library(TraMineR)
df2 <- structure(list(ID = structure(c(1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L,
3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L,
3L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L,
4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L,
4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L,
4L, 4L, 4L, 4L, 4L), .Label = c("1", "2", "3", "4"), class = "factor"),
Date = structure(c(12432, 12432, 15262, 15293, 15385, 15385,
15475, 15567, 15598, 15628, 15689, 15810, 15871, 15963, 16024,
16054, 16144, 16205, 16297, 16389, 12432, 12492, 12614, 12676,
12767, 12888, 12949, 13071, 13132, 13253, 13344, 13406, 13436,
13559, 13618, 13709, 13771, 13862, 13953, 14045, 14137, 12432,
12676, 13010, 14045, 14137, 14228, 14290, 14410, 14502, 14563,
14714, 14744, 14744, 14775, 14836, 14897, 14897, 14928, 14958,
15048, 15201, 15232, 15323, 15414, 15536, 15536, 15536, 15567,
15567, 15598, 15628, 15659, 15659, 15751, 15751, 15779, 15810,
15810, 15840, 15871, 15871, 15901, 15901, 15963, 15993, 16054,
16054, 16085, 16116, 16144, 16144, 16175), class = "Date"),
st = structure(c(10L, 10L, 3L, 3L, 3L, 5L, 5L, 5L, 5L, 5L,
5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 2L, 2L, 2L, 2L, 2L,
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,
2L, 1L, 1L, 1L, 7L, 7L, 7L, 7L, 1L, 7L, 1L, 6L, 1L, 7L, 7L,
6L, 1L, 6L, 6L, 7L, 1L, 1L, 7L, 1L, 1L, 1L, 4L, 7L, 4L, 7L,
4L, 8L, 4L, 7L, 7L, 8L, 8L, 7L, 8L, 8L, 7L, 8L, 7L, 8L, 7L,
7L, 7L, 8L, 8L, 8L, 7L, 8L, 9L), .Label = c("1", "2", "3",
"4", "5", "6", "7", "8", "9", "10"), class = "factor")), class = "data.frame", row.names = c(NA,
-93L), .Names = c("ID", "Date", "st"))
我可以使用以下方法制作事件序列：
ev <- seqecreate(id = df2$ID,timestamp = df2$Date,event = df2$st)
但是，我放松了日期（我认为这对于绘图来说更好） . 或者我可以将 ev 某种方式转换为状态序列？
1 回答
要绘制序列，你可以做
它为您提供以下平行坐标图