Question

这是我的数据：

> head(data)

    id C1 C2 C3 B1 B2 B3 Name
    12 3  12 8  1  3  12 Agar
    14 4  11 9  5  12 14 LB
    18 7  17 6  7  14 16 YEF
    20 9  15 4  3  11 17 KAN

所以我使用reshape2包中的融合函数来重新组织我的数据。现在它看起来像是：

dt <- melt(data, measure.vars=2:7)

> head(dt)

   n    v variable value   rt
1 id Name        p    C1   1
2 12 Agar        p     3   2
3 14   LB        p     4   3
4 18  YEF        p     7   6
5 20  KAN        p     9   3
6 id Name        u    C2   1

我对我的数据进行了一些计算，现在还有一个额外的列。我们称之为“rt”。我现在想用一个额外的列将我的数据转换为之前的“状态”。你知道任何有用的功能吗？

dput(dt)
structure(list(n = structure(c(5L, 1L, 2L, 3L, 4L, 5L, 1L, 2L, 
3L, 4L, 5L, 1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 
4L, 5L, 1L, 2L, 3L, 4L), class = "factor", .Label = c("12", "14", 
"18", "20", "id")), v = structure(c(4L, 1L, 3L, 5L, 2L, 4L, 1L, 
3L, 5L, 2L, 4L, 1L, 3L, 5L, 2L, 4L, 1L, 3L, 5L, 2L, 4L, 1L, 3L, 
5L, 2L, 4L, 1L, 3L, 5L, 2L), class = "factor", .Label = c("Agar", 
"KAN", "LB", "Name", "YEF")), variable = structure(c(1L, 1L, 
1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 
4L, 4L, 5L, 5L, 5L, 5L, 5L, 6L, 6L, 6L, 6L, 6L), .Label = c("p", 
"u", "k", "l", "t", "h"), class = "factor"), value = c("C1", 
"3", "4", "7", "9", "C2", "12", "11", "17", "15", "C3", "8", 
"9", "6", "4", "B1", "1", "5", "7", "3", "B2", "3", "12", "14", 
"11", "B3", "12", "14", "16", "17")), .Names = c("n", "v", "variable", 
"value"), row.names = c(NA, -30L), class = "data.frame")

Answer 1

在“reshape2”世界中，melt和*cast齐头并进。

以下是melt data.frame和dcast将其恢复为原始形式的示例。您需要对数据采取类似的方法。

mydf <- data.frame(A = LETTERS[1:3], B = 1:3, C = 4:6)
mydf
#   A B C
# 1 A 1 4
# 2 B 2 5
# 3 C 3 6

library(reshape2)

mDF <- melt(mydf, id.vars="A")
mDF
dcast(mDF, A ~ variable, value.var="value")
#   A B C
# 1 A 1 4
# 2 B 2 5
# 3 C 3 6

在dcast步骤中，将~之前的项目视为“id”变量，将之后的项目视为生成的列名称。 value.var应该是值将填充由id变量和列名创建的结果“网格”的列。

如何从reshape2包中“退回”融化功能？

1 个答案: