R 中從多個數據框中提取相同的列並將它們綁定到一個新的數據框中

Question

我正在尋找一行代碼，它允許我從多個數據框中提取同名的列並將它們綁定到一個 dataframe 中。我還希望在 dataframe 之后命名新的 dataframe 中的每一列來自。

下面是我一直在使用可重現數據的代碼。 我一直在嘗試 do.call 但是我無法讓它工作：

Asset   <- structure(c(63.281303433027, 63.3979720475464, 63.6714334032718, 
            62.9559893597375, 63.0078420773017, 62.8893215800121, 31.6989860237732, 
            31.8357167016359, 31.4779946798687, 31.5039210386508, 31.4446607900061, 
            31.0492838185792, 63.3979720475464, 63.6714334032718, 62.9559893597375, 
            63.0078420773017, 62.8893215800121, 62.0985676371584), 
            class = c("xts","zoo"), .indexCLASS = "Date", .indexTZ = "UTC", tclass = "Date", tzone = "UTC", 
            index = structure(c(1550534400, 1550620800, 1550707200, 1550793600, 1551052800, 1551139200),tzone = "UTC", tclass = "Date"), .Dim = c(6L, 3L), 
            .Dimnames = list(NULL, c("Beginning.Value", "Unit.Price", "Ending.Value")))

Register<- structure(c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 212.156319855224, 
            213.718845942538, 211.63547782612, 211.809091835821, 211.63547782612, 
            207.989583622389),
            class = c("xts", "zoo"), .indexCLASS = "Date", .indexTZ = "UTC", tclass = "Date", tzone = "UTC", 
            index = structure(c(1550534400,1550620800, 1550707200, 1550793600, 1551052800, 1551139200), tzone = "UTC", tclass = "Date"), .Dim = c(6L, 3L), 
            .Dimnames = list(NULL, c("Amount", "Taxes", "Ending.Value")))

Ledger<- structure(c(0.994402284972246, 1.00685740995534, 0.991497559782253, 
            1.00156143848816, 1.00071020618011, 0.995451606923588, 161.592601088027, 
            160.688051756542, 161.789955602362, 160.414346177021, 160.664823311196, 
            160.778928461638, 160.688051756542, 161.789955602362, 160.414346177021, 
            160.664823311196, 160.778928461638, 160.04764269659), class = c("xts", "zoo"), .indexCLASS = "Date", .indexTZ = "UTC", tclass = "Date", tzone = "UTC", 
            index = structure(c(1550534400, 1550620800, 1550707200, 1550793600, 1551052800, 1551139200), tzone = "UTC", tclass = "Date"), .Dim = c(6L, 3L), 
            .Dimnames = list(NULL, c("Discount_Proxy", "Beginning.Value","Ending.Value")))

dfs <- data.frame(c("Ledger","Registry","Ledger"))
names(dfs) <- "Data Frame"

Values <- do.call('cbind', list(dfs[,1]$Ending.Value))

Answer 1

一個基於 tidyverse 的想法在被問到這個問題時可能不存在，但應該可以很好地擴展。 唯一需要指定數據集名稱的地方是將數據框捆綁到一個列表中； 在返回數據幀列表的情況下（例如從目錄中讀取文件），您甚至不需要在那里進行。 tibble::lst是基本list的包裝器，它按原始 object 名稱命名值。 {{ ds }}:= Ending.Value是一種簡潔的表示法，可根據變量ds中的值動態重命名Ending.Value列，在本例中為數據集的名稱。

library(dplyr)
library(xts)

tibble::lst(Asset, Register, Ledger) %>%
  purrr::map(as.data.frame) %>%
  purrr::imap(function(dat, ds) select(dat, {{ ds }} := Ending.Value)) %>%
  bind_cols()
#>               Asset Register   Ledger
#> 2019-02-19 63.39797 212.1563 160.6881
#> 2019-02-20 63.67143 213.7188 161.7900
#> 2019-02-21 62.95599 211.6355 160.4143
#> 2019-02-22 63.00784 211.8091 160.6648
#> 2019-02-25 62.88932 211.6355 160.7789
#> 2019-02-26 62.09857 207.9896 160.0476

Answer 2

如果您不介意在列表中命名data.frames：

list_ls <- list("Asset" = Asset, "Register" = Register, "Ledger" = Ledger)

foo <- do.call(cbind, lapply(list_ls, function(x) x$Ending.Value))

test <- cbind(Asset$Ending.Value, Register$Ending.Value, Ledger$Ending.Value)
colnames(test) <- c("Asset", "Register", "Ledger")

length(which(foo != test))