr)
    library(data.table)
    library(sqldf)
    
    set.seed(123)
    N <- 1e5
    d1 <- data.frame(x=sample(N,N), y1=rnorm(N))
    d2 <- data.frame(x=sample(N,N), y2=rnorm(N))
    
    g1 <- sample(1:1000, N, replace = TRUE)
    g2<- sample(1:1000, N, replace = TRUE)
    d <- data.frame(d1, g1, g2)
    
    library(rbenchmark)
    
    benchmark(replications = 1, order = "elapsed",
       merge = merge(d1, d2),
       plyr = join(d1, d2),
       data.table = { 
          dt1 <- data.table(d1, key = "x")
          dt2 <- data.table(d2, key = "x")
          data.frame( dt1[dt2,list(x,y1,y2=dt2$y2)] )
          },
       sqldf = sqldf(c("create index ix1 on d1(x)",
          "select * from main.d1 join d2 using(x)"))
    )
    
    set.seed(123)
    N <- 1e5
    g1 <- sample(1:1000, N, replace = TRUE)
    g2<- sample(1:1000, N, replace = TRUE)
    d <- data.frame(x=sample(N,N), y=rnorm(N), g1, g2)
    
    benchmark(replications = 1, order = "elapsed",
       aggregate = aggregate(d[c("x", "y")], d[c("g1", "g2")], mean), 
       data.table = {
          dt <- data.table(d, key = "g1,g2")
          dt[, colMeans(cbind(x, y)), by = "g1,g2"]
       },
       plyr = ddply(d, .(g1, g2), summarise, avx = mean(x), avy=mean(y)),
       sqldf = sqldf(c("create index ix on d(g1, g2)",
          "select g1, g2, avg(x), avg(y) from main.d group by g1, g2"))
    )
    

    The outputs from the two benchmark call comparing the merge calculations are:

    Joining by: x
            test replications elapsed relative user.self sys.self user.child sys.child
    3 data.table            1    0.34 1.000000      0.31     0.01         NA        NA
    2       plyr            1    0.44 1.294118      0.39     0.02         NA        NA
    1      merge            1    1.17 3.441176      1.10     0.04         NA        NA
    4      sqldf            1    3.34 9.823529      3.24     0.04         NA        NA
    

    The output from the benchmark call comparing the aggregate calculations are:

            test replications elapsed  relative user.self sys.self user.child sys.child
    4      sqldf            1    2.81  1.000000      2.73     0.02         NA        NA
    1  aggregate            1   14.89  5.298932     14.89     0.00         NA        NA
    2 data.table            1  132.46 47.138790    131.70     0.08         NA        NA
    3       plyr            1  212.69 75.690391    211.57     0.56         NA        NA
    
    0 讨论(0)
  • 2020-11-27 09:57

    The 132 seconds reported in Gabor's results for data.table is actually timing base functions colMeans and cbind (the memory allocation and copying induced by using those functions). There are good and bad ways of using data.table, too.

    benchmark(replications = 1, order = "elapsed", 
      aggregate = aggregate(d[c("x", "y")], d[c("g1", "g2")], mean),
      data.tableBad = {
         dt <- data.table(d, key = "g1,g2") 
         dt[, colMeans(cbind(x, y)), by = "g1,g2"]
      }, 
      data.tableGood = {
         dt <- data.table(d, key = "g1,g2") 
         dt[, list(mean(x),mean(y)), by = "g1,g2"]
      }, 
      plyr = ddply(d, .(g1, g2), summarise, avx = mean(x), avy=mean(y)),
      sqldf = sqldf(c("create index ix on d(g1, g2)",
          "select g1, g2, avg(x), avg(y) from main.d group by g1, g2"))
      ) 
    
                test replications elapsed relative user.self sys.self
    3 data.tableGood            1    0.15    1.000      0.16     0.00
    5          sqldf            1    1.01    6.733      1.01     0.00
    2  data.tableBad            1    1.63   10.867      1.61     0.01
    1      aggregate            1    6.40   42.667      6.38     0.00
    4           plyr            1  317.97 2119.800    265.12    51.05
    
    packageVersion("data.table")
    # [1] ‘1.8.2’
    packageVersion("plyr")
    # [1] ‘1.7.1’
    packageVersion("sqldf")
    # [1] ‘0.4.6.4’
    R.version.string
    # R version 2.15.1 (2012-06-22)
    

    Please note that I don't know plyr well so please do check with Hadley before relying on the plyr timings here. Also note that the data.table do include the time to convert to data.table and set the key, for fareness.


