使用两列的数据框中的百分位数

Question

也许这是一个简单的问题，但我被卡住了。

我的数据框（来自年度调查）包含几个物种的长度数据（按年份和运输）。我想获得每年，每个物种的 95 个百分位。我的数据框示例，

    structure(list(year = c(2015L, 2015L, 2015L, 2015L, 2014L, 2016L,
    2015L, 2016L, 2014L, 2016L, 2015L, 2015L, 2016L, 2016L, 2014L, 2014L,
    2014L, 2015L, 2016L, 2016L), cod_haul = structure(c(72L, 51L, 77L,
    43L, 20L, 92L, 75L, 93L, 9L, 103L, 65L, 63L, 85L, 102L, 27L, 24L,
    14L, 55L, 114L, 105L), .Label = c("N14_02", "N14_03", "N14_04",
    "N14_06", "N14_07", "N14_08", "N14_10", "N14_13", "N14_16", "N14_17",
    "N14_19", "N14_21", "N14_24", "N14_25", "N14_26", "N14_27", "N14_28",
    "N14_29", "N14_30", "N14_32", "N14_33", "N14_35", "N14_37", "N14_39",
    "N14_40", "N14_41", "N14_42", "N14_44", "N14_51", "N14_54", "N14_55",
    "N14_56", "N14_57", "N14_58", "N14_61", "N14_62", "N14_64", "N14_66",
    "N14_67", "N15_01", "N15_03", "N15_07", "N15_11", "N15_12", "N15_14",
    "N15_16", "N15_18", "N15_19", "N15_20", "N15_22", "N15_23", "N15_24",
    "N15_25", "N15_26", "N15_27", "N15_28", "N15_29", "N15_30", "N15_31",
    "N15_32", "N15_36", "N15_37", "N15_39", "N15_41", "N15_44", "N15_46",
    "N15_47", "N15_48", "N15_52", "N15_55", "N15_56", "N15_58", "N15_59",
    "N15_60", "N15_62", "N15_63", "N15_64", "N15_66", "N15_67", "N16_04",
    "N16_06", "N16_07", "N16_08", "N16_11", "N16_12", "N16_13", "N16_15",
    "N16_17", "N16_18", "N16_20", "N16_22", "N16_23", "N16_25", "N16_28",
    "N16_29", "N16_30", "N16_31", "N16_32", "N16_33", "N16_34", "N16_35",
    "N16_37", "N16_40", "N16_41", "N16_45", "N16_46", "N16_47", "N16_48",
    "N16_49", "N16_50", "N16_51", "N16_52", "N16_53", "N16_54", "N16_56",
    "N16_58", "N16_60", "N16_61", "N16_62", "N16_63", "N16_64","N16_66"),
     class = "factor"), haul = c(58L, 23L, 64L, 11L, 32L, 23L, 62L, 25L,
     16L, 40L, 44L, 39L, 12L, 37L, 42L, 39L, 25L, 27L, 54L, 45L), name =
     structure(c(2L, 23L, 11L, 2L, 19L, 15L, 18L, 16L, 3L, 21L, 16L, 21L,
     20L, 19L, 3L, 18L, 16L, 11L, 7L, 13L), .Label = c("Argentina 
     sphyraena", "Arnoglossus laterna", "Blennius ocellaris", "Boops 
     boops", "Callionymus lyra", "Callionymus maculatus", "Capros aper",
     "Cepola macrophthalma", "Chelidonichthys cuculus", "Chelidonichthys
     lucerna", "Conger conger", "Eutrigla gurnardus", "Gadiculus 
     argenteus", "Galeus melastomus", "Helicolenus dactylopterus", 
     "Lepidorhombus boscii", "Lepidorhombus whiffiagonis", "Merluccius
      merluccius", "Microchirus variegatus", "Micromesistius poutassou",
      "Phycis blennoides", "Raja clavata", "Scyliorhinus canicula", 
      "Solea solea", "Trachurus trachurus", "Trisopterus luscus"), class
      = "factor"), length = c(9L, 18L, 50L, 12L, 14L, 12L, 31L, 19L, 15L,
      16L, 26L, 48L, 23L, 10L, 16L, 24L, 12L, 46L, 75L, 13L), number =
      c(5L, 4L, 1L, 2L, 29L, 5L, 2L, 14L, 1L, 1L, 4L, 1L, 29L, 21L, 2L,
      1L, 2L, 1L, 2L, 14L)), row.names = c(NA, 20L), class = 
       "data.frame") 

尽管我尝试了几种方法，但都没有找到解决方法。

非常感谢任何建议或意见。

谢谢！

Ps：虽然这不是绝对必要的，但如果可以将百分位数作为新列添加到数据框中，那就太好了。

Answer 1

    df %>%
group_by(year) %>%
summarize(species.95 = quantile(species, 0.95)

我无法下载您的数据框，但您可以使用分位数函数找到每个物种的 95%。

Answer 2

如果我猜对了

library(tidyverse)                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                            "collector")), skip = 1L), class = "col_spec"))
df %>% 
  group_by(year, name) %>% 
  mutate(q95 = quantile(length, probs = 0.95))

或

library(data.table)
setDT(df)
df[, q95 := quantile(length, probs = 0.95), by = list(year, name)][order(name, year)]