如何通过使用data.table来提高当前使用ddply的数据清理代码的性能？

Question

我正在尝试使用ddply清理数据,但它在1.3M行上运行速度非常慢.

示例代码:

#Create Sample Data Frame
num_rows <- 10000
df <- data.frame(id=sample(1:20, num_rows, replace=T), 
                Consumption=sample(-20:20, num_rows, replace=T), 
                StartDate=as.Date(sample(15000:15020, num_rows, replace=T), origin = "1970-01-01"))
df$EndDate <- df$StartDate + 90
#df <- df[order(df$id, df$StartDate, df$Consumption),]
#Are values negative? 
# Needed for subsetting in ddply rows with same positive and negative values
df$Neg <- ifelse(df$Consumption < 0, -1, 1)
df$Consumption <- abs(df$Consumption)

我编写了一个函数来删除行,其中一行中的消耗值相同但对另一行中的消耗值为负(对于相同的id).

#Remove rows from a data frame where there is an equal but opposite consumption value
#Should ensure only one negative value is removed for each positive one. 
clean_negatives <- function(x3){
  copies <- abs(sum(x3$Neg))
  sgn <- ifelse(sum(x3$Neg) <0, -1, 1) 
  x3 <- x3[0:copies,]
  x3$Consumption <- sgn*x3$Consumption
  x3$Neg <- NULL
  x3}

然后我使用ddply应用该函数来删除数据中的这些错误行

ptm <- proc.time()
df_cleaned <- ddply(df, .(id,StartDate, EndDate, Consumption),
                    function(x){clean_negatives(x)})
proc.time() - ptm

我希望我可以使用data.table来加快速度,但我无法理解如何使用data.table来帮助.

有1.3M的行,到目前为止,我的桌面整天都在计算,但还没有完成.

Answer 1

您的问题询问data.table实施情况.所以,我在这里展示了它.您的功能也可以大大简化.你可以先获得sign通过总结Neg,然后过滤表,然后乘Consumption用sign(如下图所示).

require(data.table)
# get the data.table in dt
dt <- data.table(df, key = c("id", "StartDate", "EndDate", "Consumption"))
# first obtain the sign directly
dt <- dt[, sign := sign(sum(Neg)), by = c("id", "StartDate", "EndDate", "Consumption")]
# then filter by abs(sum(Neg))
dt.fil <- dt[, .SD[seq_len(abs(sum(Neg)))], by = c("id", "StartDate", "EndDate", "Consumption")]
# modifying for final output (line commented after Statquant's comment
# dt.fil$Consumption <- dt.fil$Consumption * dt.fil$sign
dt.fil[, Consumption := (Consumption*sign)]
dt.fil <- subset(dt.fil, select=-c(Neg, sign))

标杆

• 百万行的数据:

  #Create Sample Data Frame
  num_rows <- 1e6
  df <- data.frame(id=sample(1:20, num_rows, replace=T), 
                  Consumption=sample(-20:20, num_rows, replace=T), 
                  StartDate=as.Date(sample(15000:15020, num_rows, replace=T), origin = "1970-01-01"))
  df$EndDate <- df$StartDate + 90
  df$Neg <- ifelse(df$Consumption < 0, -1, 1)
  df$Consumption <- abs(df$Consumption)
  

• 该data.table函数:

  FUN.DT <- function() {
      require(data.table)
      dt <- data.table(df, key=c("id", "StartDate", "EndDate", "Consumption"))
      dt <- dt[, sign := sign(sum(Neg)), 
                 by = c("id", "StartDate", "EndDate", "Consumption")]
      dt.fil <- dt[, .SD[seq_len(abs(sum(Neg)))], 
                 by=c("id", "StartDate", "EndDate", "Consumption")]
      dt.fil[, Consumption := (Consumption*sign)]
      dt.fil <- subset(dt.fil, select=-c(Neg, sign))
  }
  

• 你的功能 ddply

  FUN.PLYR <- function() {
      require(plyr)
      clean_negatives <- function(x3) {
          copies <- abs(sum(x3$Neg))
          sgn <- ifelse(sum(x3$Neg) <0, -1, 1) 
          x3 <- x3[0:copies,]
          x3$Consumption <- sgn*x3$Consumption
          x3$Neg <- NULL
          x3
      }
      df_cleaned <- ddply(df, .(id, StartDate, EndDate, Consumption), 
                             function(x) clean_negatives(x))
  }
  

• 基准测试rbenchmark(仅限1次运行)

  require(rbenchmark)
  benchmark(FUN.DT(), FUN.PLYR(), replications = 1, order = "elapsed")
  
          test replications elapsed relative user.self sys.self user.child sys.child
  1   FUN.DT()            1   6.137    1.000     5.926    0.211          0         0
  2 FUN.PLYR()            1 242.268   39.477   152.855    82.881         0         0
  

我的data.table实现速度比您当前的plyr实现速度快39倍(我将其与您的实现进行比较,因为函数不同).

Note:我在函数中加载了包以获得获得结果的完整时间.此外,出于同样的原因我转换data.frame到data.table与基准函数内部键.因此,这是最低速度.