17

我在重塑大型数据框时遇到了困难。而且我过去相对幸运地避免了重塑问题，这也意味着我在这方面很糟糕。

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 reshape 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：我尝试在我的本地机器（4GB 双核 iMac 3.06Ghz）上运行它，但它一直失败：

> 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

当我有机会时，我会尝试在我们的一台更大的机器上运行它。

4

5 回答 5

21

reshape 对我来说也总是很棘手，但它似乎总是通过一些试验和错误来工作。这是我最终发现的：

> 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 melt/cast 使用的长格式。所以我重新融化了它。第二个参数（id.vars）是不融化的东西的列表。第三个参数（measure.vars）是变化的事物的列表。

然后，演员使用一个公式。波浪号左边是保持原样的东西，波浪号右边是用于调节值列的列。

或多或少...！

于 2009-10-07T19:11:41.343 回答
6

cast以 Harlan 的回答为基础 - 如果数据已经是长格式，并且在调用中指定了保存值的列，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
于 2009-10-07T22:21:31.497 回答
3

另一种选择是使用spreadfromtidyr

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

相反的动作是由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，还有dcast.data.table来自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

数据

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))
于 2014-12-29T04:22:20.343 回答
2

您也可以使用包reshape中的功能stats。我没有您的示例数据集，但它看起来像这样：

reshape(x, idvar=c("seq","detailed.name"), timevar="treatment", direction="wide")
于 2009-10-07T19:55:30.460 回答
1

如果您想使用 获得相同的结果reshape2，这是对reshape包的更快和更高效的内存重写，那么以下将起作用。

主要的变化是dcast当你想cast用 adata.frame作为输出时使用该函数。这取代cast了reshape

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
于 2012-11-27T22:31:14.897 回答