R 对列表中的数据帧行进行子集和重新组合

这里有两种替代方法.

base R approach

迭代唯一的奶牛并提取results年中该日期的每一头相关奶牛.

(cows <- unique(unlist(lapply(results, names))))
# [1] "Cow1" "Cow2" "Cow3" "Cow4"
lapply(cows, \(cow) {
    lapply(results, \(date) {
        date[[cow]]
    })
}) |> setNames(cows)

输出(输出到jsonlite::toJSON(pretty = TRUE)，因为R以可怕的格式打印嵌套列表).

{
  "Cow1": {
    "2023-08-15 00:01:00": ["NA", 65, 2, 7],
    "2023-08-15 00:16:00": ["NA", 65, 2, 7]
  },
  "Cow2": {
    "2023-08-15 00:01:00": [65, "NA", 16, 84],
    "2023-08-15 00:16:00": [65, "NA", 16, 84]
  },
  "Cow3": {
    "2023-08-15 00:01:00": [2, 16, "NA", 8],
    "2023-08-15 00:16:00": [2, 16, "NA", 8]
  },
  "Cow4": {
    "2023-08-15 00:01:00": [7, 84, 8, "NA"],
    "2023-08-15 00:16:00": [7, 84, 8, "NA"]
  }
}

dplyr approach

前bind_rows()个你的原始列表，让你pivot_longer()你的数据.第split()章又是一个名字，又是一个日期

library(dplyr)
results |>
    bind_rows(.id = "date") |>
    tidyr::pivot_longer(-date) |>
    split(~name) |>
    lapply(\(df)
        lapply(
            split(df, ~date),
            \(x) x$value
        )
    )

输出同上.

Input data

results <- list(`2023-08-15 00:01:00` = structure(list(Cow1 = c(NA, 65, 
2, 7), Cow2 = c(65, NA, 16, 84), Cow3 = c(2, 16, NA, 8), Cow4 = c(7, 
84, 8, NA)), class = "data.frame", row.names = c(NA, -4L)), `2023-08-15 00:16:00` = structure(list(
    Cow1 = c(NA, 65, 2, 7), Cow2 = c(65, NA, 16, 84), Cow3 = c(2, 
    16, NA, 8), Cow4 = c(7, 84, 8, NA)), class = "data.frame", row.names = c(NA, -4L)))