R 通过收集多个列来整理数据集？

我希望通过以下方式处理数据来整理数据集：

age gender  education       previous_comp_exp   tutorial_time   qID.1    time_taken.1   qID.2    time_taken.2   
18  Male    Undergraduate   casual gamer        62.17926        sor9     39.61206       sor8     19.4892
24  Male    Undergraduate   casual gamer        85.01288        sor9     50.92343       sor8     16.15616

为此：

age gender  education       previous_comp_exp   tutorial_time   qID      time_taken 
18  Male    Undergraduate   casual gamer        62.17926        sor9     39.61206       
18  Male    Undergraduate   casual gamer        62.17926        sor8     19.4892
24  Male    Undergraduate   casual gamer        85.01288        sor9     50.92343       
24  Male    Undergraduate   casual gamer        85.01288        sor8     16.15616

我尝试过使用

gather（）

，但我只能让它在一列中工作，而且我一直收到以下警告：

警告消息：度量变量的属性不相同； 它们将被丢弃

---

有什么想法吗？

使用

数据中的
熔化
。表
（请参见
？模式
）：

结果：

   age gender     education previous_comp_exp tutorial_time variable  qID time_taken
1:  18   Male Undergraduate      casual_gamer      62.17926        1 sor9   39.61206
2:  24   Male Undergraduate      casual_gamer      85.01288        1 sor9   50.92343
3:  18   Male Undergraduate      casual_gamer      62.17926        2 sor8   19.48920
4:  24   Male Undergraduate      casual_gamer      85.01288        2 sor8   16.15616

# A tibble: 4 x 7
    age gender     education previous_comp_exp tutorial_time   qID time_taken
  <int> <fctr>        <fctr>            <fctr>         <dbl> <chr>      <dbl>
1    18   Male Undergraduate      casual_gamer      62.17926  sor9   39.61206
2    18   Male Undergraduate      casual_gamer      62.17926  sor8   19.48920
3    24   Male Undergraduate      casual_gamer      85.01288  sor9   50.92343
4    24   Male Undergraduate      casual_gamer      85.01288  sor8   16.15616

df = structure(list(age = c(18L, 24L), gender = structure(c(1L, 1L
), .Label = "Male", class = "factor"), education = structure(c(1L, 
1L), .Label = "Undergraduate", class = "factor"), previous_comp_exp = structure(c(1L, 
1L), .Label = "casual_gamer", class = "factor"), tutorial_time = c(62.17926, 
85.01288), qID.1 = structure(c(1L, 1L), .Label = "sor9", class = "factor"), 
    time_taken.1 = c(39.61206, 50.92343), qID.2 = structure(c(1L, 
    1L), .Label = "sor8", class = "factor"), time_taken.2 = c(19.4892, 
    16.15616)), .Names = c("age", "gender", "education", "previous_comp_exp", 
"tutorial_time", "qID.1", "time_taken.1", "qID.2", "time_taken.2"
), class = "data.frame", row.names = c(NA, -2L))

或使用
tidyr
：

library(dplyr)
library(tidyr)

df %>%
  gather(variable, value, qID.1:time_taken.2) %>%
  mutate(variable = sub("\\.\\d$", "", variable)) %>%
  group_by(variable) %>%
  mutate(ID = row_number()) %>%
  spread(variable, value, convert = TRUE) %>%
  select(-ID)

结果：

   age gender     education previous_comp_exp tutorial_time variable  qID time_taken
1:  18   Male Undergraduate      casual_gamer      62.17926        1 sor9   39.61206
2:  24   Male Undergraduate      casual_gamer      85.01288        1 sor9   50.92343
3:  18   Male Undergraduate      casual_gamer      62.17926        2 sor8   19.48920
4:  24   Male Undergraduate      casual_gamer      85.01288        2 sor8   16.15616

# A tibble: 4 x 7
    age gender     education previous_comp_exp tutorial_time   qID time_taken
  <int> <fctr>        <fctr>            <fctr>         <dbl> <chr>      <dbl>
1    18   Male Undergraduate      casual_gamer      62.17926  sor9   39.61206
2    18   Male Undergraduate      casual_gamer      62.17926  sor8   19.48920
3    24   Male Undergraduate      casual_gamer      85.01288  sor9   50.92343
4    24   Male Undergraduate      casual_gamer      85.01288  sor8   16.15616

df = structure(list(age = c(18L, 24L), gender = structure(c(1L, 1L
), .Label = "Male", class = "factor"), education = structure(c(1L, 
1L), .Label = "Undergraduate", class = "factor"), previous_comp_exp = structure(c(1L, 
1L), .Label = "casual_gamer", class = "factor"), tutorial_time = c(62.17926, 
85.01288), qID.1 = structure(c(1L, 1L), .Label = "sor9", class = "factor"), 
    time_taken.1 = c(39.61206, 50.92343), qID.2 = structure(c(1L, 
    1L), .Label = "sor8", class = "factor"), time_taken.2 = c(19.4892, 
    16.15616)), .Names = c("age", "gender", "education", "previous_comp_exp", 
"tutorial_time", "qID.1", "time_taken.1", "qID.2", "time_taken.2"
), class = "data.frame", row.names = c(NA, -2L))

在base R中，您可以使用功能强大的
重塑
在一行语句中将数据从宽格式转换为长格式：

   reshape(dx,direction="long",
        varying=list(grep("qID",colnames(dx)),
                     grep("time_taken",colnames(dx))),
        v.names=c("qID","time_taken"))

     age gender     education previous_comp_exp tutorial_time time  qID time_taken id
1.1  18   Male Undergraduate      casual_gamer      62.17926    1 sor9   39.61206  1
2.1  24   Male Undergraduate      casual_gamer      85.01288    1 sor9   50.92343  2
1.2  18   Male Undergraduate      casual_gamer      62.17926    2 sor8   19.48920  1
2.2  24   Male Undergraduate      casual_gamer      85.01288    2 sor8   16.15616  2

这不是一个错误，这是一个警告，让您知道堆叠的两列具有不同的属性（可能它们都是因子，但具有不同的级别），因此这些属性会被删除到输出中。考虑到你有一对需要叠放的柱子，这可能有助于从宽到长的重塑。非常好地利用了熔体@用户，在Tidy方法中，年龄变量消失了。知道为什么吗？@stenfeio谢谢你能听到！我实际上读错了数据。由于我使用的是
read.table
，
casual
和
gamer
被视为单独的列，第一列作为行名。它没有抛出错误，因为列数恰好匹配。看我的编辑我不知道图案的功能，线条很好@用户在看到您的解决方案之前，我有一个问题：您可以在
spread
中使用
convert=TRUE
，而不是稍后在
mutate
中手动设置类型。我想您也错误地读取了数据。您可以在我的回答中使用新的
dput
。@用户答对了。现在修好了。