我有以下数据集：10分钟的实例来自一个Feed，该Feed显示给定时间站点的自行车数量。我的目标是找出每个station_id
> dim(sample) [1] 6 73 > sample station_id 2014-08-18 20:00:00 2014-08-18 20:10:00 2014-08-18 20:20:00 1 1 1 0 0 2 2 18 18 19 3 3 5 4 4 4 4 21 20 20 5 5 9 10 8 6 6 9 9 9 2014-08-18 20:30:00 2014-08-18 20:40:00 2014-08-18 20:50:00 2014-08-18 21:00:00 1 2 1 1 0 2 18 18 21 22 3 4 4 4 4 4 21 22 25 23 5 5 7 3 0 6 9 9 9 9 2014-08-18 21:10:00 2014-08-18 21:20:00 2014-08-18 21:30:00 2014-08-18 21:40:00 1 0 4 2 1 2 21 22 22 21 3 4 6 3 3 4 25 24 24 24 5 0 0 0 0 6 9 9 10 10 2014-08-18 21:50:00 2014-08-18 22:00:00 2014-08-18 22:10:00 2014-08-18 22:20:00 1 3 2 3 2 2 21 22 18 16 3 4 5 6 4 4 25 22 21 21 5 4 0 0 3 6 10 10 11 11 2014-08-18 22:30:00 2014-08-18 22:40:00 2014-08-18 22:50:00 2014-08-18 23:00:00 1 2 3 8 11 2 15 14 16 17 3 4 4 4 4 4 23 21 20 20 5 0 1 0 0 6 11 12 12 12 2014-08-18 23:10:00 2014-08-18 23:20:00 2014-08-18 23:30:00 2014-08-18 23:40:00 1 11 13 14 15 2 17 15 10 3 3 5 5 5 4 4 22 19 19 17 5 1 0 2 3 6 12 12 12 12 2014-08-18 23:50:00 2014-08-19 00:00:00 2014-08-19 00:10:00 2014-08-19 00:20:00 1 16 16 17 19 2 2 2 2 1 3 5 6 5 5 4 18 19 19 20 5 3 0 2 6 6 12 12 12 12 2014-08-19 00:30:00 2014-08-19 00:40:00 2014-08-19 00:50:00 2014-08-19 01:00:00 1 20 21 22 23 2 1 4 3 3 3 3 3 3 3 4 20 19 20 19 5 0 1 2 5 6 12 12 12 12 2014-08-19 01:10:00 2014-08-19 01:20:00 2014-08-19 01:30:00 2014-08-19 01:40:00 1 22 23 23 21 2 3 4 4 4 3 4 4 4 4 4 21 24 23 24 5 0 3 2 1 6 12 12 12 12 2014-08-19 01:50:00 2014-08-19 02:00:00 2014-08-19 02:10:00 2014-08-19 02:20:00 1 22 23 22 22 2 3 1 1 1 3 4 4 4 4 4 24 22 21 22 5 0 1 2 2 6 12 12 12 12 2014-08-19 02:30:00 2014-08-19 02:40:00 2014-08-19 02:50:00 2014-08-19 03:00:00 1 20 22 22 22 2 1 1 1 1 3 4 4 4 4 4 22 24 24 23 5 2 3 3 3 6 12 12 12 12 2014-08-19 03:10:00 2014-08-19 03:20:00 2014-08-19 03:30:00 2014-08-19 03:40:00 1 22 22 21 21 2 1 1 1 1 3 4 4 4 4 4 23 23 23 22 5 4 3 3 3 6 12 12 12 12 2014-08-19 03:50:00 2014-08-19 04:00:00 2014-08-19 04:10:00 2014-08-19 04:20:00 1 21 21 21 21 2 1 1 1 1 3 4 4 4 4 4 22 22 23 23 5 2 3 3 3 6 12 12 12 12 2014-08-19 04:30:00 2014-08-19 04:40:00 2014-08-19 04:50:00 2014-08-19 05:00:00 1 21 21 21 21 2 1 1 1 1 3 4 4 4 4 4 23 23 23 23 5 5 4 4 4 6 12 12 12 12 2014-08-19 05:10:00 2014-08-19 05:20:00 2014-08-19 05:30:00 2014-08-19 05:40:00 1 21 21 21 21 2 1 1 1 1 3 4 4 4 4 4 24 24 24 24 5 4 4 4 4 6 12 12 12 12 2014-08-19 05:50:00 2014-08-19 06:00:00 2014-08-19 06:10:00 2014-08-19 06:20:00 1 21 21 21 22 2 1 1 1 1 3 4 4 4 4 4 24 24 24 24 5 4 4 4 3 6 12 12 12 12 2014-08-19 06:30:00 2014-08-19 06:40:00 2014-08-19 06:50:00 2014-08-19 07:00:00 1 22 22 22 22 2 1 1 1 1 3 5 5 5 5 4 24 24 24 24 5 3 4 4 4 6 12 12 12 12 2014-08-19 07:10:00 2014-08-19 07:20:00 2014-08-19 07:30:00 2014-08-19 07:40:00 1 22 22 22 22 2 6 5 5 5 3 5 5 5 5 4 24 24 24 24 5 4 4 4 4 6 12 12 12 12 2014-08-19 07:50:00 1 22 2 5 3 5 4 24 5 4 6 12
