我遇到了重塑大型数据帧的困难。我过去一直很幸运地避免重塑问题，这也意味着我很糟糕。
我当前的数据框看起来像这样：
unique_id seq response detailed.name treatment
a N1 123.23 descr. of N1 T1
a N2 231.12 descr. of N2 T1
a N3 231.23 descr. of N3 T1
...
b N1 343.23 descr. of N1 T2
b N2 281.13 descr. of N2 T2
b N3 901.23 descr. of N3 T2
...
我想：
seq detailed.name T1 T2
N1 descr. of N1 123.23 343.23
N2 descr. of N2 231.12 281.13
N3 descr. of N3 231.23 901.23
我查看了整形包，但我不确定如何将处理因子转换为单独的列名。
谢谢！
编辑：我尝试在我的本地计算机上运行此程序（4GB双核iMac 3.06Ghz）并且它始终失败：
> d.tmp.2 <- cast(d.tmp, `SEQ_ID` + `GENE_INFO` ~ treatments)
Aggregation requires fun.aggregate: length used as default
R(5751) malloc: *** mmap(size=647168) failed (error code=12)
*** error: can't allocate region
*** set a breakpoint in malloc_error_break to debug
当我有机会的时候，我会尝试在我们更大的机器上运行它。
答案 0 :(得分：18)
> x
unique_id seq response detailed.name treatment
1 a N1 123.23 dN1 T1
2 a N2 231.12 dN2 T1
3 a N3 231.23 dN3 T1
4 b N1 343.23 dN1 T2
5 b N2 281.13 dN2 T2
6 b N3 901.23 dN3 T2
> x2 <- melt(x, c("seq", "detailed.name", "treatment"), "response")
> x2
seq detailed.name treatment variable value
1 N1 dN1 T1 response 123.23
2 N2 dN2 T1 response 231.12
3 N3 dN3 T1 response 231.23
4 N1 dN1 T2 response 343.23
5 N2 dN2 T2 response 281.13
6 N3 dN3 T2 response 901.23
> cast(x2, seq + detailed.name ~ treatment)
seq detailed.name T1 T2
1 N1 dN1 123.23 343.23
2 N2 dN2 231.12 281.13
3 N3 dN3 231.23 901.23
您的原始数据已经是长格式，但不是融合/强制转换使用的长格式。所以我重新融化了。第二个参数（id.vars）是不易融化的事物列表。第三个参数（measure.vars）是变化的事物列表。
然后，演员使用公式。波浪号的左侧是保持不变的东西，波浪号的右侧是用于调节值列的列。
或多或少......！
答案 1 :(得分：6)
基于Harlan的答案 - 如果数据已经是长格式，并且在cast调用中指定了列保持值，则可以避免重熔步骤。
> x <- read.table(textConnection(" unique_id seq response detailed.name treatment
+ 1 a N1 123.23 dN1 T1
+ 2 a N2 231.12 dN2 T1
+ 3 a N3 231.23 dN3 T1
+ 4 b N1 343.23 dN1 T2
+ 5 b N2 281.13 dN2 T2
+ 6 b N3 901.23 dN3 T2"))
>
> cast(x, seq + detailed.name ~ treatment, value = "response")
seq detailed.name T1 T2
1 N1 dN1 123.23 343.23
2 N2 dN2 231.12 281.13
3 N3 dN3 231.23 901.23
答案 2 :(得分：3)
另一种选择是使用spread
tidyr
library(tidyr)
Wide1 <- spread(x[-1], treatment, response)
Wide1
# seq detailed.name T1 T2
#1 N1 dN1 123.23 343.23
#2 N2 dN2 231.12 281.13
#3 N3 dN3 231.23 901.23
相反的行动由gather
gather(Wide1, detailed.name, response, T1:T2)
# seq detailed.name detailed.name response
#1 N1 dN1 T1 123.23
#2 N2 dN2 T1 231.12
#3 N3 dN3 T1 231.23
#4 N1 dN1 T2 343.23
#5 N2 dN2 T2 281.13
#6 N3 dN3 T2 901.23
此外，来自dcast.data.table
data.table
library(data.table)
dcast.data.table(setDT(x), seq + detailed.name~treatment,
value.var='response')
# seq detailed.name T1 T2
#1: N1 dN1 123.23 343.23
#2: N2 dN2 231.12 281.13
#3: N3 dN3 231.23 901.23
x <- structure(list(unique_id = structure(c(1L, 1L, 1L, 2L, 2L, 2L
), .Label = c("a", "b"), class = "factor"), seq = structure(c(1L,
2L, 3L, 1L, 2L, 3L), .Label = c("N1", "N2", "N3"), class = "factor"),
response = c(123.23, 231.12, 231.23, 343.23, 281.13, 901.23
), detailed.name = structure(c(1L, 2L, 3L, 1L, 2L, 3L), .Label = c("dN1",
"dN2", "dN3"), class = "factor"), treatment = structure(c(1L,
1L, 1L, 2L, 2L, 2L), .Label = c("T1", "T2"), class = "factor")), .Names =
c("unique_id", "seq", "response", "detailed.name", "treatment"), class =
"data.frame", row.names = c(NA, -6L))
答案 3 :(得分：2)
您还可以使用reshape包中的stats功能。我没有您的样本数据集，但它看起来像这样：
reshape(x, idvar=c("seq","detailed.name"), timevar="treatment", direction="wide")
答案 4 :(得分：1)
如果您希望使用reshape2获得相同的结果，这是reshape包的更快且更高效的内存重写，那么以下内容将起作用。
主要更改是当您希望dcast输出cast时使用data.frame功能。这取代了cast
reshape功能
library(reshape2)
x = read.table(text = "unique_id seq response detailed.name treatment
a N1 123.23 dN1 T1
a N2 231.12 dN2 T1
a N3 231.23 dN3 T1
b N1 343.23 dN1 T2
b N2 281.13 dN2 T2
b N3 901.23 dN3 T2",
sep = "", header = TRUE)
x
y <- dcast(x, seq + detailed.name ~ treatment, value.var = "response")
y
# seq detailed.name T1 T2
# 1 N1 dN1 123.23 343.23
# 2 N2 dN2 231.12 281.13
# 3 N3 dN3 231.23 901.23
# EDIT to show how to return to the original data set:
melt(y, id.vars=c('seq', 'detailed.name'), variable.name='T', value.name='response')
# seq detailed.name T response
# 1 N1 dN1 T1 123.23
# 2 N2 dN2 T1 231.12
# 3 N3 dN3 T1 231.23
# 4 N1 dN1 T2 343.23
# 5 N2 dN2 T2 281.13
# 6 N3 dN3 T2 901.23