我有一个看起来像这样的data.frame：
> 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
我想通过id重塑所有四个非id变量。换句话说，我想要colnames：
surveyNum pio1 pio2 pio3 caremgmt1 caremgmt2 caremgmt3 prev1 prev2 prev3 price1 price2 price3
我可以为一个变量做到这一点：
> 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
但是，当我尝试一些时，它完全失败了：
> 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
任何提示？我可以使用base（stats）reshape命令，但我真的试图摆脱它，因为它会导致过多的手动头皮创伤......
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")
答案 0 :(得分：13)
我认为问题是ff.df尚未充分熔化。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，在投射时，投射公式右侧指示的变量决定了投射结果中出现的列。通过仅指示.id，我相信您要求cast以某种方式将所有这些值向量填充到三列--1,2和3中。将数据一直向下熔化会创建variable列，可让您指定.id和variable向量的组合应定义投射数据框的列。
（对不起，如果我是重复/迂腐！我也试图为自己解决这个问题）
答案 1 :(得分：7)
您可以使用reshape R中的base函数执行此操作。列排序有点奇怪，但可以轻松修复。
reshape(ff.df, direction = 'wide', idvar = "surveyNum", timevar = '.id')
答案 2 :(得分：4)
您可以使用dcast的开发版data.table中的 library(data.table)
ff.cast <- dcast(setDT(ff.df), surveyNum~.id,
value.var=c('pio', 'caremgmt', 'prev', 'price'))
head(ff.cast,3)
# surveyNum 1_pio 2_pio 3_pio 1_caremgmt 2_caremgmt 3_caremgmt 1_prev 2_prev
#1: 1 2 2 2 2 1 1 1 2
#2: 2 2 1 2 1 2 2 2 2
#3: 3 1 2 1 1 2 1 2 1
# 3_prev 1_price 2_price 3_price
#1: 2 2 6 3
#2: 1 1 5 5
#3: 2 2 5 2
来执行此操作，即v1.9.5
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