在每个组内创建一个序号
Create a sequential number within each group
我需要根据另一列创建一个数字序列
我有这个数据框：
head(df)
id date lc lon lat gap_days gap
1 20162.03 2003-10-19 14:33:00 Tagging -39.370 -18.480 NA <NA>
2 20162.03 2003-10-21 12:19:00 1 -38.517 -18.253 1.90694444 gap
3 20162.03 2003-10-21 13:33:00 1 -38.464 -18.302 0.05138889 no
4 20162.03 2003-10-21 16:38:00 A -38.461 -18.425 0.12847222 no
5 20162.03 2003-10-21 18:50:00 A -38.322 -18.512 0.09166667 no
6 20162.03 2003-10-23 10:33:00 B -38.674 -19.824 1.65486111 gap
我根据 gap_days 栏在“差距”栏中指出了超过一天的差距。
现在，我需要拆分我的数据。每个间隙序列将是一个新的单独帧。
所以，如果我有ID 20162.03，并且这个id有一个或多个空位，这个序列将根据空位的数量进行拆分。
为此，我将使用包 move 和函数 burst" 和 split
但是，为此我需要创建一个新列，其中包含一系列数字，表示新的 ID 分隔，如 (seq 列):
id date lc lon lat gap_days gap seq
1 20162.03 2003-10-19 14:33:00 Tagging -39.370 -18.480 NA <NA> 1
2 20162.03 2003-10-21 12:19:00 1 -38.517 -18.253 1.90694444 gap 1
3 20162.03 2003-10-21 13:33:00 1 -38.464 -18.302 0.05138889 no 1
4 20162.03 2003-10-21 16:38:00 A -38.461 -18.425 0.12847222 no 1
5 20162.03 2003-10-21 18:50:00 A -38.322 -18.512 0.09166667 no 1
6 20162.03 2003-10-23 10:33:00 B -38.674 -19.824 1.65486111 gap 2
7 20162.03 2003-10-23 17:52:00 B -38.957 -19.511 0.30486111 no 2
8 20162.03 2003-11-02 08:14:00 B -42.084 -24.071 9.59861111 gap 3
9 20162.03 2003-11-02 09:36:00 A -41.999 -24.114 0.05694444 no 3
10 20687.03 2003-10-27 17:02:00 Tagging -39.320 -18.460 NA <NA> 4
11 20687.03 2003-10-27 19:44:00 2 -39.306 -18.454 0.11250000 no 4
12 20687.03 2003-10-27 21:05:00 1 -39.301 -18.458 0.05625000 no 4
但是，如您所见，我有一系列“缺口”和“否”，还有 NA。
我找不到解决办法。
有人有解决办法吗？
编辑：
structure(list(id = c("20162.03", "20162.03", "20162.03", "20162.03",
"20162.03", "20162.03", "20162.03", "20162.03", "20162.03", "20687.03",
"20687.03", "20687.03"), date = structure(c(1066573980, 1066738740,
1066743180, 1066754280, 1066762200, 1066905180, 1066931520, 1067760840,
1067765760, 1067274120, 1067283840, 1067288700), class = c("POSIXct",
"POSIXt"), tzone = "GMT"), lc = structure(c(4L, 1L, 1L, 2L, 2L,
3L, 3L, 3L, 2L, 4L, 6L, 1L), .Label = c("1", "A", "B", "Tagging",
"0", "2", "3", "N", "P", "Z"), class = "factor"), lon = c(-39.37,
-38.517, -38.464, -38.461, -38.322, -38.674, -38.957, -42.084,
-41.999, -39.32, -39.306, -39.301), lat = c(-18.48, -18.253,
-18.302, -18.425, -18.512, -19.824, -19.511, -24.071, -24.114,
-18.46, -18.454, -18.458), gap_days = c(NA, 1.90694444444444,
0.0513888888888889, 0.128472222222222, 0.0916666666666667, 1.65486111111111,
0.304861111111111, 9.59861111111111, 0.0569444444444444, NA,
0.1125, 0.05625), gap = c(NA, "gap", "no", "no", "no", "gap",
"no", "gap", "no", NA, "no", "no")), row.names = c(NA, 12L), class = "data.frame")
使用 Base R 的简单解决方案：
df$seq <- ave(sapply(df$gap, identical, "gap"), df$id, FUN = cumsum)
df
#> id date lc lon lat gap_days gap seq
#> 1 20162.03 2003-10-19 14:33:00 Tagging -39.370 -18.480 NA <NA> 0
#> 2 20162.03 2003-10-21 12:19:00 1 -38.517 -18.253 1.90694444 gap 1
#> 3 20162.03 2003-10-21 13:33:00 1 -38.464 -18.302 0.05138889 no 1
#> 4 20162.03 2003-10-21 16:38:00 A -38.461 -18.425 0.12847222 no 1
#> 5 20162.03 2003-10-21 18:50:00 A -38.322 -18.512 0.09166667 no 1
