需要帮助重塑R上的数据

Question

我正在寻找一种重塑这种数据集的方法：

mydata<-data.frame(var=rep(c("A","B","C"),each=3),code=rep(c("x","y","z"),3),yearA=1:9,yearB=10:18,yearC=20:28)

例如：

var code    yearA   yearB   yearC
A    x      1       10      20
A    y      2       11      21
A    z      3       12      22
B    x      4       13      23
B    y      5       14      24
B    z      6       15      25
C    x      7       16      26
C    y      8       17      27
C    z      9       18      28

到此：

code    year    var.A    var.B    var.C
x       yearA    1       4        7
x       yearB    10      13       16
x       yearC    20      23       26
y       yearA    4       5        8
y       yearB    13      14       17
y       yearC    23      24       27
z       yearA    3       6        9
z       yearB    12      15       18
z       yearC    22      25       28

我尝试融化，重塑。。但结果不是我想要的。知道吗？

thks

Answer 1

library(reshape2)
mydata.melt <- melt(mydata)
mydata.dcast <- dcast(mydata.melt, code+variable~var)
mydata.dcast

使用重新整形2进行回答。

Answer 2

这是Hadley新的tidyr(重塑2的替代和部分替代品(包的另一个解决方案

library("tidyr")
library("dplyr")
mydata <- data.frame(var=rep(c("A","B","C"), each=3), 
                     code=rep(c("x","y","z"),3),
                     yearA=1:9, yearB=10:18, yearC=20:28)
mydata %>% 
  gather(year, value, yearA:yearC) %>%
  mutate(var = paste0("var", ".", var)) %>%
  spread(var, value)    

Answer 3

答案与Jot eN几乎相同，但在显示recast可以做什么方面略有不同。请记住，通常不会做melt();recast()，因为后者包含前者。

 >mfoo<-melt(mydata)
 >mfoo
   var code variable value
1    A    x    yearA     1
2    A    y    yearA     2
3    A    z    yearA     3
4    B    x    yearA     4
5    B    y    yearA     5
6    B    z    yearA     6
7    C    x    yearA     7
8    C    y    yearA     8
9    C    z    yearA     9
10   A    x    yearB    10
11   A    y    yearB    11
12   A    z    yearB    12
13   B    x    yearB    13
14   B    y    yearB    14
15   B    z    yearB    15
16   C    x    yearB    16
17   C    y    yearB    17
18   C    z    yearB    18
19   A    x    yearC    20
20   A    y    yearC    21
21   A    z    yearC    22
22   B    x    yearC    23
23   B    y    yearC    24
24   B    z    yearC    25
25   C    x    yearC    26
26   C    y    yearC    27
27   C    z    yearC    28
>recast(mfoo,code+variable~var)
Using var, code, variable as id variables
$data
      [,1] [,2] [,3]
 [1,]    1    4    7
 [2,]   10   13   16
 [3,]   20   23   26
 [4,]    2    5    8
 [5,]   11   14   17
 [6,]   21   24   27
 [7,]    3    6    9
 [8,]   12   15   18
 [9,]   22   25   28
$labels
$labels[[1]]
  code variable
1    x    yearA
2    x    yearB
3    x    yearC
4    y    yearA
5    y    yearB
6    y    yearC
7    z    yearA
8    z    yearB
9    z    yearC
$labels[[2]]
  var
1   A
2   B
3   C

所以你所要做的就是cbind，前两个列表元素。不要气馁：melt和recast需要一些时间才能适应。我总是要重新教自己如何组织formula以获得所需的输出。

Answer 4

这是笨重的base R重塑：

d <- read.table(text='var code    yearA   yearB   yearC
A    x      1       10      20
A    y      2       11      21
A    z      3       12      22
B    x      4       13      23
B    y      5       14      24
B    z      6       15      25
C    x      7       16      26
C    y      8       17      27
C    z      9       18      28', header=TRUE, stringsAsFactors=FALSE)

long <- reshape(d, dir='long', varying=list(3:5), idvar=c('code', 'var'), 
                timevar='year', v.names='v', times=c('A', 'B', 'C'))
reshape(long, idvar=c('code', 'year'), timevar='var')
#       code year v.A v.B v.C
# x.A.A    x    A   1   4   7
# y.A.A    y    A   2   5   8
# z.A.A    z    A   3   6   9
# x.A.B    x    B  10  13  16
# y.A.B    y    B  11  14  17
# z.A.B    z    B  12  15  18
# x.A.C    x    C  20  23  26
# y.A.C    y    C  21  24  27
# z.A.C    z    C  22  25  28