我有以下10个案例的样本数据，其中三个重复测量两个因变量“Rapport”和“STRS”：
structure(list(SubID = structure(1:10, .Label = c("1", "2", "3",
"4", "5", "6", "7", "8", "9", "10", "11", "12", "13", "14", "15",
"16", "17", "18", "19", "20", "21", "22", "23", "24", "25", "26",
"27", "28", "29", "30", "31", "32", "33", "34", "35", "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", "62", "63", "64", "65", "66", "67", "68", "69", "70",
"71", "72", "73", "74", "75", "76", "77", "78", "79", "80", "81",
"82", "83", "84"), class = "factor"), Gender = structure(c(3L,
2L, 3L, 2L, 3L, 2L, 3L, 2L, 3L, 2L), .Label = c("#NULL!", "1",
"2"), class = "factor"), Age = structure(c(5L, 3L, 2L, 2L, 3L,
5L, 5L, 2L, 2L, 3L), .Label = c("#NULL!", "10", "11", "8", "9"
), class = "factor"), Rapport.1 = structure(c(22L, 25L, 19L,
10L, 18L, 19L, 20L, 20L, 21L, 16L), .Label = c("#NULL!", "1.1",
"1.85", "2.45", "2.5", "2.55", "2.6", "2.75", "2.8", "2.85",
"2.9", "2.95", "3.2", "3.25", "3.3", "3.35", "3.4", "3.45", "3.5",
"3.55", "3.6", "3.65", "3.7", "3.75", "3.8", "3.85", "3.9", "3.95"
), class = "factor"), Rapport.2 = structure(c(29L, 31L, 27L,
17L, 9L, 26L, 24L, 21L, 30L, 32L), .Label = c("#NULL!", "1.25",
"1.4", "1.6", "1.95", "2.05", "2.3", "2.35", "2.45", "2.5", "2.65",
"2.7", "2.75", "2.8", "2.85", "3", "3.05", "3.1", "3.15", "3.2",
"3.35", "3.4", "3.45", "3.5", "3.55", "3.6", "3.65", "3.7", "3.75",
"3.8", "3.85", "3.9", "3.95", "4"), class = "factor"), Rapport.3 = structure(c(32L,
35L, 22L, 22L, 5L, 25L, 30L, 21L, 25L, 34L), .Label = c("#NULL!",
"1.35", "1.45", "1.6", "1.75", "1.85", "1.9", "1.95", "2.05",
"2.1", "2.25", "2.3", "2.35", "2.4", "2.45", "2.6", "2.75", "2.8",
"2.9", "2.95", "3", "3.05", "3.1", "3.2", "3.25", "3.3", "3.35",
"3.4", "3.45", "3.5", "3.55", "3.6", "3.7", "3.75", "3.8", "3.85"
), class = "factor"), STRS.1 = structure(c(33L, 10L, 8L, 18L,
29L, 22L, 7L, 28L, 37L, 26L), .Label = c("#NULL!", "100", "102",
"103", "104", "106", "107", "108", "109", "110", "111", "112",
"113", "114", "115", "116", "117", "118", "119", "120", "122",
"123", "124", "125", "126", "127", "128", "129", "132", "133",
"69", "71", "73", "85", "88", "89", "92", "97", "99"), class = "factor"),
STRS.2 = structure(c(37L, 19L, 9L, 22L, 21L, 22L, 16L, 16L,
42L, 31L), .Label = c("#NULL!", "100", "101", "103", "104",
"105", "106", "107", "108", "110", "111", "113", "114", "115",
"116", "117", "118", "119", "120", "121", "122", "123", "124",
"125", "126", "127", "128", "129", "131", "132", "136", "137",
"138", "139", "158", "63", "76", "80", "91", "94", "95",
"98", "99"), class = "factor"), STRS.3 = structure(c(31L,
