我知道在这里从宽到长的问题已经被问了太多次了，但是我不知道如何将以下内容转换为长格式。射击我甚至对SO进行了2次重复测量，问了一个广泛到长期的问题。我对无法转换数据感到沮丧。我该如何打开（可变顺序无关紧要）：
id trt work.T1 play.T1 talk.T1 total.T1 work.T2 play.T2 talk.T2 total.T2
1 x1.1 cnt 0.34434350 0.7841665 0.1079332 0.88803151 0.64836951 0.87954320 0.7233519 0.5630988
2 x1.2 tr 0.06132255 0.8426960 0.3338658 0.04685878 0.23478670 0.19711687 0.5164015 0.7617968
3 x1.3 tr 0.36897981 0.1834721 0.3241316 0.76904051 0.07629721 0.06945971 0.4118995 0.7452974
4 x1.4 tr 0.40759356 0.5285396 0.5654258 0.23022542 0.92309504 0.15733957 0.4132653 0.7078273
5 x1.5 cnt 0.91433676 0.7029476 0.2031782 0.31518412 0.14721669 0.33345678 0.7620444 0.9868082
6 x1.6 tr 0.88870525 0.9132728 0.2197045 0.28266959 0.82239037 0.18006177 0.2591765 0.4516309
7 x1.7 cnt 0.98373218 0.2591739 0.6331153 0.71319565 0.41351839 0.14648269 0.7631898 0.1182174
8 x1.8 tr 0.47719528 0.7926248 0.3525205 0.86213792 0.61252061 0.29057544 0.9824048 0.2386353
9 x1.9 tr 0.69350823 0.6144696 0.8568732 0.10632352 0.06812050 0.93606889 0.6701190 0.4705228
10 x1.10 cnt 0.42574646 0.7006205 0.9507216 0.55032776 0.90413220 0.10246047 0.5899279 0.3523231
到这个：
id trt time work play talk total
1 x1.1 cnt 1 0.34434350 0.78416653 0.1079332 0.88803151
2 x1.2 tr 1 0.06132255 0.84269599 0.3338658 0.04685878
3 x1.3 tr 1 0.36897981 0.18347215 0.3241316 0.76904051
4 x1.4 tr 1 0.40759356 0.52853960 0.5654258 0.23022542
5 x1.5 cnt 1 0.91433676 0.70294755 0.2031782 0.31518412
6 x1.6 tr 1 0.88870525 0.91327276 0.2197045 0.28266959
7 x1.7 cnt 1 0.98373218 0.25917392 0.6331153 0.71319565
8 x1.8 tr 1 0.47719528 0.79262477 0.3525205 0.86213792
9 x1.9 tr 1 0.69350823 0.61446955 0.8568732 0.10632352
10 x1.10 cnt 1 0.42574646 0.70062053 0.9507216 0.55032776
11 x1.1 cnt 2 0.64836951 0.87954320 0.7233519 0.56309884
12 x1.2 tr 2 0.23478670 0.19711687 0.5164015 0.76179680
13 x1.3 tr 2 0.07629722 0.06945971 0.4118995 0.74529740
14 x1.4 tr 2 0.92309504 0.15733957 0.4132653 0.70782726
15 x1.5 cnt 2 0.14721669 0.33345678 0.7620444 0.98680824
16 x1.6 tr 2 0.82239038 0.18006177 0.2591765 0.45163091
17 x1.7 cnt 2 0.41351839 0.14648269 0.7631898 0.11821741
18 x1.8 tr 2 0.61252061 0.29057544 0.9824048 0.23863532
19 x1.9 tr 2 0.06812050 0.93606889 0.6701190 0.47052276
20 x1.10 cnt 2 0.90413220 0.10246047 0.5899279 0.35232307
数据集
id <- paste('x', "1.", 1:10, sep="")
set.seed(10)
DF <- data.frame(id, trt=sample(c('cnt', 'tr'), 10, T), work.T1=runif(10),
play.T1=runif(10), talk.T1=runif(10), total.T1=runif(10),
work.T2=runif(10), play.T2=runif(10), talk.T2=runif(10),
total.T2=runif(10))
先感谢您！
编辑：我使用
set.seed时发生了一些棘手的事情（肯定是我做的错误）。上面的实际数据不是使用set.seed(10)时将获得的数据。我保留该错误的原因是历史准确性，它实际上并不会影响人们提供的解决方案。 最佳答案
这非常接近，更改列名应该在您的技能范围内：
reshape(DF,
varying=c(work= c(3, 7), play= c(4,8), talk= c(5,9), total= c(6,10) ),
direction="long")
编辑：添加几乎是一个确切的解决方案的版本：
reshape(DF, varying=list(work= c(3, 7), play= c(4,8), talk= c(5,9), total= c(6,10) ),
v.names=c("Work", "Play", "Talk", "Total"),
# that was needed after changed 'varying' arg to a list to allow 'times'
direction="long",
times=1:2, # substitutes number for T1 and T2
timevar="times") # to name the time col
关于r - 每次从宽到长的多种测量，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/9684671/