describe我的函式有以下問題psych：我想描述資料框的選定變數，然后用 and 洗掉一些subset結果select。這似乎只適用于資料框，但我得到了一個describe類。對我來說，有時它似乎有效，有時它不起作用，我認為這實際上是不可能的。然而，確實它作業了幾次，我已經可以保存輸出，安排得很好，完全符合我想要的樣子。但現在它再次回傳該類的錯誤describe無法轉換為資料框。我看到的問題可能是我得到了一個串列串列（至少環境是這樣說的）。由于我是一個完全的編程新手，所以我無法解決這個問題，即使在搜索了如何轉換這個類之后，我也只是不明白。
Descriptives = describe(NumericData[5:44], na.rm = TRUE, interp = FALSE,
skew = TRUE, ranges = TRUE, trim = .1, type = 3,
check = TRUE, fast = NULL, quant = c(.25, .50, .75),
IQR = FALSE)
Descriptives = as.data.frame(Descriptives)
Descriptives = subset(Descriptives, select = -c(vars, median, trimmed, mad, range))
colnames(Descriptives) = c("N", "MEAN", "SD", "MIN", "MAX", "SKEW", "KURTOSIS", "SE", "Q1", "MEDIAN", "Q3")
Descriptives = round(Descriptives, digits = 4)
options(max.print = 1000)
print(as.data.frame(Descriptives))
write.table(Descriptives, file = "Descriptives.txt", sep = ",")
uj5u.com熱心網友回復：
好的，所以錯誤現在發生在每一行（第一個塊 = 第 1 行），describe 函式實際上作業正常，但是我無法選擇結果變數的子集（第 3 行），這里它說引數子集丟失并且或命名左側的列（第 4 行），這告訴我試圖為小于二維的物件設定列名。此外，round（第 5 行）回傳一個錯誤，指出不能將數值引數用于非數學函式。當我想列印之前出現在第 2 行中的錯誤“無法強制類 '“describe”' 到 data.frame”時，現在出現。它作業了一次，最近至少沒有選擇結果變數的子集，但現在沒有任何作業，我不明白為什么..代碼保持不變。
我使用的資料集的 2 行：
structure(list(Age = c(24, 23, 44, 48, 35, 56, 64, 29, 20, 62,
35, 31, 32, 60, 57, 66, 46, 18, 52, 63, 64, 35, 54, 58, 61, 52,
52, 33, 49, 28, 22, 27, 40, 53, 18, 19, 43, 44, 26, 28, 38, 18,
50, 45, 23, 38, 50, 36, 72, 62, 33, 28, 29, 42, 48, 42, 29, 70,
27, 33, 22, 62, 67, 20, 32, 22, 32, 67, 29, 55, 49, 19, 52, 20,
30, 24, 18, 24, 23, 22, 19, 20, 29, 22, 20, 19, 21, 18, 22, 22,
18, 24, 22, 24, 19, 25, 24, 25, 20, 21, 23, 39, 60, 53, 47, 48,
40, 29, 24, 27, 21, 21, 27, 22, 20, 23, 36, 22, 25, 27, 66, 54,
54, 64, 49, 40), FTND = c(5, 7, 0, 6, 0, 6, 0, NA, 3, 4, 0, 7,
NA, 0, 4, 3, 4, 1, 0, 6, 0, 5, 0, NA, NA, 3, 0, 2, NA, 0, 0,
0, NA, NA, NA, NA, NA, 4, 0, 10, NA, NA, 8, NA, 3, 7, 0, 0, 5,
2, 0, 6, 7, 0, 4, 2, 0, NA, 0, 0, 0, 0, 0, 0, 4, 0, 0, NA, 3,
NA, NA, NA, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, NA, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0)), row.names = c(NA,
-126L), class = "data.frame")
我用 describe 得到的資料集是一個串列串列，其中不包含結果變數，而是我原始資料集的所有變數。同樣通過使用 (dput(Descriptives[5:6]) 它應該列印變數 age 和 FTND 而不是 EmQ （實際上是變數/第 9 行）：
structure(list(EmQ = structure(list(descript = "EmQ", units = NULL,
format = NULL, counts = c(n = "97", missing = "29", distinct = "39",
Info = "0.998", Mean = "44.04", Gmd = "12.33", `.05` = "23.6",
`.10` = "27.6", `.25` = "38.0", `.50` = "45.0", `.75` = "50.0",
`.90` = "58.4", `.95` = "60.2"), values = list(value = c(19,
21, 22, 24, 25, 27, 28, 30, 32, 33, 34, 36, 37, 38, 39, 40,
41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55,
56, 57, 58, 59, 60, 61, 64, 66), frequency = structure(c(2,
2, 1, 2, 2, 1, 1, 2, 2, 2, 3, 1, 3, 1, 2, 2, 3, 4, 3, 4,
8, 7, 3, 6, 2, 4, 3, 1, 1, 1, 1, 4, 2, 1, 2, 3, 2, 2, 1), .Dim = 39L)),
extremes = c(L1 = 19, L2 = 21, L3 = 22, L4 = 24, L5 = 25,
H5 = 59, H4 = 60, H3 = 61, H2 = 64, H1 = 66)), class = "describe"),
EmQ10 = structure(list(descript = "EmQ10", units = NULL,
format = NULL, counts = c(n = "108", missing = "18",
distinct = "19", Info = "0.993", Mean = "10.16", Gmd = "4.346",
`.05` = "4.0", `.10` = "5.7", `.25` = "7.0", `.50` = "10.0",
`.75` = "13.0", `.90` = "15.0", `.95` = "16.0"), values = list(
value = c(2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13,
14, 15, 16, 17, 18, 19, 20), frequency = structure(c(1,
3, 4, 3, 5, 13, 10, 11, 10, 11, 5, 11, 5, 6, 5, 2,
1, 1, 1), .Dim = 19L)), extremes = c(L1 = 2, L2 = 3,
L3 = 4, L4 = 5, L5 = 6, H5 = 16, H4 = 17, H3 = 18, H2 = 19,
H1 = 20)), class = "describe")), descript = "NumericData[5:44]", dimensions = c(126L,
2L), class = "describe")
我通過 describeBy 獲得的資料。還有一個串列，串列 1（？）中只有 2 個組，即對照組和患者，還包含所需的結果變數，如修剪，中位數作為我猜的屬性）：
structure(list(NULL, NULL), .Dim = 2L, .Dimnames = list(Group = c(NA_character_,
NA_character_)))
很抱歉，很長的帖子，我不知道如何把它更好..
uj5u.com熱心網友回復：
這是我曾經得到并想再次得到的：
N MEAN SD MIN MAX SKEW KURTOSIS SE Q1 MEDIAN Q3
Age 126 36.254 15.6578 18 72 0.6067 -0.9925 1.3949 22.25 30.5 49
FTND 107 1.2617 2.3121 0 10 1.7475 2.0378 0.2235 0 0 1.5
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