將相同數據幀的列表融合到 R 中的一個 dataframe

Question

我有九個相同數據框的waterlevels列表，其中包含有關壓力計 K01 到 K09 的信息。 這些數據幀中的每一個都包含 96860 行，它們都有 13 個變量，除了一個有 21 個變量（這些額外的列並不重要）。

$K01
           dateandtime seconds pressure_kPa temperature_.C baropressure_kpa barocompensation_kpa waterlevel_cm lengthcables_cm
1  2019-11-05 00:00:00       0      111.482         13.261           98.854               12.628      128.7697             490
2  2019-11-05 00:15:00     900      111.506         13.261           98.883               12.623      128.7188             490
3  2019-11-05 00:30:00    1800      111.511         13.261           98.872               12.639      128.8819             490
4  2019-11-05 00:45:00    2700      111.544         13.261           98.898               12.646      128.9533             490
5  2019-11-05 01:00:00    3600      111.536         13.313           98.913               12.623      128.7188             490 

我想使用以下代碼將此列表融合為一個 dataframe 僅包含一個變量barocompensation_kpa和dateandtime變量：

waterlevels_all <- melt(waterlevels, id.vars=c("dateandtime", "barocompensation_kpa"))

但隨后生成的 dataframe waterlevels_all有 10364020 個觀測值，而只有這個 df 應該有 871740 行 (9*96860)。 所以它必須有 3 個變量： dateandtime 、 barocompensation_kpa和L1是相應 dataframe 的名稱（= 滲壓計的名稱 - K01 到 K09）。

> head(waterlevels_all)
        dateandtime barocompensation_kpa variable value  L1
1 2019-11-05 00:00:00         17.55336  seconds     0 K01
2 2019-11-05 00:15:00         17.55489  seconds   900 K01
3 2019-11-05 00:30:00         17.55703  seconds  1800 K01
4 2019-11-05 00:45:00         17.55347  seconds  2700 K01
5 2019-11-05 01:00:00         17.55540  seconds  3600 K01
6 2019-11-05 01:15:00         17.56050  seconds  4500 K01

這怎么可能？ 如何獲得正確的 dataframe？

Answer 1

以下是tidyverse的一些選項 - 使用imap循環list ， select感興趣的列，並使用列表元素名稱（ .y ）創建新列“L1”

library(dplyr)
library(purrr)
imap_dfr(waterlevels, ~ .x %>% 
   select(dateandtime, barocompensation_kpa) %>%
   mutate(L1 = .y))

或者使用map ，只需select列並使用.id創建新列（ _dfr通過綁定列表元素返回單個數據集）

map_dfr(waterlevels, ~ .x %>% 
     select(dateandtime, barocompensation_kpa), .id = "L1")

Answer 2

嘗試lapply與隨后的Map和rbind 。

lapply(lst, subset, select=c('dateandtime', 'barocompensation_kpa')) |>
  Map(f=`[<-`, 'L1', value=names(lst)) |>
  # c(make.row.names=FALSE) |>  ## optional, to reset row names
  do.call(what=rbind)
#       dateandtime barocompensation_kpa  L1
# K01.1           0                    0 K01
# K01.2           0                    0 K01
# K01.3           0                    0 K01
# K02.1           0                    0 K02
# K02.2           0                    0 K02
# K02.3           0                    0 K02
# K03.1           0                    0 K03
# K03.2           0                    0 K03
# K03.3           0                    0 K03

---

數據：

lst <- list(K01 = structure(list(dateandtime = c(0, 0, 0), seconds = c(0, 
0, 0), pressure_kPa = c(0, 0, 0), temperature_.C = c(0, 0, 0), 
    baropressure_kpa = c(0, 0, 0), barocompensation_kpa = c(0, 
    0, 0), waterlevel_cm = c(0, 0, 0), lengthcables_cm = c(0, 
    0, 0)), row.names = c(NA, -3L), class = "data.frame"), K02 = structure(list(
    dateandtime = c(0, 0, 0), seconds = c(0, 0, 0), pressure_kPa = c(0, 
    0, 0), temperature_.C = c(0, 0, 0), baropressure_kpa = c(0, 
    0, 0), barocompensation_kpa = c(0, 0, 0), waterlevel_cm = c(0, 
    0, 0), lengthcables_cm = c(0, 0, 0)), row.names = c(NA, -3L
), class = "data.frame"), K03 = structure(list(dateandtime = c(0, 
0, 0), seconds = c(0, 0, 0), pressure_kPa = c(0, 0, 0), temperature_.C = c(0, 
0, 0), baropressure_kpa = c(0, 0, 0), barocompensation_kpa = c(0, 
0, 0), waterlevel_cm = c(0, 0, 0), lengthcables_cm = c(0, 0, 
0), X9 = c(0, 0, 0), X10 = c(0, 0, 0), X11 = c(0, 0, 0), X12 = c(0, 
0, 0), X13 = c(0, 0, 0), X14 = c(0, 0, 0), X15 = c(0, 0, 0), 
    X16 = c(0, 0, 0), X17 = c(0, 0, 0), X18 = c(0, 0, 0), X19 = c(0, 
    0, 0), X20 = c(0, 0, 0), X21 = c(0, 0, 0)), class = "data.frame", row.names = c(NA, 
-3L)))