    This answer has been updated since originally answered in Dec 2010. The previous benchmark results are below. Please see revision history of this answer to see what changed.

                  test replications elapsed   relative user.self sys.self
    4   data.tableBest            1   0.532   1.000000     0.488    0.020
    7            sqldf            1   2.059   3.870301     2.041    0.008
    3 data.tableBetter            1   9.580  18.007519     9.213    0.220
    1        aggregate            1  14.864  27.939850    13.937    0.316
    2  data.tableWorst            1 152.046 285.800752   150.173    0.556
    6 plyrwithInternal            1 198.283 372.712406   189.391    7.665
    5             plyr            1 225.726 424.296992   208.013    8.004
    
    0 讨论(0)
  • 2020-11-27 10:00

    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)
    
    0 讨论(0)
  • 2020-11-27 10:01

    Thought it would be interesting to post a benchmark with dplyr in the mix: (had a lot of things running)

                test replications elapsed relative user.self sys.self
    5          dplyr            1    0.25     1.00      0.25     0.00
    3 data.tableGood            1    0.28     1.12      0.27     0.00
    6          sqldf            1    0.58     2.32      0.57     0.00
    2  data.tableBad            1    1.10     4.40      1.09     0.01
    1      aggregate            1    4.79    19.16      4.73     0.02
    4           plyr            1  186.70   746.80    152.11    30.27
    
    packageVersion("data.table")
    [1] ‘1.8.10’
    packageVersion("plyr")
    [1] ‘1.8’
    packageVersion("sqldf")
    [1] ‘0.4.7’
    packageVersion("dplyr")
    [1] ‘0.1.2’
    R.version.string
    [1] "R version 3.0.2 (2013-09-25)"
    

    Just added:

    dplyr = summarise(dt_dt, avx = mean(x), avy = mean(y))
    

    and setup the data for dplyr with a data table:

    dt <- tbl_dt(d)
    dt_dt <- group_by(dt, g1, g2)
    

    Updated: I removed data.tableBad and plyr and nothing but RStudio open (i7, 16GB ram).

    With data.table 1.9 and dplyr with data frame:

                test replications elapsed relative user.self sys.self
    2 data.tableGood            1    0.02      1.0      0.02     0.00
    3          dplyr            1    0.04      2.0      0.04     0.00
    4          sqldf            1    0.46     23.0      0.46     0.00
    1      aggregate            1    6.11    305.5      6.10     0.02
    

    With data.table 1.9 and dplyr with data table:

                test replications elapsed relative user.self sys.self
    2 data.tableGood            1    0.02        1      0.02     0.00
    3          dplyr            1    0.02        1      0.02     0.00
    4          sqldf            1    0.44       22      0.43     0.02
    1      aggregate            1    6.14      307      6.10     0.01
    
    packageVersion("data.table")
    [1] '1.9.0'
    packageVersion("dplyr")
    [1] '0.1.2'
    

    For consistency here is the original with all and data.table 1.9 and dplyr using a data table:

                test replications elapsed relative user.self sys.self
    5          dplyr            1    0.01        1      0.02     0.00
    3 data.tableGood            1    0.02        2      0.01     0.00
    6          sqldf            1    0.47       47      0.46     0.00
    1      aggregate            1    6.16      616      6.16     0.00
    2  data.tableBad            1   15.45     1545     15.38     0.01
    4           plyr            1  110.23    11023     90.46    19.52
    

    I think this data is too small for the new data.table and dplyr :)

    Larger data set:

    N <- 1e8
    g1 <- sample(1:50000, N, replace = TRUE)
    g2<- sample(1:50000, N, replace = TRUE)
    d <- data.frame(x=sample(N,N), y=rnorm(N), g1, g2)
    

    Took around 10-13GB of ram just to hold the data before running the benchmark.

    Results:

                test replications elapsed relative user.self sys.self
    1          dplyr            1   14.88        1      6.24     7.52
    2 data.tableGood            1   28.41        1     18.55      9.4
    

    Tried a 1 billion but blew up ram. 32GB will handle it no problem.


    [Edit by Arun] (dotcomken , could you please run this code and paste your benchmarking results? Thanks).

    require(data.table)
    require(dplyr)
    require(rbenchmark)
    
    N <- 1e8
    g1 <- sample(1:50000, N, replace = TRUE)
    g2 <- sample(1:50000, N, replace = TRUE)
    d <- data.frame(x=sample(N,N), y=rnorm(N), g1, g2)
    
    benchmark(replications = 5, order = "elapsed", 
      data.table = {
         dt <- as.data.table(d) 
         dt[, lapply(.SD, mean), by = "g1,g2"]
      }, 
      dplyr_DF = d %.% group_by(g1, g2) %.% summarise(avx = mean(x), avy=mean(y))
    ) 
    

    As per Arun's request here the output of what you provided me to run:

            test replications elapsed relative user.self sys.self
    1 data.table            5   15.35     1.00     13.77     1.57
    2   dplyr_DF            5  137.84     8.98    136.31     1.44
    

    Sorry for the confusion, late night got to me.

    Using dplyr with data frame seems to be the less efficient way to process summaries. Is this methods to compare the exact functionality of data.table and dplyr with their data structure methods included? I'd almost prefer to separate that as most data will need to be cleaned before we group_by or create the data.table. It could be a matter of taste but I think the most important part is how efficiently the data can be modeled.

    0 讨论(0)
提交回复
热议问题