最终的输出应该看起来像这样（当然， 列的实际数量自然是73，但实例中的数量太多了。
> reshape2 station_id 2014-08-18 20:00:00 2014-08-18 20:10:00 2014-08-18 20:20:00 1 1 1 -1 0 2 2 18 0 1 3 3 5 -1 0 4 4 21 -1 0 5 5 9 1 -2 6 6 9 0 0 2014-08-18 20:30:00 2014-08-18 20:40:00 1 2 -1 2 -1 0 3 0 0 4 1 1 5 -3 2 6 0 0
如何循环R中的所有列？
以下是一小部分数据
> dput(sample)
structure(list(station_id = 1:6, `2014-08-18 20:00:00` = c(1L,
18L, 5L, 21L, 9L, 9L), `2014-08-18 20:10:00` = c(0L, 18L, 4L,
20L, 10L, 9L), `2014-08-18 20:20:00` = c(0L, 19L, 4L, 20L, 8L,
9L), `2014-08-18 20:30:00` = c(2L, 18L, 4L, 21L, 5L, 9L), `2014-08-18 20:40:00` = c(1L,
18L, 4L, 22L, 7L, 9L), `2014-08-18 20:50:00` = c(1L, 21L, 4L,
25L, 3L, 9L), `2014-08-18 21:00:00` = c(0L, 22L, 4L, 23L, 0L,
9L)), .Names = c("station_id", "2014-08-18 20:00:00", "2014-08-18 20:10:00", "2014-08-18 20:20:00", "2014-08-18 20:30:00", "2014-08-18 20:40:00", "2014-08-18 20:50:00", "2014-08-18 21:00:00"), row.names = c(NA,
6L), class = "data.frame")
答案 0 :(得分：2)
更短更快的是：
> set.seed(100)
> df = data.frame(id = 1:10, replicate(8,sample(0:1,10,rep=TRUE)))
> df
id X1 X2 X3 X4 X5 X6 X7 X8
1 1 0 1 1 0 0 0 0 0
2 2 0 1 1 1 1 0 1 0
3 3 1 0 1 0 1 0 1 1
4 4 0 0 1 1 1 0 1 1
5 5 0 1 0 1 1 1 0 1
6 6 0 1 0 1 0 0 0 1
7 7 1 0 1 0 1 0 0 1
8 8 0 0 1 1 1 0 0 1
9 9 1 0 1 1 0 1 0 1
10 10 0 1 0 0 0 0 1 0
> cbind(df[,1:2],df[,3:ncol(df)] - df[,2:(ncol(df) - 1)])
id X1 X2 X3 X4 X5 X6 X7 X8
1 1 0 1 0 -1 0 0 0 0
2 2 0 1 0 0 0 -1 1 -1
3 3 1 -1 1 -1 1 -1 1 0
4 4 0 0 1 0 0 -1 1 0
5 5 0 1 -1 1 0 0 -1 1
6 6 0 1 -1 1 -1 0 0 1
7 7 1 -1 1 -1 1 -1 0 1
8 8 0 0 1 0 0 -1 0 1
9 9 1 -1 1 0 -1 1 -1 1
10 10 0 1 -1 0 0 0 1 -1
答案 1 :(得分：0)
如果它是你想要的循环，下面是一种方法。我确信有更优雅的方式....
> set.seed(100)
> df = data.frame(id = 1:10, replicate(8,sample(0:1,10,rep=TRUE)))
> df
id X1 X2 X3 X4 X5 X6 X7 X8
1 1 0 1 1 0 0 0 0 0
2 2 0 1 1 1 1 0 1 0
3 3 1 0 1 0 1 0 1 1
4 4 0 0 1 1 1 0 1 1
5 5 0 1 0 1 1 1 0 1
6 6 0 1 0 1 0 0 0 1
7 7 1 0 1 0 1 0 0 1
8 8 0 0 1 1 1 0 0 1
9 9 1 0 1 1 0 1 0 1
10 10 0 1 0 0 0 0 1 0
> df1 = df
> for(i in 3:ncol(df1)) df1[,i] = df[,i]-df[,(i-1)]
> df1
id X1 X2 X3 X4 X5 X6 X7 X8
1 1 0 1 0 -1 0 0 0 0
2 2 0 1 0 0 0 -1 1 -1
3 3 1 -1 1 -1 1 -1 1 0
4 4 0 0 1 0 0 -1 1 0
5 5 0 1 -1 1 0 0 -1 1
6 6 0 1 -1 1 -1 0 0 1
7 7 1 -1 1 -1 1 -1 0 1
8 8 0 0 1 0 0 -1 0 1
9 9 1 -1 1 0 -1 1 -1 1
10 10 0 1 -1 0 0 0 1 -1
答案 2 :(得分：0)
如下：
t(apply(sample[,-1], 1, diff))