11L, 19L, 23L, 22L, 13L, 17L, 17L, 34L, 29L), .Label = c("#NULL!",
"102", "104", "105", "106", "107", "108", "109", "110", "111",
"112", "114", "117", "118", "119", "120", "122", "123", "124",
"125", "126", "127", "128", "129", "130", "131", "132", "133",
"134", "135", "66", "70", "75", "81", "85", "87", "88", "94",
"98"), class = "factor")), .Names = c("SubID", "Gender",
"Age", "Rapport.1", "Rapport.2", "Rapport.3", "STRS.1", "STRS.2",
"STRS.3"), row.names = c(NA, 10L), class = "data.frame")
我尝试在重塑中使用“融化”功能，在tidyr中使用“聚集”功能，但两者都产生一个列，其中变量名称为“Rapport”和“STRS”堆叠，另一列具有其值。我无法弄清楚如何为“Rapport”值生成单个列，为“STRS”值生成另一列，以便我可以使用随机效应模型（注意：我遗漏了其他人口统计变量和协变量）。任何对这两个功能的帮助都将非常感激。
teachermelt <- melt(TeacherW,
id.vars=c("SubID", "Gender","Age"),
measure.vars=c("Rapport.1", "Rapport.2", "Rapport.3", "STRS.1","STRS.2","STRS.3" ),
variable.name="Rapport","STRS",
value.name="Rapport","STRS)
teachertidy <- gather(TeacherW, Rapport, STRS, Rapport.1:STRS.3)
我终于能够使用这个“重塑”功能获得longform，这看起来很简单，但我不确定在这样做时我是否需要注意什么：
Teacherl<-reshape(TeacherW, varying = 4:9, sep = ".", idvar="SubID", direction = 'long')
View(Teacherl)
答案 0 :(得分：1)
我很难确定这是否是你想要的，但这是一个
从df开始：
SubID Gender Age Rapport.1 Rapport.2 Rapport.3 STRS.1 STRS.2 STRS.3
1 1 2 9 3.65 3.75 3.6 73 76 66
2 2 1 11 3.8 3.85 3.8 110 120 112
3 3 2 10 3.5 3.65 3.05 108 108 124
4 4 1 10 2.85 3.05 3.05 118 123 128
5 5 2 11 3.45 2.45 1.75 132 122 127
Tidyr sol'n：library(dplyr)
library(tidyr)
df %>%
unite(one,contains("1")) %>% # unite all columns that contain '1' with default sep = "_" into single new column named "one"
unite(two, contains("2")) %>%
unite(three, contains("3")) %>%
gather(replicate,values,one:three) %>% # gather all columns between that named "one" and that named "three" (inclusive) into two new columns: a key column (named "replicate") and a value column (named "values")
separate(values,c("Rapport","STRS"),sep = "_") # separate the column named "values" into two new columns named "Rapport" and "STRS" according to the separator "_".
给出：
SubID Gender Age replicate Rapport STRS
1 1 2 9 one 3.65 73
2 2 1 11 one 3.8 110
3 3 2 10 one 3.5 108
4 4 1 10 one 2.85 118
5 5 2 11 one 3.45 132
6 1 2 9 two 3.75 76
7 2 1 11 two 3.85 120
8 3 2 10 two 3.65 108
9 4 1 10 two 3.05 123
10 5 2 11 two 2.45 122
11 1 2 9 three 3.6 66
12 2 1 11 three 3.8 112
13 3 2 10 three 3.05 124
14 4 1 10 three 3.05 128
15 5 2 11 three 1.75 127
你要求的是什么（我认为）是收集Rapport和STRS cols，但是根据他们的提名（.1，{{ 1}}，.2）。为了整理你：
.3将链接变量合并为一列（形成变量unite()，one，two）。在此之后你可以
three和gather()） replicate这些列。最后，
values separate()变量返回其组成变量values和Rapport。我认为这里适当的“整洁”数据结构将是:(只是为了安全）
STRS