dplyr：计算data.frame中多列的levels百分比并将其转换为wide

Question

数据

df <- data.frame(id=c(rep("site1", 3), rep("site2", 8), rep("site3", 9), rep("site4", 15)),
                 major_rock = c("greywacke",    "mudstone", "gravel",   "greywacke",    "gravel",   "mudstone", "gravel", "mudstone", "mudstone",   
                                "conglomerate", "gravel", "mudstone",   "greywacke","conglomerate", "gravel",   "gravel",   "greywacke","gravel",   
                                "greywacke",    "gravel",   "mudstone", "greywacke",    "gravel", "gravel", "gravel",   "conglomerate", "greywacke",
                                "coquina",  "gravel",   "gravel",   "greywacke",    "gravel",   "mudstone","mudstone",  "gravel"),
                 minor_rock = c("sandstone mudstone basalt chert limestone",  "limestone",   "sand silt clay", "sandstone mudstone basalt chert limestone",
                                "sand silt clay", "sandstone conglomerate coquina tephra", NA, "limestone",  "mudstone sandstone coquina limestone",
                                "sandstone mudstone limestone",  "sand loess silt",  "sandstone conglomerate coquina tephra", "sandstone mudstone basalt chert limestone",
                                "sandstone mudstone limestone", "sand loess silt", "loess silt sand", "sandstone mudstone conglomerate chert limestone basalt",
                                "sand silt clay",  "sandstone mudstone conglomerate", "loess sand silt", "sandstone conglomerate coquina tephra", "sandstone mudstone basalt chert limestone",
                                "sand loess silt", "sand silt clay", "loess silt sand",  "sandstone mudstone limestone", "sandstone mudstone conglomerate chert limestone basalt",
                                "limestone", "loess sand silt",  NA, "sandstone mudstone conglomerate", "sandstone siltstone mudstone limestone silt lignite", "limestone",
                                "mudstone sandstone coquina limestone", "mudstone tephra loess"),
                 area_ha = c(1066.68,   7.59,   3.41,   4434.76,    393.16, 361.69, 306.75, 124.93, 95.84,  9.3,    8.45,   4565.89,    2600.44,    2198.52,    
                             2131.71,   2050.09,    1640.47,    657.09, 296.73, 178.12, 10403.53,   8389.2,  8304.08,   3853.36,    2476.36,    2451.25,    
                             1640.47,   1023.02,    532.94, 385.68, 296.73, 132.45, 124.93, 109.12, 4.87))

我想要什么？

我需要准备df 进行另一项分析，该分析要求每个站点只有一行。所以在最终的data.framedf_fin中，每个站点都会有major_rock和minor_rock中级别的比例，列名（变量）将是major_rock和minor_rock的级别。

我可以为每个变量（major_rock 和 minor_rock）执行此操作，然后像下面这样组合它们

我做了什么？

对于major_rock

library(tidyverse)

df_major_rock <- df %>% 
  dplyr::select(-minor_rock) %>% 
  dplyr::group_by(id, major_rock) %>% 
  dplyr::summarise(total_area = sum(area_ha)) %>% 
  dplyr::group_by(id) %>% 
  dplyr::mutate(percent_major = total_area/sum(total_area) * 100) %>% 
  dplyr::select(-total_area) %>% 
  tidyr::spread(major_rock, percent_major)

> df_major_rock
Source: local data frame [4 x 6]
Groups: id [4]

      id conglomerate  coquina     gravel greywacke   mudstone
* <fctr>        <dbl>    <dbl>      <dbl>     <dbl>      <dbl>
1  site1           NA       NA  0.3164205  98.97929  0.7042907
2  site2    0.1621656       NA 12.3517842  77.32960 10.1564462
3  site3   13.4720995       NA 30.7432536  27.80577 27.9788787
4  site4    6.1085791 2.549393 39.0992422  25.73366 26.5091274

minor_rock 也一样

df_minor_rock <- df %>% 
  dplyr::select(-major_rock) %>% 
  dplyr::group_by(id, minor_rock) %>% 
  dplyr::summarise(total_area = sum(area_ha)) %>% 
  dplyr::group_by(id) %>% 
  dplyr::mutate(percent_minor = total_area/sum(total_area) * 100)%>% 
  dplyr::select(-total_area) %>% 
  tidyr::spread(minor_rock, percent_minor)

> df_minor_rock
Source: local data frame [4 x 15]
Groups: id [4]

      id limestone `loess sand silt` `loess silt sand` `mudstone sandstone coquina limestone` `mudstone tephra loess` `sand loess silt`
* <fctr>     <dbl>             <dbl>             <dbl>                                  <dbl>                   <dbl>             <dbl>
1  site1 0.7042907                NA                NA                                     NA                      NA                NA
2  site2 2.1784240                NA                NA                              1.6711771                      NA          0.147344
3  site3        NA          1.091484         12.562550                                     NA                      NA         13.062701
4  site4 2.8607214          1.328100          6.171154                              0.2719299              0.01213617         20.693984
# ... with 8 more variables: `sand silt clay` <dbl>, `sandstone conglomerate coquina tephra` <dbl>, `sandstone mudstone basalt chert
#   limestone` <dbl>, `sandstone mudstone conglomerate` <dbl>, `sandstone mudstone conglomerate chert limestone basalt` <dbl>, `sandstone
#   mudstone limestone` <dbl>, `sandstone siltstone mudstone limestone silt lignite` <dbl>, `<NA>` <dbl>

然后，我将两个 data.frames 连接在一起（df_major_rock 和 df_minor_rock），因此最终 data.frame df_fin 将只有 4 个观察值（每个站点一行），变量将是major_rock 和 minor_rock

df_fin <- df_major_rock %>% 
  dplyr::right_join(., df_minor_rock, by="id")

问题

df_fin 正是我想要的。但是，在这个可重现的示例中，我只展示了 2 个变量（major_rock 和 minor_rock），我必须创建两个不同的 data.frames 来获取每个变量的级别比例，然后将它们连接在一起以获得最终输出 df_fin .在我的实际数据中，除了major_rock 和minor_rock 之外，我还有许多变量，我还想获得每个站点的级别比例。我认为应该有比我更直接或更短的方法。任何建议将不胜感激？

Answer 1

您可以通过使用data.table::dcast 将其缩短一点，这会将您的数据分散到列中。然后，您可以使用 rowSums 一步计算百分比。虽然可能有更好的方法来做到这一点，但我在循环中为每一列包装了这种方法：

df_fin  <- data.frame(id = unique(df$id))
myColumns <- setdiff(colnames(df)[-1], "area_ha")

for (name in myColumns){
  dcastFormula <- paste0("id ~ ", name)
  tempdf <- data.table::dcast(df, dcastFormula, sum)
  tempdf[,-1] <-  tempdf[,-1]/rowSums(tempdf[,-1],na.rm = TRUE)*100
  df_fin  <- left_join(df_fin , tempdf, by ="id")
}

与往常一样，可能还有其他几种方法可以做到这一点，但这是一个比您的起点更简单的示例。此外，它可能需要根据您的其他列和/或您希望如何聚合它们进行修改。