在R中重塑铸造和熔炼

2 回答

• 0

  也许你对 tidyverse 替代感兴趣

  library(tidyverse)
  df %>%
      gather(Year, val, -Variables, -Country) %>%
      spread(Variables, val)
  #  Country Year Happiness Power
  #1  France 2000      1872  1213
  #2  France 2001      2345  1234
  #3      UK 2000      2234  1726
  #4      UK 2001      9082  6433
  

  或使用 reshape2::melt 和 reshape2::dcast

  reshape2::dcast(
      reshape2::melt(df, id.vars = c("Country", "Variables"), variable.name = "Year"),
      Country + Year ~ Variables)
  #        Country Year Happiness Power
  #1  France 2000      1872  1213
  #2  France 2001      2345  1234
  #3      UK 2000      2234  1726
  #4      UK 2001      9082  6433
  

  或（相同）使用 data.table::melt 和 data.table::dcast

  data.table::dcast(
      data.table::melt(df, id.vars = c("Country", "Variables"), variable.name = "Year"), 
      Country + Year ~ Variables)
  #  Country Year Happiness Power
  #1  France 2000      1872  1213
  #2  France 2001      2345  1234
  #3      UK 2000      2234  1726
  #4      UK 2001      9082  6433
  

  在性能/运行时方面，我认为 data.table 或 tidyr 解决方案是最有效的 . 您可以通过在某些较大的样本数据上运行 microbenchmark 来进行检查 .

  ---

  样本数据

  df <-read.table(text =
      "  Variables Country  2000  2001
  1 Power     France   1213  1234
  2 Happiness France   1872  2345
  3 Power     UK       1726  6433
  4 Happiness UK       2234  9082", header = T)
  colnames(df)[3:4] <- c("2000", "2001")
  

  ---

  基准分析

  根据（略微）更大的 78x22 样本数据集对四种方法进行分析后得出结果 .

  set.seed(2017)
  df <- data.frame(
      Variables = rep(c("Power", "Happiness", "something_else"), 26),
      Country = rep(LETTERS[1:26], each = 3),
      matrix(sample(10000, 20 * 26 * 3), nrow = 26 * 3))
  colnames(df)[3:ncol(df)] <- 2000:2019
  
  library(microbenchmark)
  library(tidyr)
  
  res <- microbenchmark(
      reshape2 = {
          reshape2::dcast(
              reshape2::melt(df, id.vars = c("Country", "Variables"), variable.name = "Year"),
              Country + Year ~ Variables)
      },
      tidyr = {
          df %>%
              gather(Year, val, -Variables, -Country) %>%
              spread(Variables, val)
      },
      datatable = {
          data.table::dcast(
              data.table::melt(df, id.vars = c("Country", "Variables"), variable.name = "Year"),
              Country + Year ~ Variables)
      },
      reshape = {
          reshape::cast(reshape::melt(df), Country + variable ~ Variables)
      }
  )
  res
  #Unit: milliseconds
  #      expr       min        lq      mean    median        uq       max neval
  #  reshape2  3.088740  3.449686  4.313044  3.919372  5.112560  7.856902   100
  #     tidyr  4.482361  4.982017  6.215872  5.771133  6.931964 28.293377   100
  # datatable  3.179035  3.511542  4.861192  4.040188  5.123103 46.010810   100
  #   reshape 27.371094 30.226222 32.425667 32.504644 34.118499 41.286803   100
  
  library(ggplot2)
  autoplot(res)
  

  

  回复于 2024-05-09T01:02:02+08:00
• 2

  如上所述，我强烈建议使用 tidyr 而不是 reshape ，或者至少使用 reshape2 而不是 reshape ，因为它修复了 reshape 的许多性能问题 .

  在 reshape 本身，你必须首先融化 datas

  > cast(melt(datas), Country + variable ~ Variables)
  Using Variables, Country as id variables
    Country variable Happiness Power
  1  France    y2000      1872  1213
  2  France    y2001      2345  1234
  3      UK    y2000      2234  1726
  4      UK    y2001      9082  6433
  

  然后根据需要重命名和转换列 .

  在 reshape2 中，代码是相同的，但您可以使用 dcast 而不是 cast . tidyr ，如@Maurits Evers上面的解决方案是一个更好的解决方案，大多数开发已从 reshape2 转移到 tidyverse

  回复于 2024-05-09T01:02:02+08:00