我有一个像这样的长格式数据框
> df2
id t treat value
1 1 1 0 5.718226
2 1 2 1 5.954669
3 1 3 0 4.485165
4 2 1 1 6.616181
5 2 2 0 4.521301
6 2 3 1 6.955451
7 3 1 0 3.851682
8 3 2 1 6.907178
9 3 3 1 6.274501
10 4 1 1 6.092860
11 4 2 0 4.327431
12 4 3 0 6.019627
我想要宽格式的输出：
id treat.1 value.1 treat.2 value.2 treat.3 value.3
1 1 0 5.718226 1 5.954669 0 4.485165
2 2 1 6.616181 0 4.521301 1 6.955451
3 3 0 3.851682 1 6.907178 1 6.274501
4 4 1 6.092860 0 4.327431 0 6.019627
我只能这样做：
l2.2 <- with(df2, split(df2, t))
l2.3 <- lapply(seq_along(l2.2), function(x) {
names(l2.2[[x]])[3:4] <- paste0(names(l2.2[[x]])[3:4], ".", names(l2.2)[x])
l2.2[[x]][-2]
})
Reduce(function(x, y)
merge(x, y, all=TRUE, by=intersect(names(x), names(y))), l2.3)
对我来说，这似乎太复杂了，但是我无法简化它。我想学习如何使用reshape()，aggregate()或data.table::dcast()执行此操作。我接近了，但我无法弄清楚到底是什么：
重塑：
df2.1 <- reshape(df2, timevar="t", idvar=c("id", "treat"), direction="wide")
df2.2 <- reshape(df2.1, timevar="treat", idvar="id", direction="wide")
> df2.2[, c(1, 2, 5, 3, 6, 4, 7)]
id value.1.0 value.1.1 value.2.0 value.2.1 value.3.0 value.3.1
1 1 5.718226 NA NA 5.954669 4.485165 NA
4 2 NA 6.616181 4.521301 NA NA 6.955451
7 3 3.851682 NA NA 6.907178 NA 6.274501
10 4 NA 6.092860 4.327431 NA 6.019627 NA
data.table：
> data.table::dcast(df2, id + treat ~ t)
id treat 1 2 3
1 1 0 5.718226 NA 4.485165
2 1 1 NA 5.954669 NA
3 2 0 NA 4.521301 NA
4 2 1 6.616181 NA 6.955451
5 3 0 3.851682 NA NA
6 3 1 NA 6.907178 6.274501
7 4 0 NA 4.327431 6.019627
8 4 1 6.092860 NA NA
我也尝试了aggregate()次失败。
有人可以告诉我如何在基数R和data.table::dcast()中执行此操作吗？
编辑：
与这个简单的示例相反，我在数据中还有其他一些id和随时间变化的变量，请参见下面的数据中的df3。
@markus的reshape()解决方案有效时，dcast()引发错误：
> reshape(df3, timevar="t", idvar=c("id.1", "id.2"), direction="wide")
id.1 id.2 treat.1 value.1.1 value.2.1 treat.2 value.1.2 value.2.2 treat.3 value.1.3 value.2.3
1 1 1 0 5.718226 0.4297986 1 5.954669 -1.4007124 0 4.485165 -1.1741134
4 2 1 1 6.616181 -1.0516253 0 4.521301 1.5686463 1 6.955451 -0.4306961
7 3 2 0 3.851682 -0.3046341 1 6.907178 -0.6669521 1 6.274501 -0.2582921
10 4 2 1 6.092860 1.3231503 0 4.327431 1.6899552 0 6.019627 -0.4263450
> dcast(df3, id ~ t, value.var = c("treat", "value"))
Error in .subset2(x, i, exact = exact) : subscript out of bounds
In addition: Warning message:
In if (!(value.var %in% names(data))) { :
the condition has length > 1 and only the first element will be used
数据：
df2 <- structure(list(id = c(1L, 1L, 1L, 2L, 2L, 2L, 3L, 3L, 3L, 4L, 4L, 4L),
t = c(1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3),
treat = c(0, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 0),
value = c(5.7182263351077, 5.95466888906233,
4.48516458093838, 6.61618146498587,
4.52130082895974, 6.95545080306353,
3.85168235272874, 6.90717809069993,
6.27450118041287, 6.09285968998526,
4.32743136605772, 6.01962658742754)),
row.names = c(NA, -12L), class = "data.frame")
df3 <- structure(list(id.1 = c(1L, 1L, 1L, 2L, 2L, 2L, 3L, 3L, 3L, 4L, 4L, 4L),
id.2 = c(1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L),
t = c(1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3),
treat = c(0, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 0),
value.1 = c(5.7182263351077, 5.95466888906233, 4.48516458093838,
6.61618146498587, 4.52130082895974, 6.95545080306353,
3.85168235272874, 6.90717809069993, 6.27450118041287,
6.09285968998526, 4.32743136605772, 6.01962658742754),
value.2 = c(0.429798618225556, -1.40071244216677, -1.17411337999421,
-1.05162530600877, 1.56864628901013, -0.430696055983823,
-0.304634116536374, -0.666952117966313, -0.258292124936937,
1.32315028276158, 1.68995518578212, -0.426345031174389)),
class = "data.frame", row.names = c(NA, -12L))
答案 0 :(得分：4)
使用dcast中的data.table，您需要将列treat和value指定为value.var。
library(data.table)
setDT(df2)
dcast(df2, id ~ t, value.var = c("treat", "value"))
# id treat_1 treat_2 treat_3 value_1 value_2 value_3
#1: 1 0 1 0 5.718226 5.954669 4.485165
#2: 2 1 0 1 6.616181 4.521301 6.955451
#3: 3 0 1 1 3.851682 6.907178 6.274501
#4: 4 1 0 0 6.092860 4.327431 6.019627
reshape将以这种方式工作
reshape(df2, idvar = "id", timevar = "t", direction = "wide")
# id treat.1 value.1 treat.2 value.2 treat.3 value.3
#1 1 0 5.718226 1 5.954669 0 4.485165
#4 2 1 6.616181 0 4.521301 1 6.955451
#7 3 0 3.851682 1 6.907178 1 6.274501
#10 4 1 6.092860 0 4.327431 0 6.019627