使用CAST重塑多个变量 [英] Reshaping several variables wide with cast

问题描述

我有一个数据框，如下所示：

> head(ff.df)
  .id pio caremgmt prev price surveyNum
1   1   2        2    1     2         1
2   1   2        1    2     1         2
3   1   1        1    2     2         3
4   1   2        2    1     5         4
5   1   1        1    1     3         5
6   1   1        2    2     4         6

我想按id重塑所有四个非id变量。换句话说，我想要连名：

surveyNum pio1 pio2 pio3 caremgmt1 caremgmt2 caremgmt3 prev1 prev2 prev3 price1 price2 price3

我可以对单个变量执行此操作：

> cast( ff.df, surveyNum~.id, value=c("pio"))
   surveyNum 1 2 3
1          1 2 2 2
2          2 2 1 2
3          3 1 2 1
4          4 2 1 1
5          5 1 2 2
6          6 1 2 1
7          7 1 1 2
8          8 2 2 1
9          9 1 1 2
10        10 1 1 1
11        11 2 2 1
12        12 1 2 2
13        13 1 1 1
14        14 2 1 1
15        15 1 2 1
16        16 2 1 2
17        17 1 2 2
18        18 2 1 2
19        19 1 2 2
20        20 2 2 2
21        21 2 1 1
22        22 1 2 1
23        23 2 1 1
24        24 2 1 2

但当我试了几次后，它就完全失败了：

> cast( ff.df, surveyNum~.id, value=c("pio","caremgmt","prev","price"))
Error in data.frame(data[, c(variables), drop = FALSE], result = data$value) : 
  arguments imply differing number of rows: 72, 0
In addition: Warning message:
In names(data) == value :
  longer object length is not a multiple of shorter object length

有什么提示吗？我可以使用base(Stats)reshape命令，但我真的在尝试摆脱它，因为它会造成太多的手动头皮创伤，因为拔头发会造成太多创伤...

ff.df <- structure(list(.id = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 3L, 3L, 
3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 
3L, 3L, 3L, 3L, 3L, 3L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 
    2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L), pio = structure(c(2L, 
2L, 1L, 2L, 1L, 1L, 1L, 2L, 1L, 1L, 2L, 1L, 1L, 2L, 1L, 2L, 1L, 
2L, 1L, 2L, 2L, 1L, 2L, 2L, 2L, 2L, 1L, 1L, 2L, 1L, 2L, 1L, 2L, 
1L, 1L, 2L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 1L, 1L, 1L, 2L, 2L, 
1L, 2L, 1L, 2L, 2L, 1L, 2L, 1L, 1L, 2L, 2L, 1L, 1L, 2L, 1L, 2L, 
1L, 2L, 2L, 1L, 2L, 1L, 1L), .Label = c("1", "2"), class = "factor"), 
    caremgmt = structure(c(2L, 1L, 1L, 2L, 1L, 2L, 2L, 1L, 1L, 
    2L, 2L, 2L, 2L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 1L, 1L, 2L, 
    1L, 2L, 1L, 2L, 1L, 1L, 2L, 2L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 
    1L, 2L, 1L, 2L, 1L, 1L, 2L, 1L, 2L, 1L, 2L, 2L, 2L, 2L, 1L, 
    1L, 2L, 1L, 2L, 1L, 1L, 1L, 1L, 2L, 1L, 2L, 2L, 2L, 1L, 1L, 
    1L, 2L, 2L), .Label = c("1", "2"), class = "factor"), prev = structure(c(1L, 
    2L, 2L, 1L, 1L, 2L, 1L, 2L, 2L, 1L, 2L, 2L, 2L, 2L, 1L, 1L, 
    1L, 2L, 2L, 1L, 2L, 1L, 1L, 1L, 2L, 1L, 2L, 2L, 1L, 1L, 1L, 
    2L, 1L, 1L, 2L, 2L, 2L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 1L, 1L, 
    2L, 2L, 2L, 2L, 1L, 2L, 1L, 1L, 2L, 1L, 1L, 1L, 2L, 1L, 2L, 
    1L, 2L, 1L, 1L, 1L, 2L, 2L, 1L, 2L, 2L, 2L), .Label = c("1", 
    "2"), class = "factor"), price = structure(c(2L, 1L, 2L, 
    5L, 3L, 4L, 1L, 5L, 4L, 3L, 1L, 2L, 6L, 6L, 5L, 4L, 6L, 3L, 
    5L, 6L, 3L, 1L, 2L, 4L, 3L, 5L, 2L, 5L, 4L, 5L, 6L, 6L, 4L, 
    6L, 4L, 1L, 2L, 3L, 1L, 2L, 2L, 5L, 1L, 6L, 1L, 3L, 4L, 3L, 
    6L, 5L, 5L, 4L, 4L, 2L, 2L, 2L, 6L, 3L, 1L, 4L, 4L, 5L, 1L, 
    3L, 6L, 1L, 3L, 5L, 1L, 3L, 6L, 2L), .Label = c("1", "2", 
    "3", "4", "5", "6"), class = "factor"), surveyNum = c(1L, 
    2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 11L, 12L, 13L, 14L, 
    15L, 16L, 17L, 18L, 19L, 20L, 21L, 22L, 23L, 24L, 1L, 2L, 
    3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 11L, 12L, 13L, 14L, 15L, 
    16L, 17L, 18L, 19L, 20L, 21L, 22L, 23L, 24L, 1L, 2L, 3L, 
    4L, 5L, 6L, 7L, 8L, 9L, 10L, 11L, 12L, 13L, 14L, 15L, 16L, 
    17L, 18L, 19L, 20L, 21L, 22L, 23L, 24L)), .Names = c(".id", 
"pio", "caremgmt", "prev", "price", "surveyNum"), row.names = c(NA, 
-72L), class = "data.frame")

推荐答案

我认为问题是ff.df还没有充分熔化。试试这个：

library(reshape)

# Melt it down
ff.melt <- melt(ff.df, id.var = c("surveyNum", ".id"))

# Note the new "variable" column, which will be combined
# with .id to make each column header
head(ff.melt)

  surveyNum .id variable value
1         1   1      pio     2
2         2   1      pio     2
3         3   1      pio     1
4         4   1      pio     2
5         5   1      pio     1
6         6   1      pio     1

# Cast it out - note that .id comes after variable in the formula;
# I think the only effect of that is that you get "pio_1" instead of "1_pio"
ff.cast <- cast(ff.melt, surveyNum ~ variable + .id)

head(ff.cast)

  surveyNum pio_1 pio_2 pio_3 caremgmt_1 caremgmt_2 caremgmt_3 prev_1 prev_2 prev_3 price_1 price_2 price_3
1         1     2     2     2          2          1          1      1      2      2       2       6       3
2         2     2     1     2          1          2          2      2      2      1       1       5       5
3         3     1     2     1          1          2          1      2      1      2       2       5       2
4         4     2     1     1          2          2          2      1      2      2       5       4       5
5         5     1     2     2          1          2          1      1      1      1       3       4       4
6         6     1     2     1          2          1          1      2      1      1       4       2       5

这对你有用吗？

基本上，在强制转换时，转换公式右侧指示的变量决定了将出现在强制转换结果中的列。通过仅指示.id，我相信您是在要求cast以某种方式将所有这些值的矢量塞进三列-1、2和3。一直向下熔化数据将创建variable列，该列允许您指定.id和variable矢量的组合应该定义强制转换数据框的列。

(如果我重复/迂腐，很抱歉！我也在试着自己解决这个问题)

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