#> 6 20162.03 2003-10-23 10:33:00 B -38.674 -19.824 1.65486111 gap 2
#> 7 20162.03 2003-10-23 17:52:00 B -38.957 -19.511 0.30486111 no 2
#> 8 20162.03 2003-11-02 08:14:00 B -42.084 -24.071 9.59861111 gap 3
#> 9 20162.03 2003-11-02 09:36:00 A -41.999 -24.114 0.05694444 no 3
#> 10 20687.03 2003-10-27 17:02:00 Tagging -39.320 -18.460 NA <NA> 0
#> 11 20687.03 2003-10-27 19:44:00 2 -39.306 -18.454 0.11250000 no 0
#> 12 20687.03 2003-10-27 21:05:00 1 -39.301 -18.458 0.05625000 no 0
然后拆分：
split(df, list(df$id, df$seq), drop = TRUE)
#> $`20162.03.0`
#> id date lc lon lat gap_days gap seq
#> 1 20162.03 2003-10-19 14:33:00 Tagging -39.37 -18.48 NA <NA> 0
#>
#> $`20687.03.0`
#> id date lc lon lat gap_days gap seq
#> 10 20687.03 2003-10-27 17:02:00 Tagging -39.320 -18.460 NA <NA> 0
#> 11 20687.03 2003-10-27 19:44:00 2 -39.306 -18.454 0.11250 no 0
#> 12 20687.03 2003-10-27 21:05:00 1 -39.301 -18.458 0.05625 no 0
#>
#> $`20162.03.1`
#> id date lc lon lat gap_days gap seq
#> 2 20162.03 2003-10-21 12:19:00 1 -38.517 -18.253 1.90694444 gap 1
#> 3 20162.03 2003-10-21 13:33:00 1 -38.464 -18.302 0.05138889 no 1
#> 4 20162.03 2003-10-21 16:38:00 A -38.461 -18.425 0.12847222 no 1
#> 5 20162.03 2003-10-21 18:50:00 A -38.322 -18.512 0.09166667 no 1
#>
#> $`20162.03.2`
#> id date lc lon lat gap_days gap seq
#> 6 20162.03 2003-10-23 10:33:00 B -38.674 -19.824 1.6548611 gap 2
#> 7 20162.03 2003-10-23 17:52:00 B -38.957 -19.511 0.3048611 no 2
#>
#> $`20162.03.3`
#> id date lc lon lat gap_days gap seq
#> 8 20162.03 2003-11-02 08:14:00 B -42.084 -24.071 9.59861111 gap 3
#> 9 20162.03 2003-11-02 09:36:00 A -41.999 -24.114 0.05694444 no 3
我需要根据另一列创建一个数字序列
我有这个数据框：
head(df)
id date lc lon lat gap_days gap
1 20162.03 2003-10-19 14:33:00 Tagging -39.370 -18.480 NA <NA>
2 20162.03 2003-10-21 12:19:00 1 -38.517 -18.253 1.90694444 gap
3 20162.03 2003-10-21 13:33:00 1 -38.464 -18.302 0.05138889 no
4 20162.03 2003-10-21 16:38:00 A -38.461 -18.425 0.12847222 no
5 20162.03 2003-10-21 18:50:00 A -38.322 -18.512 0.09166667 no
6 20162.03 2003-10-23 10:33:00 B -38.674 -19.824 1.65486111 gap
我根据 gap_days 栏在“差距”栏中指出了超过一天的差距。 现在，我需要拆分我的数据。每个间隙序列将是一个新的单独帧。
所以，如果我有ID 20162.03，并且这个id有一个或多个空位，这个序列将根据空位的数量进行拆分。
为此，我将使用包 move 和函数 burst" 和 split
但是，为此我需要创建一个新列，其中包含一系列数字，表示新的 ID 分隔，如 (seq 列):
id date lc lon lat gap_days gap seq
1 20162.03 2003-10-19 14:33:00 Tagging -39.370 -18.480 NA <NA> 1
2 20162.03 2003-10-21 12:19:00 1 -38.517 -18.253 1.90694444 gap 1
3 20162.03 2003-10-21 13:33:00 1 -38.464 -18.302 0.05138889 no 1
4 20162.03 2003-10-21 16:38:00 A -38.461 -18.425 0.12847222 no 1
5 20162.03 2003-10-21 18:50:00 A -38.322 -18.512 0.09166667 no 1
6 20162.03 2003-10-23 10:33:00 B -38.674 -19.824 1.65486111 gap 2
7 20162.03 2003-10-23 17:52:00 B -38.957 -19.511 0.30486111 no 2
8 20162.03 2003-11-02 08:14:00 B -42.084 -24.071 9.59861111 gap 3
9 20162.03 2003-11-02 09:36:00 A -41.999 -24.114 0.05694444 no 3
10 20687.03 2003-10-27 17:02:00 Tagging -39.320 -18.460 NA <NA> 4
11 20687.03 2003-10-27 19:44:00 2 -39.306 -18.454 0.11250000 no 4
12 20687.03 2003-10-27 21:05:00 1 -39.301 -18.458 0.05625000 no 4
但是，如您所见，我有一系列“缺口”和“否”，还有 NA。 我找不到解决办法。 有人有解决办法吗？
编辑：
