如何在R语言DataFrame中按年份和物种计算95分位数并添加为新列

Hey there! Let's get this sorted for you. You need to compute the 95th percentile of length for each species within each year, then append that value as a new column to your original data frame. The dplyr package (from the tidyverse) makes this straightforward and readable.

Step 1: Load Required Package #

First, make sure you have dplyr installed and loaded:

# Install if you haven't already
install.packages("dplyr")
library(dplyr)

Step 2: Prepare Your Data #

Let's assign your sample data to a data frame (I'll call it survey_data):

survey_data <- 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")

Step 3: Calculate 95th Percentile and Add to Data Frame #

Use group_by() to group your data by year and name, then mutate() to create a new column with the 95th percentile value. The quantile() function handles the calculation, and we add na.rm = TRUE to skip any missing length values (just in case your full dataset has gaps):

survey_data_with_p95 <- survey_data %>%
  group_by(year, name) %>%
  mutate(p95_length = quantile(length, probs = 0.95, na.rm = TRUE)) %>%
  ungroup()

Step 4: Check the Result #

You can view the first few rows to verify the new column is correctly added:

head(survey_data_with_p95)

Alternative: Base R Method #

If you prefer not to use tidyverse packages, you can use the ave() function in base R to achieve the same result:

survey_data$p95_length <- ave(survey_data$length, 
                             survey_data$year, 
                             survey_data$name, 
                             FUN = function(x) quantile(x, probs = 0.95, na.rm = TRUE))

Key Notes #

• The grouping is year + species (name) because you want the 95th percentile per species within each year.
• na.rm = TRUE is crucial—if your dataset has any missing length values, this ensures the calculation doesn't fail.
• The ungroup() step in the dplyr method is optional but good practice to avoid accidentally working with grouped data later on.

内容的提问来源于stack exchange，提问作者Juan Carlos