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sample <- 
structure(list(GB05 = structure(c(22L, 34L, 26L, 2L), .Dim = 4L, .Dimnames = structure(list(
    c("98", "100", "102", "106")), .Names = ""), class = "table"), 
    GB18 = structure(c(8L, 14L, 70L), .Dim = 3L, .Dimnames = structure(list(
        c("173", "175", "177")), .Names = ""), class = "table"), 
    GB06 = structure(c(2L, 16L, 48L, 10L, 10L, 6L), .Dim = 6L, .Dimnames = structure(list(
        c("234", "238", "240", "242", "244", "246")), .Names = ""), class = "table"), 
    GB27 = structure(c(2L, 28L, 2L, 2L, 4L, 3L, 2L, 2L, 15L, 
    17L, 4L, 5L), .Dim = 12L, .Dimnames = structure(list(c("145", 
    "147", "149", "151", "156", "159", "165", "167", "169", "171", 
    "173", "175")), .Names = ""), class = "table"), GB24 = structure(c(2L, 
    4L, 41L, 10L, 6L, 2L, 14L, 2L, 3L), .Dim = 9L, .Dimnames = structure(list(
        c("240", "241", "242", "243", "244", "247", "249", "251", 
        "253")), .Names = ""), class = "table"), GB28 = structure(c(30L, 
    22L, 2L, 10L, 2L, 4L, 2L, 2L, 2L), .Dim = 9L, .Dimnames = structure(list(
        c("363", "364", "365", "367", "371", "377", "380", "384", 
        "390")), .Names = ""), class = "table"), GB15 = structure(c(12L, 
    16L, 43L, 2L, 3L, 4L), .Dim = 6L, .Dimnames = structure(list(
        c("142", "144", "146", "147", "148", "152")), .Names = ""), class = "table"), 
    GB02 = structure(c(6L, 78L, 4L), .Dim = 3L, .Dimnames = structure(list(
        c("194", "197", "200")), .Names = ""), class = "table"), 
    GB10 = structure(c(2L, 36L, 7L, 1L, 16L, 20L), .Dim = 6L, .Dimnames = structure(list(
        c("124", "126", "128", "130", "132", "134")), .Names = ""), class = "table"), 
    GB14 = structure(c(3L, 6L, 7L, 37L, 7L), .Dim = 5L, .Dimnames = structure(list(
        c("181", "184", "187", "193", "196")), .Names = ""), class = "table")), .Names = c("GB05", 
"GB18", "GB06", "GB27", "GB24", "GB28", "GB15", "GB02", "GB10", 
"GB14"))

このリスト内の名前をこのラップリー呼び出しでプロット タイトルとして表示するにはどうすればよいですか (プロットごとに 1 つのタイトル)。

dev.new()
par(mfrow=c(2,5))
lapply(sample,function(x) plot(x,main=names[x]))

main=names[x] が機能しない理由がわかりません。

4

1 に答える 1

21

lapply代わりに、リストアイテムの名前に使用してください。

lapply(names(afn), function(x) plot(afn[[x]], main=x))

ここに画像の説明を入力してください

名前が取得されない理由を確認するには、runlapply(afn, function(x) names(x))を実行すると各リストアイテムの名前などが表示されます。同じことを試してlapply(names(afn), function(x) x)、違いを比較してください。次に、を使用して名前で1つのリストアイテムを抽出できることを思い出して[[ください。

于 2013-02-09T16:52:55.383 に答える