我有一些看起来像这样的数据
[[1]]
[1] "" "2019" "2020" "2021" "2022" "2023"
[[2]]
[1] "" "2019" "2020" "2021" "2022" "2023"
[[3]]
[1] "" "2020" "2021" "2022" "2023"
和
[[1]]
X1 X2 X3 X4 X5 X6
1 Ventas Cliente 98716127 1477902 1970536 2955805 4433708
2 Revenue\n€ 987,161 1,477,902 1,970,536 2,955,805 4,433,708
3 Expenditure\n€ 890,623 1,337,031 1,732,873 2,569,122 3,829,434
4 Profit (Loss)\n€ 96,538 140,871 237,663 386,683 604,274
[[2]]
X1 X2 X3 X4 X5 X6
1 Number of registers that pay 0 5 200 2000 10000
2 Revenue\n€ 0 0 13,000 300,000 1,700,000
3 Expenditure\n€ 0 500 22,000 500,000 2,000,000
4 Profit (Loss)\n€ 0 -500 -9,000 -200,000 -300,000
[[3]]
X1 X2 X3 X4 X5
1 Ventas espacios publicitarios -- -- -- --
2 Revenue\n€ 2,000 150,000 1,200,000 5,000,000
3 Expenditure\n€ 5,500 250,000 600,000 1,500,000
4 Profit (Loss)\n€ -3,500 -100,000 600,000 3,500,000
我想通过列表myNames重命名每个列表数据框的列。
预期的输出（对于第一个列表项）：
"" "2019" "2020" "2021" "2022" "2023"
1 Ventas Cliente 98716127 1477902 1970536 2955805 4433708
2 Revenue\n€ 987,161 1,477,902 1,970,536 2,955,805 4,433,708
3 Expenditure\n€ 890,623 1,337,031 1,732,873 2,569,122 3,829,434
4 Profit (Loss)\n€ 96,538 140,871 237,663 386,683 604,274
此数据框列表现在具有myNames中相应的列名。
数据：
列名：
myNames <- list(c("", "2019", "2020", "2021", "2022", "2023"), c("", "2019",
"2020", "2021", "2022", "2023"), c("", "2020", "2021", "2022",
"2023"), c("", "2021", "2022", "2023"), c("", "2017", "2018",
"2019", "2020", "2021", "2022", "2023"), c("", "2019", "2020",
"2021", "2022", "2023"))
表格：
myTables <- list(structure(list(X1 = c("Ventas Cliente", "Revenue\n\200",
"Expenditure\n\200", "Profit (Loss)\n\200"), X2 = c("98716127",
"987,161", "890,623", "96,538"), X3 = c("1477902", "1,477,902",
"1,337,031", "140,871"), X4 = c("1970536", "1,970,536", "1,732,873",
"237,663"), X5 = c("2955805", "2,955,805", "2,569,122", "386,683"
), X6 = c("4433708", "4,433,708", "3,829,434", "604,274")), class = "data.frame", row.names = c(NA,
-4L)), structure(list(X1 = c("Number of registers that pay",
"Revenue\n\200", "Expenditure\n\200", "Profit (Loss)\n\200"),
X2 = c(0L, 0L, 0L, 0L), X3 = c(5L, 0L, 500L, -500L), X4 = c("200",
"13,000", "22,000", "-9,000"), X5 = c("2000", "300,000",
"500,000", "-200,000"), X6 = c("10000", "1,700,000", "2,000,000",
"-300,000")), class = "data.frame", row.names = c(NA, -4L
)), structure(list(X1 = c("Ventas espacios publicitarios", "Revenue\n\200",
"Expenditure\n\200", "Profit (Loss)\n\200"), X2 = c("--", "2,000",
"5,500", "-3,500"), X3 = c("--", "150,000", "250,000", "-100,000"
), X4 = c("--", "1,200,000", "600,000", "600,000"), X5 = c("--",
"5,000,000", "1,500,000", "3,500,000")), class = "data.frame", row.names = c(NA,
-4L)), structure(list(X1 = c("Acuerdos I+D, licencias y royalties",
"Revenue\n\200", "Expenditure\n\200", "Profit (Loss)\n\200"),
