我在r，a1和a2中有两个动物园对象，我想合并，然后填写缺失的值。我想用na.approx()填充a1的缺失值，我想用na.locf()填充a2的缺失值。我怎么能做到这一点？
一个例子：
a1 <- zoo(data.frame(b=c(1,3,4,6),c=c(2,6,8,12)),c(1,3,4,6))
b c 1 1 2 3 3 6 4 4 8 6 6 12
a2 <- zoo(data.frame(d=c(1,0)),c(2,5))
d 2 1 5 0
a3 <- merge(a1,a2)
b c d 1 1 2 NA 2 NA NA 1 3 3 6 NA 4 4 8 NA 5 NA NA 0 6 6 12 NA
现在我想进入a4：
b c d 1 1 2 NA 2 2 4 1 3 3 6 1 4 4 8 1 5 5 10 0 6 6 12 0
但a4 <- na.locf(a3)或a4 <- na.approx(a3)都将所有列视为相同。我该如何单独处理这些列？或者我可以在合并之前解决这个问题？
提前致谢
**编辑：现在包含真实数据**
为了更详细地说明，这里有一些真实数据：
> dput(a5)
structure(c(133.7, NA, 133.345, NA, 134.2, NA, 135.721, 136.456,
136.677, 137.347, 138.324, NA, 139.086, 139.622, 140.475, NA,
141.179, 141.652, 141.811, 125.901, NA, 125.965, NA, 127.402,
NA, 128.529, 128.797, 129.267, 130.08, 130.831, NA, 131.313,
132.008, 132.85, NA, 133.416, 133.842, 133.986, NA, 0, NA, 1,
NA, 0, NA, NA, NA, NA, NA, 1, NA, NA, NA, 0, NA, NA, NA), .Dim = c(19L,
3L), .Dimnames = list(NULL, c("sup", "ret", "gas")), index = structure(c(1387242143,
1387242156, 1387242158, 1387242169, 1387242173, 1387242186, 1387242188,
1387242203, 1387242218, 1387242233, 1387242248, 1387242252, 1387242263,
1387242278, 1387242293, 1387242305, 1387242308, 1387242323, 1387242338
), class = c("POSIXct", "POSIXt")), class = "zoo")
> a5
sup ret gas
2013-12-16 19:02:23 133.700 125.901 NA
2013-12-16 19:02:36 NA NA 0
2013-12-16 19:02:38 133.345 125.965 NA
2013-12-16 19:02:49 NA NA 1
2013-12-16 19:02:53 134.200 127.402 NA
2013-12-16 19:03:06 NA NA 0
2013-12-16 19:03:08 135.721 128.529 NA
2013-12-16 19:03:23 136.456 128.797 NA
2013-12-16 19:03:38 136.677 129.267 NA
2013-12-16 19:03:53 137.347 130.080 NA
2013-12-16 19:04:08 138.324 130.831 NA
2013-12-16 19:04:12 NA NA 1
2013-12-16 19:04:23 139.086 131.313 NA
2013-12-16 19:04:38 139.622 132.008 NA
2013-12-16 19:04:53 140.475 132.850 NA
2013-12-16 19:05:05 NA NA 0
2013-12-16 19:05:08 141.179 133.416 NA
2013-12-16 19:05:23 141.652 133.842 NA
2013-12-16 19:05:38 141.811 133.986 NA
Stu的解决方案并不能保持前两列数据的准确性。我希望前两列中的NA用na.approx()进行插值，最后一列用最后一次观察结转locf()进行插值。
> na.locf(ceiling(na.approx(a5)))
sup ret gas
2013-12-16 19:02:23 134 126 NA
2013-12-16 19:02:36 134 126 0
2013-12-16 19:02:38 134 126 1
2013-12-16 19:02:49 134 128 1
2013-12-16 19:02:53 135 128 1
2013-12-16 19:03:06 136 129 0
2013-12-16 19:03:08 136 129 1
2013-12-16 19:03:23 137 129 1
2013-12-16 19:03:38 137 130 1
2013-12-16 19:03:53 138 131 1
2013-12-16 19:04:08 139 131 1
2013-12-16 19:04:12 139 131 1
2013-12-16 19:04:23 140 132 1
2013-12-16 19:04:38 140 133 1
2013-12-16 19:04:53 141 133 1
2013-12-16 19:05:05 142 134 0
2013-12-16 19:05:08 142 134 0
2013-12-16 19:05:23 142 134 0
2013-12-16 19:05:38 142 134 0
**以上不是我需要的**
再次感谢
**编辑 - 显示我正在寻找的结果的解决方案**
> a6 <- cbind(na.approx(a5[,c("sup","ret")]),na.locf(a5[,c("gas")]))
> a6
sup ret na.locf(a5[, c("gas")])
2013-12-16 19:02:23 133.7000 125.9010 NA
2013-12-16 19:02:36 133.3923 125.9565 0
2013-12-16 19:02:38 133.3450 125.9650 0
2013-12-16 19:02:49 133.9720 127.0188 1
2013-12-16 19:02:53 134.2000 127.4020 1
2013-12-16 19:03:06 135.5182 128.3787 0
2013-12-16 19:03:08 135.7210 128.5290 0
2013-12-16 19:03:23 136.4560 128.7970 0
2013-12-16 19:03:38 136.6770 129.2670 0
2013-12-16 19:03:53 137.3470 130.0800 0
2013-12-16 19:04:08 138.3240 130.8310 0
2013-12-16 19:04:12 138.5272 130.9595 1
2013-12-16 19:04:23 139.0860 131.3130 1
2013-12-16 19:04:38 139.6220 132.0080 1
2013-12-16 19:04:53 140.4750 132.8500 1
2013-12-16 19:05:05 141.0382 133.3028 0
2013-12-16 19:05:08 141.1790 133.4160 0
2013-12-16 19:05:23 141.6520 133.8420 0
2013-12-16 19:05:38 141.8110 133.9860 0
仍然需要处理新对象的名称，但这很容易。
答案 0 :(得分：1)
使用column-bind函数，同时将每列分组到所需的na.fill函数中：
a4 <- cbind(na.approx(a3[,1:2]), na.locf(a3[,3]))
它只有一行代码，您可以根据需要重新排列列。