使用 for 循环在 R 中创建数据框列表

Question

我有一个包含 26 个数据框的列表。我正在使用下面的代码来创建一个向量。有没有办法可以使用 for 循环对所有 26 个数据帧应用相同的唯一函数，并在每个数据帧中创建一个包含 26 个数据帧的新列表？每个数据框将根据该数据框创建不同的向量。

(unique(paste(list[[1]]$row, list[[1]]$col, sep=""))

> (unique(paste(list[[1]]$row, list[[1]]$col, sep="")))
 [1] "A1"  "A2"  "A3"  "A5"  "A6"  "A7"  "A8"  "A9"  "A10" "A11" "A12" "B1"  "B2"  "B3"  "B4"  "B5"  "B6"  "B7"  "B8"  "B9" 
[21] "B10" "B11" "B12" "C1"  "C2"  "C3"  "C4"  "C5"  "C6"  "C7"  "C8"  "C9"  "C10" "C11" "C12" "D1"  "D2"  "D3"  "D4"  "D5" 
[41] "D6"  "D7"  "D8"  "D9"  "D10" "D11" "D12" "E1"  "E2"  "E3"  "E4"  "E5"  "E6"  "E7"  "E8"  "E9"  "E10" "E11" "E12" "F1" 
[61] "F2"  "F3"  "F4"  "F5"  "F6"  "F7"  "F8"  "F9"  "F10" "F11" "F12" "G1"  "G2"  "G3"  "G4"  "G5"  "G6"  "G7"  "G8"  "G9" 
[81] "G10" "G11" "G12" "H1"  "H2"  "H3"  "H4"  "H5"  "H6"  "H7"  "H8"  "H9"  "H10" "H11" "H12"

所以我想要的结果是......

[[1]]
[1] "A1"  "A2"  "A3"  "A5"  "A6"  "A7"  "A8"  "A9"  "A10" "A11" "A12" "B1"  "B2"  "B3"  "B4"  "B5"  "B6"  "B7"  "B8"  "B9" 
[21] "B10" "B11" "B12" "C1"  "C2"  "C3"  "C4"  "C5"  "C6"  "C7"  "C8"  "C9"  "C10" "C11" "C12" "D1"  "D2"  "D3"  "D4"  "D5" 
[41] "D6"  "D7"  "D8"  "D9"  "D10" "D11" "D12" "E1"  "E2"  "E3"  "E4"  "E5"  "E6"  "E7"  "E8"  "E9"  "E10" "E11" "E12" "F1" 
[61] "F2"  "F3"  "F4"  "F5"  "F6"  "F7"  "F8"  "F9"  "F10" "F11" "F12" "G1"  "G2"  "G3"  "G4"  "G5"  "G6"  "G7"  "G8"  "G9" 
[81] "G10" "G11" "G12" "H1"  "H2"  "H3"  "H4"  "H5"  "H6"  "H7"  "H8"  "H9"  "H10" "H11" "H12"

[[2]]
[1] "A1"  "A2"  "A3"  "A5"  "A6"  "A7"  "A8"  "A9"  "A10" "A11" "A12" "B1"  "B2"  "B3"  "B4"  "B5"  "B6"  "B7"  "B8"  "B9" 
[21] "B10" "B11" "B12" "C1"  "C2"  "C3"  "C4"  "C5"  "C6"  "C7"  "C8"  "C9"  "C10" "C11" "C12" "D1"  "D2"  "D3"  "D4"  "D5" 
[41] "D6"  "D7"  "D8"  "D9"  "D10" "D11" "D12" "E1"  "E2"  "E3"  "E4"  "E5"  "E6"  "E7"  "E8"  "E9"  "E10" "E11" "E12" "F1" 
[61] "F2"  "F3"  "F4"  "F5"  "F6"  "F7"  "F8"  "F9"  "F10" "F11" "F12" "G1"  "G2"  "G3"  "G4"  "G5"  "G6"  "G7"  "G8"  "G9" 
[81] "G10" "G11" "G12" "H1"  "H2"  "H3"  "H4"  "H5"  "H6"  "H7"  "H8"  "H9"  "H10" "H11" "H12"

[[3]] 
[1] "E10" "E11" "E12" "F1"  "F2"  "F3"  "F4"  "F5"  "F6"  "F7"  "F8"  "F9"  "F10" "F11" "F12" "G1"  "G2"  "G3"  "G4"  "G5" 
[21] "G6"  "G7"  "G8"  "G9"  "G10" "G11" "G12" "H1"  "H2"  "H3"  "H4"  "H5"  "H6"  "H7"  "H8"  "H9"  "H10" "H11" "H12"

etc etc until [[26]]

澄清

每个数据框包含以下列表包含的任何或所有元素（以下列表有 96 个元素）：

list(c(paste0(rep(LETTERS[1:8], each=12), rep(1:12, 8))))

所以我需要一些能够遍历我的 26 个数据框列表并告诉我每个数据框包含的元素的东西。因为某些数据框可能没有全部 96 个元素，所以我展示了 [[3]] 在我想要的结果示例中只有 .. 40 个元素。希望现在更清楚..

Answer 1

类似：

lapply(list, function(x) unique(do.call(paste0, x)))

应该这样做。 do.call 调用它的第一个参数（在本例中为函数paste0），参数设置为do.call 的第二个参数。由于x 是一个数据框，它是一个列表，所以效果很好。这是示例输出：

[[1]]
[1] "G18" "J20" "N12" "U11" "E1" 

[[2]]
[1] "F10" "E14" "Q18" "I7"  "X13"

[[3]]
[1] "Y8"  "F1"  "P7"  "C15" "Z6" 

[[4]]
[1] "M14" "O16" "L2"  "E13" "S7" 

[[5]]
[1] "V16" "Q1"  "S9"  "M13" "L12"

以及我使用的数据：

set.seed(1)
list <- replicate(5, data.frame(row=sample(LETTERS, 5), col=sample(1:20, 5)), s=F)
lapply(list, function(x) do.call(paste0, x))

Answer 2

我不明白你的数据结构，所以请原谅我的假数据集。

您也可以在lapply 函数中使用transform。

df1 <- data.frame(a = c(1, 2, 3),
                  b = c(4, 5, 6))

df2 <- data.frame(a = c(5, 6, 7),
                  b = c(8, 9, 10))

dfList <- list(df1 = df1, df2 = df2)

lapply(dfList, transform, c = a - b)