structure(list(id = c("20162.03", "20162.03", "20162.03", "20162.03",
"20162.03", "20162.03", "20162.03", "20162.03", "20162.03", "20687.03",
"20687.03", "20687.03"), date = structure(c(1066573980, 1066738740,
1066743180, 1066754280, 1066762200, 1066905180, 1066931520, 1067760840,
1067765760, 1067274120, 1067283840, 1067288700), class = c("POSIXct",
"POSIXt"), tzone = "GMT"), lc = structure(c(4L, 1L, 1L, 2L, 2L,
3L, 3L, 3L, 2L, 4L, 6L, 1L), .Label = c("1", "A", "B", "Tagging",
"0", "2", "3", "N", "P", "Z"), class = "factor"), lon = c(-39.37,
-38.517, -38.464, -38.461, -38.322, -38.674, -38.957, -42.084,
-41.999, -39.32, -39.306, -39.301), lat = c(-18.48, -18.253,
-18.302, -18.425, -18.512, -19.824, -19.511, -24.071, -24.114,
-18.46, -18.454, -18.458), gap_days = c(NA, 1.90694444444444,
0.0513888888888889, 0.128472222222222, 0.0916666666666667, 1.65486111111111,
0.304861111111111, 9.59861111111111, 0.0569444444444444, NA,
0.1125, 0.05625), gap = c(NA, "gap", "no", "no", "no", "gap",
"no", "gap", "no", NA, "no", "no")), row.names = c(NA, 12L), class = "data.frame")
使用 Base R 的简单解决方案：
df$seq <- ave(sapply(df$gap, identical, "gap"), df$id, FUN = cumsum)
df
#> id date lc lon lat gap_days gap seq
#> 1 20162.03 2003-10-19 14:33:00 Tagging -39.370 -18.480 NA <NA> 0
#> 2 20162.03 2003-10-21 12:19:00 1 -38.517 -18.253 1.90694444 gap 1
#> 3 20162.03 2003-10-21 13:33:00 1 -38.464 -18.302 0.05138889 no 1
#> 4 20162.03 2003-10-21 16:38:00 A -38.461 -18.425 0.12847222 no 1
#> 5 20162.03 2003-10-21 18:50:00 A -38.322 -18.512 0.09166667 no 1
#> 6 20162.03 2003-10-23 10:33:00 B -38.674 -19.824 1.65486111 gap 2
#> 7 20162.03 2003-10-23 17:52:00 B -38.957 -19.511 0.30486111 no 2
#> 8 20162.03 2003-11-02 08:14:00 B -42.084 -24.071 9.59861111 gap 3
#> 9 20162.03 2003-11-02 09:36:00 A -41.999 -24.114 0.05694444 no 3
#> 10 20687.03 2003-10-27 17:02:00 Tagging -39.320 -18.460 NA <NA> 0
#> 11 20687.03 2003-10-27 19:44:00 2 -39.306 -18.454 0.11250000 no 0
#> 12 20687.03 2003-10-27 21:05:00 1 -39.301 -18.458 0.05625000 no 0
然后拆分：
split(df, list(df$id, df$seq), drop = TRUE)
#> $`20162.03.0`
#> id date lc lon lat gap_days gap seq
#> 1 20162.03 2003-10-19 14:33:00 Tagging -39.37 -18.48 NA <NA> 0
#>
#> $`20687.03.0`
#> id date lc lon lat gap_days gap seq
#> 10 20687.03 2003-10-27 17:02:00 Tagging -39.320 -18.460 NA <NA> 0
#> 11 20687.03 2003-10-27 19:44:00 2 -39.306 -18.454 0.11250 no 0
#> 12 20687.03 2003-10-27 21:05:00 1 -39.301 -18.458 0.05625 no 0
#>
#> $`20162.03.1`
#> id date lc lon lat gap_days gap seq
#> 2 20162.03 2003-10-21 12:19:00 1 -38.517 -18.253 1.90694444 gap 1
#> 3 20162.03 2003-10-21 13:33:00 1 -38.464 -18.302 0.05138889 no 1
#> 4 20162.03 2003-10-21 16:38:00 A -38.461 -18.425 0.12847222 no 1
#> 5 20162.03 2003-10-21 18:50:00 A -38.322 -18.512 0.09166667 no 1
#>
#> $`20162.03.2`
#> id date lc lon lat gap_days gap seq
#> 6 20162.03 2003-10-23 10:33:00 B -38.674 -19.824 1.6548611 gap 2
#> 7 20162.03 2003-10-23 17:52:00 B -38.957 -19.511 0.3048611 no 2
#>
#> $`20162.03.3`
#> id date lc lon lat gap_days gap seq
#> 8 20162.03 2003-11-02 08:14:00 B -42.084 -24.071 9.59861111 gap 3
#> 9 20162.03 2003-11-02 09:36:00 A -41.999 -24.114 0.05694444 no 3