使用多个因变量重复测量重塑与tidyr

I have the following sample data of 10 cases with three repeated measures for two dependent variables "Rapport" and "STRS":

我有以下10个案例的样本数据,其中有两个因变量“Rapport”和“STRS”的三个重复测量:

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")

I tried to use the "melt" function in reshape and the "gather" function in tidyr but both produce one column with the variable names "Rapport" and "STRS" stacked and another column with their values. I haven't been able to figure out how to produce a single column for the "Rapport" values and another column for the "STRS" values so that I can use a random effects model (note:I left out the other demograpic variables and covariates). Any help with these two functions would be much appreciated.

我试图在重塑中使用“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”,而另一列具有其值。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)。任何有关这两个功能的帮助将不胜感激。

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)

I was finally able to obtain the longform using this "reshape" function, which seems quite simple but I'm not sure if there's anything I need to be aware of when doing it this way:

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)

1 个解决方案

#1

I'ts difficult to be sure if this is what you want but here's a

我很难确定这是否是你想要的,但这里是一个

Starting with df:

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 "_".

which gives:

   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

explanation:

What you are asking for (i think) is to gather both Rapport and STRS cols but linked according to their nominations (.1,.2,.3). To tidy this up you:

你要求的(我认为)是收集Rapport和STRS cols,但根据他们的提名(.1,.2,.3)进行链接。为了整理你:

• unite() the linked variables together into one column for each (forming varibles one, two, three). After this you can

  unite()将链接变量一起合并为一列(形成一个,两个,三个变量)。在此之后你可以

• gather() these columns according to a key-value pair (here replicate and values). Lastly,

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

• separate() the values variable back into its constituent variables Rapport and STRS.

将values变量()变回其组成变量Rapport和STRS。

N.B.

I think the appropriate "tidy" data structure here would be: (just to be safe)

我认为这里适当的“整洁”数据结构将是:(只是为了安全)

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