lapplyの呼び出しでプロットタイトルとしてリスト名を追加するR

Question

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lapply呼び出しでプロットタイトルとして表示されるリストの名前を取得するにはどうすればよいですか（各プロットごとに1つのタイトル）？

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

なぜmain = names [x]が機能しないのか分かりません。 を表示するには

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

はあなたの名前を取得されていませんなぜ、実行lapply(afn, function(x) names(x))はあなたに名それぞれのを与える：代わりに、リスト項目の名前に

Answer 1

使用lapplyリスト項目などlapply(names(afn), function(x) x)と同じように試して、その違いを比較してください。次に、[[を使用して単一のリスト項目を名前で抽出できることを忘れないでください。