X2 = c("--", "720,013", "711,153", "8,860"), X3 = c("--",
"3,897,696", "1,043,255", "2,854,441"), X4 = c("--", "8,752,244",
"1,063,784", "7,688,460")), class = "data.frame", row.names = c(NA,
-4L)), structure(list(X1 = c("0", "Revenue\n$", "Expenditure\n$",
"Profit (Loss)\n$"), X2 = c(0L, 0L, 0L, 0L), X3 = c(0L, 0L, 0L,
0L), X4 = c(0L, 0L, 0L, 0L), X5 = c("1", "116,288", "73,920",
"42,368"), X6 = c("2", "1,153,119", "96,096", "1,057,023"), X7 = c("3",
"10,397,610", "124,925", "10,272,685"), X8 = c("4", "92,822,419",
"162,403", "92,660,016")), class = "data.frame", row.names = c(NA,
-4L)), structure(list(X1 = c("end-to-end integration", "Revenue\n\200",
"Expenditure\n\200", "Profit (Loss)\n\200"), X2 = c(0L, 0L, 0L,
0L), X3 = c("8", "20,000", "37,000", "-17,000"), X4 = c("22",
"431,000", "451,000", "-20,000"), X5 = c("37", "927,000", "523,000",
"404,000"), X6 = c("52", "1,386,000", "630,000", "756,000")), class = "data.frame", row.names = c(NA,
-4L)))
答案 0 :(得分：3)
您可以在setNames内尝试Map
Map(function(x, y) setNames(x,y), myTables, myNames)
或更简单的一个（感谢@jogo）
Map(setNames, myTables, myNames)
答案 1 :(得分：2)
您可以使用Map：
Map(function(x, y) {names(x) <- y;x}, myTables, myNames)
与purrr的map2类似：
purrr::map2(myTables, myNames, ~{names(.x) <- .y;.x})
答案 2 :(得分：2)
您还可以使用：
#Code
New <- mapply(function(x,y) setNames(x,y),myTables,myNames)
输出：
[[1]]
2019 2020 2021 2022 2023
1 Ventas Cliente 98716127 1477902 1970536 2955805 4433708
2 Revenue\n€ 987,161 1,477,902 1,970,536 2,955,805 4,433,708
3 Expenditure\n€ 890,623 1,337,031 1,732,873 2,569,122 3,829,434
4 Profit (Loss)\n€ 96,538 140,871 237,663 386,683 604,274
[[2]]
2019 2020 2021 2022 2023
1 Number of registers that pay 0 5 200 2000 10000
2 Revenue\n€ 0 0 13,000 300,000 1,700,000
3 Expenditure\n€ 0 500 22,000 500,000 2,000,000
4 Profit (Loss)\n€ 0 -500 -9,000 -200,000 -300,000
[[3]]
2020 2021 2022 2023
1 Ventas espacios publicitarios -- -- -- --
2 Revenue\n€ 2,000 150,000 1,200,000 5,000,000
3 Expenditure\n€ 5,500 250,000 600,000 1,500,000
4 Profit (Loss)\n€ -3,500 -100,000 600,000 3,500,000
[[4]]
2021 2022 2023
1 Acuerdos I+D, licencias y royalties -- -- --
2 Revenue\n€ 720,013 3,897,696 8,752,244
3 Expenditure\n€ 711,153 1,043,255 1,063,784
4 Profit (Loss)\n€ 8,860 2,854,441 7,688,460
[[5]]
2017 2018 2019 2020 2021 2022 2023
1 0 0 0 0 1 2 3 4
2 Revenue\n$ 0 0 0 116,288 1,153,119 10,397,610 92,822,419
3 Expenditure\n$ 0 0 0 73,920 96,096 124,925 162,403
4 Profit (Loss)\n$ 0 0 0 42,368 1,057,023 10,272,685 92,660,016
[[6]]
2019 2020 2021 2022 2023
1 end-to-end integration 0 8 22 37 52
2 Revenue\n€ 0 20,000 431,000 927,000 1,386,000
3 Expenditure\n€ 0 37,000 451,000 523,000 630,000
4 Profit (Loss)\n€ 0 -17,000 -20,000 404,000 756,000
答案 3 :(得分：2)
我们可以使用lapply
lapply(seq_along(myTables), function(i) setNames(myTables[[i]], myNames[[i]]))