重塑与 tidyr 用于具有多个因变量的重复测量

Question

我有以下 10 个案例的样本数据，对两个因变量“Rapport”和“STRS”进行了三个重复测量：

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 reshape 中使用“melt”函数，在 tidyr 中使用“gather”函数，但两者都生成一列，其中变量名称为“Rapport”和“STRS”堆叠，另一列包含它们的值。我无法弄清楚如何为“Rapport”值生成单列，为“STRS”值生成另一列，以便我可以使用随机效应模型（注意：我省略了其他人口统计变量和协变量）。对于这两个功能的任何帮助将不胜感激。

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”功能获得长格式，这看起来很简单，但我不确定这样做时是否需要注意什么：

Teacherl<-reshape(TeacherW, varying = 4:9, sep = ".", idvar="SubID", direction = 'long')
View(Teacherl)

Answer 1

我很难确定这是否是您想要的，但这是一个

以df开头：

  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

Tidyrsol'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

解释：

您要求的（我认为）是收集 Rapport 和 STRS 列，但根据他们的提名链接（.1,.2,.3） .整理一下：

• unite() 将链接的变量合并为一列（形成变量one、two、three）。之后你可以

• gather() 这些列根据键值对（此处为 replicate 和 values）。最后，

• separate() 将 values 变量返回到其组成变量 Rapport 和 STRS。

注意

我认为这里适当的“整洁”数据结构是：（为了安全起见）

df %>%
  gather(key, value, -SubID,-Gender,-Age) %>% 
  separate(key, into = c("var","idx"), sep="\\.")